Electro Chemistry

7/16/2019 Electro Chemistry

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Reconstruct Your Chemistry With Prince Sir



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The branch of chemistry which deals with the study of relationship betweenelectrical energy

and chemical energy and interconversion of one form of energy into an-other is called

electrochemistry.

Conductors :- The substances which allow the passage of electricity through them are called conductors.

Insulators / Non-conductors :- The substances which donot allow the passage of electricity through them are called insulators.

Conductors are broadly of 2 types

(a) Metallic conductors :- These are the metallic substances which allow the passage of electricity through them without undergoing any chemical change.e.g. copper, silver, aluminium etc. Besides metals,

(b) Electrolytes :- These are the substances which allow the passage of electricity through their molten state or through their aqueous sol.’s and alsoundergo chemical decomposition at the same time.

Some minerals & non-metal graphite also show conduction due to the movementof electrons. They are collectively called electronic conductors.

Fac t o rs a f f e c t ing e l e c t r o lyt i c c onduc t i on( i ) T he in ter - ion ic attra ction :- These are the attractive interactions between the ions furnished by the electrolyte in solution and are referred to as solute

solute interaction. Larger the interionic attraction lesser is freedom of movementand smaller will be the electrical conductivity.

( i i ) Solv ation o f io ns :-These are the attractive interactions between the ionsof the electrolyte and molecules of the solvent and called as solute solventinteractions. Larger the solute solvent interactions greater is the extent of solvationand lower will be the electrical conductivity of the solution.

CHEMISTRY

ELECTRO CHEMISTRY

THEORY SHEET

I IT



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( i i i ) Viscos ity of solvent :- It refers to the attractive interactions between thesolvent molecules called as solvent solvent interactions. Larger the solvent - solventinteractions larger will be the viscosity and more will be the hinderance to themovement of ions and hence lower will be the electrical conductivity.

( iv ) Tem p. :- As the temp. of electrolyte solution is increased, the effect of above factors diminishes and kinetic energy of the ions increases. Thus electricalconductance of electrolytic solution increases.

Terms re lat ed t o e l e c t r i cal conduct iv i ty1 . Res is tance :- Every conducting material offers some obstruction to the flowof electricity which is called resistance. Ohm’s law relates resistance (R) to thecurrent strength.

Ohm’s law :- It is obeyed by metallic as well as electrolytic conductors. It states that“the potential difference across the conductors is direct ly proportional tothe current f l owing through i t ”

i.e. Potential difference current

or Potential difference

Current= Constant

Mathematically :-V

I= R or I = V

R

2 . Res is tivi ty :- Resistance (R) of any conductor is directly proportional to thelength (l) of the conductor and inversely proportional to its area (a) of crosssection :-

R l

aor R =

l

aunits ® W cm.

Where (rho) is the constant of proportionality & called as specific resistance or resistivity. It is given as

Ra

l. If l = 1 cm., a = 1 cm2 then R = and

hence Resistivity may be defined as “resistance offered by the conductor of 1 cmlength with area of cross section equal to 1 cm 2.

3 . Conductance (G) :- It is the measure of the ease with which current flows through the conductor . It may be defined as the reciprocal of resistance. It isdenoted by G.

R R

R

1 4

3



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G = 1

Runit ® –1 or mho.

4 . Specific conductance or conductivity () :- It is the reciprocal of specificresistance. It may also be defined as the conductance of one centimeter cube of

the conductor. It is represented by Greek letter Kappa (k)k =

1 1

R

l

a unit ®–1 cm–1.

5 . Molar conductivity ( m ) :- It may be defined as the conductance of a solutioncontaining 1g molecule or 1 mol of electrolyte such that the entire solution is placed between two electrodes one centimeter apart. It is denoted by the symbol

m . It is related to conductivity (k) as

m = V or m = K

C V :- Vol. , C :- conc.

If C of sol is in mol L –1 then 1 g mol of electrolyte is present in 1000

C cm3 of sol.

Thus m = k Units :- Ohm–1 cm2 mol–1 / s cm2 mol–1.

Expe r imenta l Measur ement o f e l e c t r i ca l c onduc t iv i ty

Arrangement o f Wheats t one b r idge : - As conductance is reciprocal of its resistance, if resistance of solution is

known, its conductance can be easily calculated. Resistance can be measuredwith the help of wheatstone bridge. The resistances R 1, R 3 & R 4 are so adjusted

that a null point is obtained. The null point is indicated by detector At null point

R 2 =3

41

R

R R



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Knowing R 1, R 4 & R 3, R 2 can be calculated & its reciprocal gives the conductanceof the solutionNote:- D.C. current can not be used inthe experiment because -

(i) Change in the concentration of the solution occurs due to electrolysis whichwill change the resistaance.(ii) Polaristaion of the electrodes which also changes the resistanxce.Thus an A.C. current is used.Var ia t i on o f c onduc t iv i ty and Molar c onduc t iv i ty w i th c onc ent ra t i on1 . Variation of (K) with conc. :- K (conductivity) refers to the conductance of one centimeter cube of the electrolyte. With the se in dilution, the no. of ions present in one cm3 of the sol. ses. Thus conductivity of the electrolytic solution

also ses as the dilution ses.

2 . Var ia ti on o f molar c onduc tiv ity w i th conc . :-( a ) S tron g E lectroly te :- These are completely ionised in their aqueoussolution e.g. KCl, NaOH etc.

( b ) W eak Electroly tes :- They ionise in aqueous sol. to the smaller extent or we can say they have low degree of ionisation, but ses with dilution e.g.NH 4OH, CH 3COOH etc.

1 . Molar conductivity of electrolytes generally ses with dilution.

2 . Rela tiv e : se in the value of m for strong electrolyte is quite small ascompared to that for weak electrolytes.



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As shown in the graph, molar conductivity of strong electrolytes attains a definitelimiting value when the conc. approaches zero or dilution approaches . It is

denoted by m & can be obtained from graph. In case of weak electrolytes such

as CH 3COOH, molar conductivity at dilution cannot be obtained.It canhowever be obtained by using Kohlrausch’s law.

Exp lanat i on : -1 . F o r w ea k e l ec tr o ly t es :- On dilution, the degree of ionisation of the weakelectrolyte ses, there by sing the value of m . Thus when limiting value of molar conductivity is reached, the degree of ionisation of weak electrolytes approachesunity i.e . whole of solute dissociate into ions. Thus a (degree of dissociation)Ostwald d i lu t i on law .

m

m

Molar conductivity at given conc.

Molar conductivity at dilution

2 . For strong electrolyte :- For strong electrolyte, the no. of ions in the sol. donotses because these are almost completely ionised in sol. at all conc’s. However in conc. sol. of strong electrolytes, inter ionic interactions are quite strong whichreduce the speed of ions and hence m .

Thus on sing the dilution, ions move apart and inter ionic interactions

ses and

m

ses.

m &

m

for strong electrolytes are related by Debye HuckelOnsager eq.

m = m – b c

b = const. which depends upon viscosity & dielectric const of solventc = conc. of sol.

Kohl rausch ’s l aw



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It states that “ at infinite dilution, when the dissociation of electrolyte iscomplete, each ion makes a definite contribution towards the molar conductivityof electrolyte, irrespective of the nature of the other ion with which it is associated.”

Molar conductivity at dilution can be expressed as the sum of the

contribution of its individual ion. If &

represent the molar conductivity of cation & anion respectively at dilution,

m v v

+ & – represent the no. of +ve and –ve ions furnished by electrolyte.

Appl i ca t i ons : -1. Calculation of molar conductivities m

of weak electrolytes at dilution.

2. Calculation of degree of dissociation of weak electrolytes

=

M c

m

3 . Ca l cu la t ion o f d i s s oc ia t i on c ons tant o f weak e l e ct r o lyt es

K =C

2

1

Also =

mc

m

mc

m

mc

m

k C

( / )

/

2

1

Thus K =C m

c

m mcm

2

( )

4 . De t e rminat i on o f th e s o lub il i ty o f s par ingly s o lub l e sa lt s

m

K

Molarity

1000

m

K

Solubility

1000

or Solubility =k

m

1000

Ionic salts like AgCl, BaSO4 etc. dissolve in H 2O to a very small extent. They aresparingly soluble salts.

5 . Ca l cu la ti on o f I on ic Pr oduc t o f H2O

K [ ][ ]H OH



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It is found that ionic conductances H + and OH – at dilution are

H H cm cm0

O 349 8 198 5

1 2 1 2. , . –

By Kohlrausch’s Law

H H OH 2

00 0 0 = 349.8 + 198.5 = 548.7 1 2

cm

Specific conductance of pure water at 298 K is found to be K = 5.54 × 10 –8 W–1 cm–

1

Applying the formula

mo

K 1000

Molarity

Molarity i.e. [H + ] = [OH - ]

E x a m p l e 1 - The specific conductivity of 0.02M KCl solution at 25 0C is

2.768 x 10-3ohm-1 cm-1 . The resistance th of this solution at 25 0 C whenmeasured with a particular cell was 250.2 ohm. The resistance of 0.01M CuSO4

solution at 25 0 C measured with same cell was 8331 ohm. Calculate the molar conductivity of CuSO4 solution.

Solu.- Cell constant = sp. conductance of KCl/ conductance of KCl= = 0.692

Now for 0.01M CuSO4 solution

Sp. conductivity = cell constant x Conductance = 0.692 = 8.312 x10-5

Molar conductivity = sp. conductivity = 8.312 x 10-5 = 8.312 x 10-5 ohm-

1cm2mol-1

Ex2- The conductivity of a saturated solution of AgCl is found to be1.86x 10-6ohm- 1 cm-1 and that of water is 610-8ohm-1 cm-1 .Calculate thesolubility of AgCl if ohm-1 cm2 eq-1.

S o l u . k AgCl = k AgCl(solution) - kWater = 1.86 10-6 - 610-8 = 1.8 10-6

S AgCl

10000

0

1000

AgCl

S

= 1.31 10-5 M

Prac t i c e Exe r c i s eQ . 1 . The equivalent conductance of an infinitely dilute solution of NH 4Cl is

150 and the ionic conductances of OH - and Cl- ions are 198 and 76



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respectively. What will be the equivalent conductance of the solution of NH 4OH at infinite dilution.If the equivalent conductance of a 0.01N solution is 9.6,what will be its degree of dissociation?

Ans- 0.0353Q . 2 . Given the equivalent conductance of sodium butyrate, sodium chloride

and hydrogen chloride as 83, 127 and 426 mho cm2 at 25 0 C respectively.Calculate the equivalent conductance of butyric acid at infinite dilution.

Ans- 382mho cm 2

Q . 3 . The molar conductivity of 0.0250 M HCOOH(aq) is 4.61 mS m2mol -1.D e t e r m i n e t h e pKa of the acid. Ans- 3 .44

Q . 4 . At 180 C, the mobilities of NH 4+ and ClO4

- ions are 6.6 10-4 and5.7 10-4 cm2 volt-1 sec-1 at infinite dilution. Calculate the equivalent conductanceof ammonium chlorate solution. A n s- 11 8.6 9 m h o cm 2 eq -1

E L E C T R O C H E M I C A L C E L L(a) It is the device in which the decrease in free energy during the spontaneousredox reaction is made to convert chemical energy into electrical energy is

called as electrochemical cell.(b) Galvanic and voltaic developed such devices and therefore these cells arealso known as Galvanic cells or voltaic cells or redox cells.

(c) The Daniel cell is a typical galvanic cell. It is designed to Make use of thespontaneous redox reaction between zinc and cupric ion to produce an electriccurrent.(d) The Daniel cell reaction represented as

Zn(S) + Cu+2

(aq)

Zn+2

(aq) + Cu(s)(e) The Daniel cell can be conventionally represented asZn(s) | Zn SO4 (aq) | salt bridge | CuSO4(aq.) | Cu (s) ,

Salt Bridge - It is an invert U-shaded tube filled with concentrationsolution of inert electrolytes such as KCl, KNO3, NH 4NO3. It filled with inertelectrolytes because for the flow of electric current to give unbroken coutaces



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all over the circuit. It prevents mechanical flow of solution but it provided free path for migration of ions to maintain an electric current through theelectrolyte solution. It prevents the accumulation of charges.

Galvanic cells : The device in which chemical energy is converted intoelectrical energy is called galvanic cell or electrochemical cell or voltaic cell.In electrochemical cell

Anode (–ve terminal) Oxidation occurs at Anode

Cathode (+ve terminal) Reduction occurs at cathode

Working of Ce l la. Zinc undergoes oxidation to form zinc ions.

Zn (s) ® Zn+2 (aq.) + 2e – (oxidation)b. The liberated electrons move towards ‘Cu’ strip.c. Cu+2 ions from electrolytic solution move towards Cu strip, pick up the e –sand get reduced to Cu atoms which are deposited at the ‘Cu’ strip

Cu+2 (aq.) + 2e – ® Cu (s) (Reduction)

Thus Zn is Anode and Cu is cathode.

Salt Br idge & i ts Funct ionSalt Bridge is a U-shaped tube containing a semisolid paste of some inert

electrolyte like KCl, KNO3, NH 4Cl etc. in agar - agar & gelatine. An inertelectrolyte is one whicha. does not react chemically with the sol. in either of the compartment.



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b. does not interfere with the net cell reaction.

Funct ions of sal t Br idge(i)It allows the flow of current by completing the circuit.(ii) It maintains electrical neutrality.(iii) It prevents accumulation of charges on electrodes & maintain flow of current.

(iv) It prevents liquid junction potential i.e. the potential difference which arisesbetween the two solutiuons when in contact with each other.

Note : - EMF is appl ied to an e lectrolytic ce l l whereas EMF is generated bya galvanic c e l l .

Repr e s enta t i on o f Galvanic c e l lLet M represent the element and Mn + its cation. Thus oxidation half cell is

represented as M/Mn + (C) [C molar conc. of ions]Reduction half cell is represented as Mn + (c) / M.

A cell is represented by writing the cathode on the RHS and anode on LHS.The two vertical lines are put between the two half cells which indicates saltbridge. The molar concentration is written in brackets after the ion.

for eg. Z n

Conventional current

Electron flow

Z Cun2

1( ) / /

( )1M Saturated / Cu

KCl M

+2

Anode (–) Cathode (+)

Salt Bridge Reduction Half.Oxidation Half

L LHSO Oxidation A AnodeN -ve terminal

E L E C T R O D E P O T E N T I A L

When a metal is placed in a solution of its ions, the metal acquires either a positive or negative charge with respect to the solution on account of this adefinite potential difference is developed between the metal and the solution.

This potential difference is called electrode potential. .Example - If a Zn plate is placed in a solution having Zn+2 ions, it becomesnegatively charged with respect to solution and thus a potential difference isset up between Zn plate and the solution.



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(i) The conversion of metal atoms into metal ions by the attractive force of polar water molecules.

M Mn+ + ne–

The metal ions go into the solution and electrons remains on the metalmaking it negatively charged.

“The tendency of the metal to change into ions is known aselectro lytic so lution pressure”

(ii) Metal ions start depositing on the metal surface leading to a positivechargeon the metal.“ This tendency of the ion is termed osmotic pressure”

The potential difference between the elctrode and the elctrolytic solution, when the electrode is in contact with the solution is called Electrode potential.

The magnitude of electrode potential depends on-(i) Nature of the elctrode(ii) Concentration of the ions in solution(iii) Temperature

Electrode potenial is of two types-1 . Ox idat i on Po t ent ia l (E

ox) 2 . R ed u c t ion P ot en t ia l (E

re d)

Eox

= - Ere d

EMF of the c e l l

It may be defined as the potential difference between the two terminals of thecell when either no or very little current is drawn from it. It is measured by potentiometer or by vaccum tube voltmeter.

EMF = Ecathode – Eanode

Standard El e c t r ode Po t ent ia lIt is not possible to measure the absolute value of the single elctrode potentialdirectly.

The common reference electrode used for this purpose is standard hydrogen electrode(SHE) or normal hydrogen electrode (NHE)whose electrode potential is arbitrarily taken to be zero. SHE have zero standard potential at all temp’s.



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It consists of a platinum foil coated with platinumblack (finely divided platinum) dipping partiallyinto an aqueous sol. in which the activity of

hydrogen ion is unity and hydrogen gas is bubbled through the sol. at 1 bar press & 298 K.

It can act as anode as well as cathode.

If SHE acts as anode then oxidation occurs as :

H 2 (g) 2H + (aq.) + 2e –

If SHE acts as cathode than reduction occurs as :-

2H + (aq.) + 2e – H 2 (g)

The electrode potentials of other electrodes aredetermined by coupling them with SHE.

Difference between EMF and Potential Difference- P.D. is the difference between the elctrode potentials of the two elctrodes under any condition while EMF is the potential generated by cell when it draws no current. EMF is the mximum voltage that a cell can deliver.



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Nernst Eq. for El ec t rode pot ent ia lFor a general reaction such asm1 A + m2 B + . . . . n1 X + n2Y + . . . .

Gibbs free energy change can be given by the equation

G = G0 + 2.303 RT log10 ....

....

21

21

mm

nn

B A

Y X

-nFE = -nFE0 + 2.303 RT Reactants

Productslog10

E = E0 -nF

RT 303.2log10Q at 25 0 C.

(This equat ion can be used for any s ingle e l e c t rode)

Mn + (aq.) + ne – M (s)

The potential of the electrode at any conc. measured with respect to standardhydrogen electrode can be represented by

E = E° – RT

nF

M

M n

ln

Converting natural logarithm to the base 10,

we get E = E0 –2.303 RT

nFlog

[M]

[Mn 1]

putting R = 8.314 JK–1 mole–1 , T = 298 K and F = 96500 coulombs, in aboveeq. we get

E = E° – 0 0591 1.log

n M n [ Since molar conc. of solid is taken as unity [M] = 1 ]

For a general e l e c t rochemical r eac t i on

a A + b B ne c C + d D.

Nernst eq. may be written as : -

E = E° – RT

nF

C D

A B

c d

a bln

E = E0 – cQnF

RT log

303.2



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Note :- If Q is inreased by 10 than EMF decreases by 59.2 mV and if Q isd ec re as ed by 10 tha n E MF in creases by 59 .2 m V.

Nens t Equat i on and Equi l i b r ium Condi t i onsat equilibrium E = 0 and Q c = Kc

E° =2 303.

log RT

n F K c

El e c t r och emica l Se r i e s : - By measuring the potential of various electrode versus hydrogen electrode

(SHE) a series of standard electrode potentials has been established.When the electrode (metal and non-metal) in contact with their ions are ar-ranged on the basis of the values of their standard reduction potentials or

standard oxidation potentials. The resulting series is called the electrochemi-cal or electromotive or activity series, of the elements.

S.NO. Reduction half cell reaction E° in volts at 25°C

1. F2

+ 2e – 2F – + 2.65

2.2

82OS + 2e – 2

4SO2 + 2.01

3. Co3+ + e – Co2+ + 1.82

4. PbO2

+ 4H+ +2

4SO + 2e – PbSO

4+ 2H

2O + 1.65

5.4

MnO + 8H+ + 5e – Mn2+ + 4H2O + 1.52

6. Au3+ + 3e – Au + 1.50

7. Cl2 + 2e – 2Cl – + 1.36

8.2

72OCr + 14H+ + 6e – 2Cr3+ + 7H

2O + 1.33

9. O2

+ 4H+ + 4e – 2H2O + 1.229

10. Br2

+ 2e – 2Br – + 1.07

11.3

NO + 4H+ +3e NO + 2H2O + 0.96

12. 2Hg2+ + 2e – 22

Hg + 0.92

13. Cu2+ + I – + e – CuI + 0.86

14. Ag+ + e – Ag + 0.799

15.2

2Hg + 2e – 2 Hg + 0.79

16. Fe3+ + e – Fe2+ + 0.77

17. I2 + 2e –

2I –

+ 0.53518. Cu+ + e – Cu + 0.53

19. Cu2+ + 2e – Cu + 0.34

20. Hg2Cl

2+ 2e – 2Hg + 2Cl – + 0.27

21. AgCl + e – Ag + Cl – + 0.222

22. Cu2+ + e – Cu+ + 0.15

23. Sn4+ + 2e – Sn2+ + 0.13

24. 2H+ + 2e – H2

0.00

25. Fe3+ + 3e – Fe – 0.036

26. Pb2+ + 2e – Pb – 0.126



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27. Sn2+ + 2e – Sn – 0.14

28. AgI + e – Ag + I – – 0.151

29. Ni2+ + 2e – Ni – 0.25

30. Co2+ + 2e – Co – 0.28

31. Cd2+ + 2e – Cd – 0.403

32. Cr3+ + e – Cr2+ - 0.41

33. Fe2+ + 2e – Fe 0.44

34. Cr3+ + 3e – Cr – 0.74

35. Zn2+ + 2e – Zn – 0.76236. 2 H

2O + 2e – H

2+ 2OH – – 0.828

37. Mn2+ + 2e – Mn – 1.18

38. Al3+ + 3e – Al – 1.66

39. H2

+ 2e – 2H – –2.25

40. Mg2+ + 2e – Mg – 2.37

41. Na+ + e – Na – 2.71

42. Ca2+ + e – Ca – 2.87

43. Ba2+ + 2e – Ba – 2.90

44. Cs+ + e – Cs – 2.92

45. K+ + e – – 2.93

46. Li+ + e – Li – 3.03

Appl i ca t i ons o f E l e c t r och emica l Se r i e s1 . Re la t iv e ox id i s ing and r educ ing powe rs o f vari ous substanc es :-

Substances with higher reduction potentials have more tendency to acceptelectrons so they are strong oxidising agents while substancess with lower reduction potentials are strong reducing agents.

2 . Ca l cu la t ion o f s tandard EMF of th e c el lE° cell = E° cathode – E° anode

3 . Pr ed i c ting f eas ib i l ity o f r edox r eac t ion

For a cell reaction to be spontaneous or feasible the E° cell is +ve reaction is feasible otherwise not.

E x . Peroxod sulphate salts (e.g. Na2S2O8) are strong oxidising agents used asbleaching agents for fats, oils and fabrics. Can oxygen gas oxidisesulphateion to peroxodisulphate ion (S2O8

2–) in acidic solution, with the O2(g) beingreduced to water Given O2(g) + 4H + (aq) + 4e– 2H 2O Eº = 1.23 V

S2O82– (aq) + 2e– 2SO4

2– (aq) Eº = 2.01 V

S o l . We want oxidation of SO42– by O2(g) which in turn is reduced to H 2O inacidic medium.

Oxdn. 2SO42– S2O8

2– + 2e– Eº = –2.01 VRedn. O2(g) + 4H + + 4e– 2H 2O Eº = +1.23 VNet. 4SO4

2– + O2(g) + 4H + 2S2O82– + 2H 2O Eºcell = –0.78 V



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The large negative values of Eºcell indicates that O2(g) will not oxidise SO42–

to S2O82– to any significant extent.

4 . Predic t ing the capabi li ty of metal to d isp lace H2

gas from ac id All metals lying above H 2 in electrochemical series can liberate H 2 gas by

reaction with acids while the metal lying below hydrogen in the electrochemicalseries cannot undergo such a reaction.

Reaction between metal m & acid to l iberate H2

gas is as fol lows : -

M + n H + (aq.) ¾® Mn + (aq.) + n

2H 2.

5 . The rmal s tab i li ty o f me ta l oxid esThe thermal stability of metal oxide depends on its electropositive nature, as

the elctropositive nature decreases from top to bottom in the electrochemical series

the thermal stability of oxides decreases . Thus the oxide of metal having high positive reduction potential are not stable towards heat.

6 . P r o d u c ts o f e le c tr o ly s i sElectrolysis :- It may be defined as a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution or molten state.

Electrolysis is conducted in an electrolytic cell which consist of two metalrods dipped in electrolytic solution. The electrode where reduction takes place iscalled cathode while electrode where oxidation takes place is called anode.

Some of the e l e c t ro lysis proc esses1 . El e c t ro lys i s o f mol t en s od ium ch l o r id e (NaCl) :-

E l e c t r o lyt i c r eac t i onNaCl (l ) Na+ (aq.) + Cl– (aq.)

Cathode : -Na + + e – Na (Reduction)

Anode : -Cl– Cl (oxidation) + e-

Cl + Cl

Cl2or 2Cl– Cl2. + 2e –

Overa l l Reac t i on : -2Na+ (aq.) + 2Cl– (aq.) 2Na (l ) + Cl2 (g)

(molten state)Thus Cl2 gas is liberated at anode.



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2 . El e ct r olys is o f an aqueous s o l. o f NaClNaCl (aq.) Na+ (aq.) + Cl– (aq.)H 2O (l ) H + (aq.) + OH – (aq.)

At cathode (–ve e l ectrode) :- Both Na+ and H + move towards cathode but asdischarge potential of H + ions is lower than that of Na+ ions, therefore H + ions

will be discharged at cathodeH + + e – H H + H H 2 (g)

At anode (+ve electrode ) :- Both Cl– and OH – move towards anode but as discharge potential of Cl– ions is lower than that of OH – ions therefore Cl– must bedischarged at anode.

Cl– – e – ClCl + Cl Cl2 (g)

Thus H 2 gas is liberated at cathode, while Cl2 gas is liberated at anode & NaOH remains in the sol.

3 . El e c tr o lys is o f aqueous CuSO 4 so l . (P t In e r t el e ct r od es ) : -CuSO4 (aq) Cu+2 (aq.) + SO4

–2 (aq.)H 2O (l) H + (aq.) + OH – (aq.)

At cathode :- Both Cu+2 and H + ions move towards cathode but as discharge potentialof Cu+2 ions is lower than that of H + ions, therefore, Cu+2 ions are discharged atcathode.

Cu+2 + 2e – Cu

At anode :- Both SO4–2 and OH – ions are present near anode but as discharge potential

of OH – ions is lower than that of SO4–2 ions therefore OH – ions are discharged at

anode.4OH – 4OH + 4e –

4OH 2H 2O (l ) + O2 (g)

. Predic t i on for the occurence of a r edox r eac t i on : -For a reaction to be spontaneous G0 < 0,since, G0 = - nFE0 , so for the a redox reaction to occur E0

cell > 0i.e. EMF of the cell must be positive.



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Question Using the standard values, predict the reaction will occur or not?(i)Fe3+(aq) & I-(aq) (ii)Ag+(aq) & Cu(s)(iii) Ag(s) & Fe3+(aq) (iv) Br 2(aq) & Fe2+(aq)

Given - V E FeFe

77.00

/ 23 , V E

I I 54.0

0

/ 2

, V E Ag Ag

80.00

/ ,

V E CuCu 34.0

0

/ 2

, V E Br Br 08.1

0

/ 2

Prac t i c e Exe r c i s e

Q . 1 . Will Fe be oxidised to Fe2+ by reaction with 1M HCl ? V E FeFe

44.00

/ 2 .

Q . 2 . )()101()2.0()(32

s Ag Ag M Mgs Mg

V E Ag Ag

8.00

/ , V E

Mg Mg37.20

/ 2

What will be the effect on EMF if concentration of Mg2+ is decreased to 0.1M.

Q . 3 . Reactions of tollen’s reagent are as follows Ag+ + e- Ag V E red 8.0

0

C6H 12O6 + H 2O C6H 12O 7 + 2H + + 2e-V E Oxd 05.0

0

Ag(NH 3)2+ + e- Ag + 2NH 3 V E red 373.0

0

(i) The value of lnK for 2Ag+ + C6H 12O6 + H 2O 2Ag + C6H 12O 7 + 2H + will be?

(ii) If on addition of NH 3, pH of solution rises to 11, then which of the followingelectrode is affected by change in pH.(a) Eox increase over E0 by 0.65V.(b) Ered increase over E0 by 0.65V.(c) Eox decrease over E0 by 0.65V.(d) Ered decrease over E0 by 0.65V.

Q . 4 . The EMF of the cells obtained by combining Zn & Cu electrodes of denielcell

with calomel elctrodes are 1.083V and -0.018V respectively at 25 0 C. If reduction potential of normal calomel electrode is 0.28V, find EMF of daniel cell.

Q . 5 . Consider the following cell

)2()1( / )1(,)1( 3423 Pt aCeCeaFeFePt

V E FeFe

77.00

/ 23 , V E CeCe

61.10

/ 34



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If an ammeter is connected between the two Pt electrodes predict the directionof flow of current. Ans- Pt(1) to Pt(2).

Q . 6 . Two students use same stock solution of ZnSO4 and a solution of CuSO4.The EMF of one cell is 0.03V higher than that of other. The concentration of

CuSO4 in the cell with higher EMF value is 0.5 M. Find out the concentration of CuSO4 in the other cell.

Q . 7 . An excess of granular Zn was added to 500 ml of 1 M Nickel nitrate tillequillibrium was established. Find out the concentration of Ni at equillibrium.

Given V E Zn Zn

75.00

/ 2 , V E Ni Ni

24.00

/ 2

The rmodynamic Tr ea tment Of Ne rns t Equat i on

(1) Gibb ’s Fre e energy and c e l l Pot ent ia lElectrical work done = - nF Ecel l .

We know that electrical work done on the system is equal to the increase in freeenergy of the system.

Hence :– G = - nF Ecell–G° = nF E°

As E°cell =RT

nF ln Kc.

Putting E°cell value in above eq.

–G° = nF RT

nF ln Kc

G° = – RT ln Kc.

or G° = – 2.303 RT log Kc.

( 2 ) H e at o f t he r ea ct io n i n t he ce ll

According to Gibbs Helmholtz equation

G = H + T PT

G

-nFE = H + T PT

nFE

)(



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H = -nFE + nFT PT

E

WherePT

E

is called the temperature coefficient of the cell

(i) If PT

E

>0, than process inside the cell is endothermic.

(ii) If PT

E

<0, than process inside the cell is exothermic.

( 3 ) E n tro py ch an ge in th e c el lG = H - T S . . . . .(1)

G = H + T PT

G

. . . . .(2)

On comparing the two equations

-T S = T PT

G

S = -

PT

G

S = nFPT

E

D I F F ER EN T T Y P E O F H A L F C EL L S

(1) Gas Ion Half Cell:- In this half cell, an inert electrode such as Pt or graphite isin contact with gas and a solution containing specified ions. for exampleHydrogen-gas-hydrogen cell

In this cell purified H 2 gas at a constant pressure is passed over Ptelectrode which is in contact with an acid solution.The half cell can act as both cathode and anode As Cathode: H

+(aq.) + e- 1/2H 2 (g) As Anode : 1/2 H 2 (g) H +(aq.) + e-

According to Nerns equation

Ecell =][

)(log

1

0591.0+

2

1

10

2

H

p H



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(2) Metal insoluble salt-anion half cell :- In this cell a metal coated with itsinsoluble salt in contact with a solution containing the anion of the insoluble salt.Follwing are some comman examples:

(i) Ag / AgCl/ Cl-

(ii) Hg- Hg2Cl2 - Cl-

(iii) Hg - HgO - OH -(iv) Hg- HgSO4 - SO4

2-

(v) Pb - PbSO4 - SO42-

Rpresentat ion of Metal insolubl e sal t -anion half c e l l :-Reduction potential - Eanion / insoluble salt / metal

Oxidation potential - Emetal / insoluble salt / anion

( i) A g - A gC l h a lf c el l: -

A S C A T H O D E AgCl(s) + e-

Ag(s) + Cl-

(aq)A S A N O D E A g(S ) + C l -(aq.) AgCl(s ) + e -

I l l u s t r a t i o n . The SRP of Ag+ /Ag electrode at 298 is 0.799 volt. K sp of AgI is8.7 10-17 . Evaluate the potential of the Ag+ /Ag electrode in a standard solutionof AgI. Also calculate the SRP of I- /AgI/Ag electrode.

SOLUTION- The point that must be noted that here a half cell is formed by coupling Ag+/Ag electrode with AgI solution, so we need to calculate the SRP of a new half cell.

Reaction at cathode AgI + e- Ag + I-

Reaction at anode Ag Ag + e -

net reaction AgI Ag + I-

Now ][Aglog1

0.05910

/ /

Ag Ag Ag Ag

E E

Ksp = [Ag+ ] [I- ] = [Ag+ ]2 = 8.7 ´ 10-17

So, [Ag+ ] = 9.3 ´ 10-9

][Aglog1

0.05910

/ /

Ag Ag Ag Ag

E E = 0.799 - 0.0591 log (9.3´10-9)

= 0.324 volt

Now, volt0.95-=log1

0.0591= sp

0 K E cell



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E0cell = EC - E A

-0.95 = Ag AgI I E

/ / - 0.799

Ag AgI I E

/ / = -0.151 Volt

( ii) H g - H g 2C l2 - C l - (Calomel e l e c t rode) : -Hg2Cl2 + 2e- 2Hg (l) + Cl-(aq.)

2 / /

o

/ / ]log[

2

0.0591-E

2222

Cl E HgCl HgCl

HgCl HgCl

( iii ) P b - P bC l2 - C l - e lectrode : -

PbCl2 + 2e- Pb(s) + 2Cl-(aq.)

Illustration 2. Given E0 = -0.268 V for Cl- /PbCl2 /Pb couple and -0.126V for the Pb2+ /Pb couple determine Ksp of PbCl2 at 25 0 C.

Solution:- First half cell PbCl2(s) + 2e- Pb(s) + 2Cl-

Second half cell Pb2+ + 2e- Pb(s)

][

1log

2

0.0591-EE

2 /

o / 22

PbPbPbPbPb

Ecell = PbPbPbPbClClE E

/ 2

/ 2 /

sp / / o

/ Pbo

2-2 / /

o / Pb

o

22 / /

o / Pb

o

2 / /

o

2 / Pb

ocell

Klog2

0.0591)-E-(E

]][Cllog[Pb2

0.0591)-E-(E

]log[2

0.0591]log[Pb

2

0.0591)-E-(E

)]log[2

0.0591-(E-

][Pb

1log

2

0.0591-EE

22

22

22

22

PbPbClClPb

PbPbClClPb

PbPbClClPb

PbPbClClPb

Cl

Cl

Ecell = 0 , so,

-0.146 = spKlog2

0.0591

Ksp = 1.56 x 10-5



24

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(3) Concentration Cells:- The cells in which electrical current is produced due to transport of a substance from higher to lower concentration. There are two types of concentration cells-

( i ) El e c t rode gas c onc ent ra ti on c el l :-

Pt, H 2 (P1) H +

(C) H 2(P2) ,PtHere, the gas is bubbled at two different partial pressures at electrode dippedin the solution of same electrolyte.Cell process

(Cathode))(2

1eH 22

- P H

(Anode)eH)(2

1 -12 P H

reaction)(Net)(2

1)(PH

2

12212 P H

From Nernst equation 2

1log2

0.0591

P

P E

For spontaniety P1>P2

I l l u s t r a t i o n The cell )()1(2 x pH H atm H Pt Normal calomel elctrode

has an EMF of 0.67 V at 25 0 C. Calculate the pH of the solution. The oxidation potential of the calomel elctrode on hydrogen scale is -0.28 V.

A N S - Applying Nernst equation for the given cell

reactants

Productslog

0591.00

n E E cellcell

anodecathodeocell E E E = 028 - 0.0

598.60591.0

28.067.0

][log

1

0591.028.067.0

2

pH

pH

H

A (ii) Electrolyte concentration cel l : - Here the concentration differencecreates the potential diffrence at the two electrodes.Zn(s) | ZnSO4 (C1) || ZnSO4 (C2) | Zn(s)

In such cells, concentration gradient arise in electrolyte solutions. Cell processmay be given as,



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e2CZnsZn 12 (Anodic process)

)C(Zn)C(Zn

)s(Zne2)C(Zn

12

22

22

(Over all process)

From Nernst equation, we have

2

1

C

Clog

F2

RT303.20E or

1

2

C

Clog

F2

RT303.2E

For spontanity of such cell reaction, C2> C1

)(),(.)(.),(),()( 2212 sPt Patmg H C aq H C aq H atmPg H sPt

2

1

C

Clog

0591.0

n E cell

( i ii ) Metal - i on - metal amalgum half c el l : -In this half cell electrode metal amalgam is placed in contact with asolution containing metal ion. Electrical contact is made by a platinum wiredipped into amalgam pool .

Mn+(aq.)(C2) + 2e- M(Hg) (C1)

2

10

)( / )( / C

Clog

0591.0

n E E

Hg M M Hg M M nn

Note:- Concentration of amalgam is represented as mass of metal per unitmass of mercury.



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E L E C T R O L Y S I SThe decomposition of electrolyte solution by passage of electric current,resulting into deposition of metals or liberation of gases at electrodes is knownas electrolysis.

E L E C T R O L Y T I C C E L LThis cell converts electrical energy into chemical energy.The entire assembly except that of the external battery isknown as the electrolytic cell

E L E C T R O D E SThe metal strip at which positive current enters is called anode ; anode is positively charged in electrolytic cell.

On the other hand, the electrode at which current leaves is called cathode.

Cathodes are negatively charged.

Anode Positive Loss of electron

or oxidation current takes place enters

Cathode Negative Gain of electron Currentor reduction leaves

takes place

F A R A D A Y ’ S L A W S O F E L E C T R O L Y S I S

1. Faraday’s First Law:- According to this law amount of substance deposited on an electrode duringelectrolysis is directy proportional to quantity of the electrical charge passed

through the electrolytic solution.If W is the mass deposited by passing Q coloumb of charge , then

W a Q or W = ZQ = Z´ I´tWhere, Z is known as electrochemical equivalent and is a characteristic of

compound deposited.

Z =96500

E So, W =96500

EIt

Note:- One Faraday is the charge carried by 1 mole of electrons.



27

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2. Faraday’s Second Law :- When same quantity of charge is passed throughdifferent electrolytes, then the masses of different substances deposited arein the ratio of their equivalent masses.

Alof masseq.

depositedAlof

Caof masseq.

depositedCaof

.

depositedNaof massmass

Naof masseq

mass

l l u s t r a t i o n sQ . 1 . A current of 3.7 ampere is passed for 6 hours between platinum electrodes in0.5 litre of a 2M solution of Ni(NO3)

2. What will be the molarity of the solution

at the end of electrolysis? What will be the molarity if Nickel electrodes are used .

Solution:- The electrode reaction is Ni2+ + 2e- Ni

amount of Ni(NO3)2 decomposed = amount of Ni deposited W=96500

EIt

For Ni(NO3)2 equivalent mass E = 1052

210

So, W = 95.8696500

360067.3105

Moles of Ni (NO3)2deposited = 414.0210

95.86

So, number of moles of Ni (NO3)2 left = 0.5´2 - 0.414 = 0.586

Molarity = M72.15.0

586.0

When nickel electrodes are used anode will dissolve and get deposited at

cathode, so the molarity of the solution remain unaffected.

Q . 2 . An aqueous solution of NaCl on electrolysis gives H 2(g), Cl2(g) and NaOH according to the reaction:

2 Cl- (aq.) + 2H 2O 2OH-(aq.) + H 2(g) + Cl2(g) A direct current of 25 ampere with a current efficiency 62% is passed through 20 L of NaCl solution (20% by mass). Write down the reactions



28

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taking place at the anode and cathode . How long will it take to produce 1Kg of Cl2?What will be the molarity of the solution with respect to hydroxideion?

S o l u t i o n : At anode : 2Cl- Cl2 + 2e-

At cathode: 2H 2O + 2e-

H 2 + 2OH -

effective current = 0.62 ´ 25 = 15.5 Ampmass of chlorine to be deposited = 1 Kg

using W =96500

EIt

hr I E

t 71.48.sec82.1753745.155.35

96500100096500W

For each mole of Cl2 , 2 moles of OH- are produced, soOH- produce = 2 ´ moles of Cl2 = 2 ´14.08 = 28.16

Molarity =20

16.28

Volume

Mole

S T O R A G E C E L L S A N D B A T T E R I E S Batteries can be classified as primary and secondary. Primary batteries can

not be returned to their original state by recharging, so when the reactants areconsumed, the battery is "dead" and must be discarded. Secondary batteries areoften called storage batteries or rechargeable batteries. The reactions in these batter-ies can be reversed; thus, the batteries can be recharged.

P R I M A R Y B A T T E R I E SD R Y C E L L S A N D A L K A L I N E B A T T E R I E S

Zinc serves as the anode, and the cathode is a graphite rod placed down thecenter of the device. These cells are often called "dry cells" because there is novisible liquid phase.the cell contains a moist paste of NH 4Cl, ZnCl2 and MnO2.The moisture is necessary because the ions present must be in a medium inwhich they can migrate from one electrode to the other. The cell generates a potential of 1.5 V using the following half-reactions:

Cathode, reductions: 2NH 4+ (aq) + 2e– 2NH 3(g) + H 2(g)

Anode, Oxidation : Zn (s) Zn2+(aq) + 2 e–



29

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M E R C U R Y C E L L SMercury cell, suitable for low current devices like hearing aids, watches, etc.consists of zinc – mercury amalgam as anode and a paste of HgO and carbonas the cathode. The electrolyte is a paste of KOH and ZnO. The electrode reactions for the cell are given below:

Anode: Zn(Hg) + 2OH – ZnO(s) + H 2O + 2e–

Cathode: HgO + H 2O + 2e– Hg(l ) + 2OH –

The overall reaction is represented byZn(Hg) + HgO(s) ZnO(s) + Hg(l )

The cell potential is approximately 1.35 V and remains constant during itslife as the overall reaction does not involve any ion in solution whoseconcentration can change during its life time.

S E C O N D A R Y B A T T E R I E S A secondary cell after use can be recharged by passing current through it in the opposite direction so that it can be used again. A good secondary cellcan undergo a large number of discharging and charging cycles. The mostimportant secondary cell is the lead storage battery.

L E A D S T O R A G E B A T T E R YIt is commonly used in automobiles and invertors. It consists of a lead anode

and a grid of lead packed with lead dioxide (PbO2 ) as cathode. A 38%solution of sulphuric acid is used as an electrolyte.Reactions when Discharging

The cell reactions when the battery is in use are given below:Anode: Pb(s) + SO4

2–(aq) PbSO4(s) + 2e–

Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e– PbSO4(s)+2H 2O(l )

i.e., overall cell reaction consisting of cathode and anode reactions is:Pb(s)+PbO2(s)+2H 2SO4(aq) 2PbSO4(s) + 2H 2O(l)

Reactions when chargingOn charging the battery the reaction is reversed and PbSO4(s) on anode andcathode is converted into Pb and PbO2, respectively.



30

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N I C K E L - C A D M I U M C E L Lnickel - cadmium cell which has longer life than the lead storage cell butmore expensive to manufacture.The chemistry of the cell utilizes the oxidation of cadmium and the

reduction of nickel (III) oxide under basic conditions.

Cathode, reduction :Ni(OH)3 (s) + H 2O(l ) + e–

Ni(OH)2(s) + OH –

(aq) Anode, Oxidation: Cd(s) + 2 OH – Cd(OH)2(s) + 2e–

The overall reaction during discharge is:Cd (s )+2Ni(OH)

3(s) CdO (s ) +2Ni(OH)

2(s ) +H

2O(l )

F U E L C E L LGalvanic cells that are designed to convert the energy of combustion of fuelslike hydrogen, methane,methanol, etc. directly into electrical energy arecalled fuel cells .

One of the most successful fuel cells uses the reaction of hydrogen withoxygen to form water . The cell was used for providing electrical power in the Apollo space programme. The water vapours produced during the reactionwere condensed and added to the drinking water supply for the astronauts.

n the cell,hydrogen and oxygen are bubbled through porous carbon electrodes intoconcentrated aqueous sodium hydroxide solution. Catalysts like finelydivided platinum or palladium metal are incorporated into the electrodes for

increasing the rate of electrode reactions. The electrode reactions are givenbelow:Cathode: O2(g) + 2H 2O(l ) + 4e– 4OH–(aq) Anode: 2H 2 (g) + 4OH–(aq) 4H 2O(l) + 4e–

Overall reaction being:

2H 2(g) + O2(g) 2 H 2O(l )

The cell runs continuously as long as the reactants are supplied. Fuel cells produce electricity with an efficiency of about 70 %.



CHEMISTRY

(ELECTROCHEMISTRY )

“ Boost your Basics ”

B Y P R I N C

E S I R

Q1. The cell reaction for thegiven cell is spontaneous if- Pt(H2, P

1) H+ (1M) | | H+ (1M) | Pt(H

2, P

2)

(A) P1

> P2

(B) P1

< P2

(C) P1

= P2

(D) P2= 1 atm

Q.2. The standard electrode potentials (reduction) of

2 / 42 / 300 and SnSnFeFe E E are 0.77 and 0.15 V Resp ectively at 25° C. The standard EMF of the

reaction

Sn4+ + 2Fe2+ Sn2+ + 2Fe3+ is -

(A) – 0.62 V (B) – 0.92 V (C) +0.31 V (D) + 0.85 V

Q.3. F2gas can’t be obtained by the electrolysis of any F – salt because-

(A) Fluorine is the strongest reducing agent

(B) Fluorine is the strongest oxidising agent

(C) Fluorine easily combine with atmospheric O2

(D) All of these

Q.4. Reverse potential is the potential app lied at which-

(A) Electrolytic cell converts into galvenic cell (B) Galvenic cell converts into electrolytic cell

(C) Cell reaction stops (D) EMF becomes zero

Q.5. Calculate the quantity of electricity (i.e. charge) delivered by a Daniel cell initially containing 1L each

of IM Cu2+ ion and 1M Zn2+ , which is operated until its potential drops to 1V. (Given :

E 0.76V;E 0.34 VZn2 / Zn

0

Cu2 / Cu

0

)

(A) 0.029 × 105 C. (B) 2.239 × 10 5 C

(C) 1.92 × 10 5 C. (D) 3.123 × 10 5 C

Q.6. Some half cell reaction & their standard potentials are given. Which combination would result in a

cell with the largest p otential ?

(i) A + e –

A –

E0

= – 0.24 V(ii) B – + e – B –2 E0 = 1.25 V

(iii) C – + 2e – C –3 E0 = – 1.25 V

(iv) D + 2e – D –2 E0 = 0.06 V

(v) E + 4e – E –4 E0 = 0.38 V

(A) i and ii (B) ii and iii (C) iv and v (D) ii and v



B Y P R I N C E

S I R

Q.7. A hydrogen electrode is immersed in a solution with pH = 0 (HCI). By how much will the potential

(reduction) change if an equivalent amount of NaOH is added to the solution. (Take p H2= 1 atm)

T = 298 K.

(A) Increases by 0.41 V (B) Increases by 59 mV

(C) Decreases by 0.41 V (D) decreases by 59 mV

Q.8. The solubility product of silver iodide is 8.3 × 10 –17 and the standard potential (reduction) of Ag, Ag+

electrode is + 0.800 volts at 25°C. The standard potential of Ag, AgI/I – electrode (reduction) from

this data is-

(A) – 0.30 V (B) + 0.15 V (C) + 0.10V (D) – 0.15 V

Q.9. EMF is an

(A) Intensive property (B) Extensive property (C) None of these (D) Can’t say

Q.10. The standard redox potentials E° of the following systems are-

System E° (volts)

(i) MnO4

– + 8H+ + 5e Mn2+ + 4H2

O 1.51

(ii) Sn4+ + 2e Sn2+ 0.15

(iii) Cr2O

7

2– + 14H+ + 6e 2Cr3+ + 7H2O 1.33

(iv) Ce4+ + e Ce3+ 1.61

The oxidising power of the various species decreases in the order

(A) Ce4+ > Cr2O7

2– > Sn4+ > MnO4 – (B) Ce4+ > MnO

4 – > Cr

2O

72– > Sn4+

(C) Cr2O

7

2– > Sn4+ > Ce4+ + MnO4

– (D) MnO4

– > Ce4+ > Sn4+ > Cr2O

7

2–

Q.11. Calculate the maximum work that can be obtained from the Daniell cell given below -

Zn(s) | Zn2+ (aq) | Cu(s). Cu 2+(aq)

Given E 0.76Vand E 0.34 VZn2 / Zn

0

Cu2 / Cu

0

(A) –212300 J (B) –202100 J (C) –513100 J (D) –232120 J

Q.12. Which is wrong about EMF ?

(A) It is the difference between electrode potential of two electrodes in an open circuit

(B) It is determined by means of a simple voltmeter

(C) The work calculated from EMF is the maximum work obtainable from a cell

(D) All are correct

Q.13. The standard reduction potentials, E°, for the half reactions are as ZnZn2+ + 2e; E° = + 0.76 V

and Fe Fe2+ + 2e; E° = + 0.41 V; the e.m.f. for the cell reaction. Fe2+ + Zn Zn2+ + Fe is-

(A) – 0.35 V (B) + 0.35 V (C) + 1.17 V (D) – 1.17 V

Q.14. Which of following cell can produce more electrical work ?

(A) Pt, H2| 0.1M NH

4Cl | | 0.1 M CH

3COOH | H

2, Pt

(B) Pt, H2| 0.1 M HCl | | 0.1 M NaOH | H

2, Pt

(C) Pt, H2| 0.1 M HCl | | 0.1 M CH

3COOK | H

2, Pt

(D) Pt, H2| 0.1 M CH

3COOK | | 0.1 M HC1|H

2, Pt



Q.15. Zn amalgam is prepared by electrolysis of aqueous ZnCl2using Hg cathode (9gm). How much current

is to be passed through ZnCl2solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by

wt. (Zn = 65.4)

(A) 5.6 amp (B) 7.2 amp (C) 8.85 amp (D) 11.2 amp

Q.16. Zn | Zn2+ (C1) | | Zn2+ (C2) | Zn. For this cell G is negative if :

(A) C1 = C2 (B) C1 < C2 (C) C2 < C1 (D) None of theseQ.17. The standard oxidation potentials of Cu/Cu2+ and Cu+ / Cu2+ are – 0.34 V and – 0.16 V respectively.

The standard electrode potential of Cu+ / Cu would be :

(A) 0.18 V (B) 0.52 V (C) 0.82 V (D) 0.49 V

Q.18. The specific conductivity of solution depends upon :

(A) Number of ions as well as mobility of ions (B) Number of ions per c.c solution

(C) Number of ions per cc as well as mobilities of ions (D) Mobilities of ions only

Q.19. Number of electrons involved in the electrodeposition of 63.5g of Cu from a solution of CuSO4is -

(A) 6.022 × 1023 (B) 3.011 × 1023 (C) 12.044 × 1023 (D) 6.022 × 1022

Q.20. Which of the following statements is correct with regard toG of cell reaction and EMF of the cell

(E) in which thereaction occurs ?

(A) Both G and E are extensive properties

(B)G is an intensive prop erty but E is an extensive propery

(C)G is an extensive property and E is an intensive prop erty

(D) Both G and E are intensive properties.

Q.21. The tempreature coefficient of the emf for the cell deg –1 . Calculate the entropy change S298K

, for

the cell reaction, Cd + 2AgCl Cd++ + 2Cl + 2Ag

(A) –105.5 JK –1 (B) –150.2 JK –1 (C) –75.5 JK –1 (D) –125.5 JK –1

Q.22. The cell Pt (H2) (1 atm) | H (pH = ?) | | I (a = 1) | AgI (s), Ag has emf, E

298= 0. The electrode

potential for the reaction AgI + e – Ag + I is – 0.151 volt. Calculate the pH value.

(A) 3.37 (B) 5.26 (C) 2.56 (D) 4.62

Q.23. The reduction potential of a hlaf-cell consisting of a Pt electrode immersed in 1.5 M Fe2+ and 0.015M

Fe3+ solution at 25° C is ( E 0.770 VFe3 / Fe2 )

(A) 0.652 V (B) 0.88 V (C) 0.710 V (D) 0.850 V

Q.24. In a concentration cell, Zn/Zn2+ (1.0 M) | | Zn2+ (0.15 M)/ Zn as the cell discharges,

(A) Reaction p roceeds to the right

(B) The two solutions approcah each other in concentration

(C) No reaction takes p lace

(D) Water gets decomposed



B

Y P R I N C E

S I R

Q.25. In a half-cell containing [Tl3+] = 0.1 M and [Tl+] = 0.01 M, the cell potential is – 1.2496 V for the

reaction Tl+ Tl3+ + 2e – . The standard reduction potential of the Tl+ /Tl3+ couple at 25° C is

(A) 1.44 V (B) 0.61 V (C) 2.44 V (D) 1.22 V

Q.26. In the following electrochemical cell, M|M+ | |X – | X, the standard reduction potentials are and

E 0.42 VX/ X

0

at 25° C. Which one of the following statements is correct ?

(A) M + X M+ + X – is spontaneous (B) M+ + X – M + X is spontaneous

(E) Ecell

= 0.77 V (D) Ecell

= – 0.77 V

Q.27. The volume of gases liberated at STP when a charge of 2F is passed through aqueous solution of

sodium phosphate- is :

(A) 11.2L (B) 44.8L (C) 33.6L (D) 22.4L

Q.28. The temperature coefficient of a standard Cd cell is( – 1.5 × 10 –5 )VK –1. The EMF of the cell at

250C is 1.018 V. During the cell operation, the temperature will -

(A) Increase (B) Decrease

(C) May increase or deacrease (D) remains constant

Q.29.

ClO3

– Cl

– 0.54 VClO

– 0.45 V 12

Cl21.07 V

1.71 V

E0

The E0 in the given diagram is,

(A) 0.5 (B) 0.6 (C) 0.7 (D) 0.8

Q.30. Calculate the useful work of the reaction

Ag(s) + 1/2Cl2(g) AgCl(s)

Given E 1.36V, E 0.22 VCl2 /Cl

0

AgCl/Ag,Cl

0

If PCl2= 1 atm and T = 298 K

(A) 110 kJ/mol (B)220 kJ/mol (C) 55 kJ/mol (D) 1000 kJ/mol

Q.31. Acetic acid has Ka= 1.8 × 10 –5 while fromic acid has K

a= 2.1 × 10 –4 . What would be the

magnitude of the emf of the cell

Pt (H2)

0.1Maceticacid

0.1M sodiumacetate

0.1M formicacid

0.1M sodiumformate

Pt (H

2) at 25° C ?

(A) 0.0315 volt (B) 0.0629 volt (C) 0.0455 volt (D) 0.0545 volt

Q.32. With t taken in seconds and I taken in Amp, the variation of I follows the equation : t 2 + I2 = 25

What amount of Ag will be electrodeposited with this current flowing in the interval 0–5 second ?

(Ag – 108)

(A) 22 mg (B) 66 mg (C) 77 mg (D) 88 mg



B Y P R

I N C E S I R

Q.33. Following cell has EMF 0.7995 V

Pt | H2

( 1 atm) | HNO3

(1 M) | | AgNO3

( 1 M) | Ag

If we add enough KCl to the Ag cell so that the final Cl –

is 1 M. Now the meacured emf of the cell is

0.222 V. The Ksp of AgCl would be :

(A) 1 × 10 –9.8 (B) 1 × 10 –19.6 (C) 2 × 10 –10 (D) 2.64 × 10 –14

Q.34. A flashlight cell has the cathodic reaction

2MnO2

(s) + Zn+2 + 2e – Zn Mn2

O4

(S)

If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the limiting reagent

MnO2is present ? [Mn = 55, O = 16]

(A) 2 × 106 sec (B) 4 × 106 sec (C) 6 × 106 sec (D) 8 × 106 sec

Q.35. A current of 0.1A was passed for 2hr through a solution of suprocyanide and 0.3745 g of copper was

deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu – 63.5)

(A) 79% (B) 39.5% (C) 63.25% (D) 6

Q.36. Which of the following changes will increase the EMF of the cell,

Co(s) CoCl2

(M1) HCl(M

2) Pt (H

2, g)

(A) Increase in the volume of the CoCl2

solution from 100 ml to 200 ml(B) Increasing the pressure of H

2(g) from 1 atm to 2 atm

(C) Increasing M2from 0.01 to 0.5 M

(D) Increasing M1from 0.01 to 0.5 M

Q.37. 100 ml of a buffer of 1 M NH3

(aq) and 1M NH4

+(aq) are placed in two voltaic cells seperately. A

current of 1.5 A is passed through both cells for 20 minutes. If only electrolysis of water takes place,

then pH of the :

(A) Left half cell will increase (B) Right half cell will increase

(C) both half cells will increase (D) bothe half cells will decrease

Q.38. A hydrogen electrode placed in a buffer solution of CH3COONa and acetic acid in the ratios x : y andy: x has electrode potential values E1andE

2respectively at 25

0

C. The pKa value of acetic acid is (E1

andE2are oxidation potentials)

(A)118.0

21 E E (B)

118.0

12 E E (C)

118.0

21 E E (D)

118.0

21 E E

Q.39. A solution containing 4.5 mM of Cr2O

7

2 -

and 15 mM of Cr3+

shows a pH of 2. calculate the potential

of half reaction. Standard potential of reactionCr2O

7

2 -

Cr3+

is 1.33 V)

(A) 2.4 V (B) 1.8 V (C) 1.07 V (D) 1.1V

Q.40. The EMF of the cell

Zn Zn2+

(0.01 M) Fe2+ (0.001 M) Fe at 298 K is 0.2905 then the value of equilibrium

constant for the cell reaction is -

(A) 0295.0

32.0

e(B) 0295.0

32.0

10(C) 0295.0

26.0

10(D) 0591.0

32.0

10

Q.41. 0.04 N solution of a weak acid has specific conductance 4.23mho cm-1. if the degree of



dissociation of acid at this dilution is0.0612, then the equivalent conductivity at infinite dilution is

.....mho cm2 eq-1

(A) 172.8 (B) 180 (C) 190 (D) 160

Multiple Choice Questions

Q.42. Consider the following graph find out the ncorrect option(s)

(A) The cell reaction is spontaneous at point C.

(B) The cell reaction is not spontaneous at A.

(C) At B cell potential is zero.

(D) The cell reaction is spontaneous at A.

Q.43 During discharging of lead storage battery, which of the following is/are true ?

(A) H2SO

4is p roduced (B) H

2O is consumed

(C) PbSO4

is formed at both electrodes (D) Density of electrolytic solution decreases

Q.44 Which of the following arrangement will produce oxygen at anode during electrolysis ?(A) Dilute H

2SO

4solution with Cu electrodes.

(B) Dilute H2SO

4solution with inert electrodes.

(C) Fused NaOH with inert electrodes.

(D) Dilute NaCl solution with inert electrodes.

Q.45 If 270.0 g of water is electrolysed during an experiment performed by miss Abhilasha with 75% current

efficiency then

(A) 168 L of O2

(g) will be evolved at anode at 1 atm & 273 K

(B) Total 504 L gases will be p roduced at 1 atm & 273 K.

(C) 336 L of H2 (g) will be evolved at anode at 1 atm & 273 K(D) 45 F electricity will be consumed

Q.46 Pick out the correct statements among the following from inspection of standard reduction potentials

(Assume standard state conditions).

2Cl (aq.) + 2e l 2Cl – (aq.)o

Cl / Cl2E = + 1.36 volt

Br2

(aq.) + 2e l 2Br – (aq.)o

Br / Br2E = + 1.09 volt

I2(s) + 2e l 2I – (aq.)

o

I / I2E = + 0.54 volt

282OS (aq.) + 2e l

24SO2 (aq.) o

SO / OS 24

282

E = + 2.00 volt

(A) Cl2

can oxidise 24SO from solution

(B) Cl2

can oxidise Br – and I – from aqueous solution

(C) 282OS can oxidise Cl – , Br – and I – from aqueous solution

(D) S2O

82– is added slowly, Br – can be reduce in presence of Cl –

A

B

C

G

e x t e n t o f r e a c t i o n



FUNDAMENTALS

GALVANIC CELL :

Representation of Cell diagrams, complete and half cell reactions :

Q.1 Write cell reaction of the following cells :

(a) Ag | Ag+ (aq) | | Cu2+ (aq) | Cu

(b) Pt | Fe2+

, Fe3+

| |4MnO , Mn

2+

, H+

| Pt(c) P t , C l

2| Cl – (aq) | | Ag+ (aq) | Ag

(d) Pt, H2

| H+ (aq) | | Cd2+ (aq) | Cd

Q.2 Write cell representation for following cells.

(a) Cd2+ (aq) + Zn (s) Zn2+ (aq) + Cd (s)

(b) 2Ag+ (aq) + H2

(g) 2H+ (aq) + 2Ag (s)

(c) 272

OCr (aq.) + 14H+ (aq) + 6Fe2+ (aq) 6Fe3+ (aq) + 2Cr3+ (aq) + 7H2O (l)

Q.3 Write half cells of each cell with following cell reactions :

(a) Zn (s) + 2H+ (aq) Zn2+ (aq) + H2 (g)(b) 2Fe3+ (aq) + Sn2+ (aq) 2Fe2+ (aq) + Sn4+ (aq)

(c) 4

MnO (aq) + 8H+ (aq) + 5Fe2+ (aq) 2Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)

(d) Pb (s) + Br2

(l) Pb2+ (aq) + 2Br – (aq)

(e) Hg2Cl

2(s) + Cu (s) Cu2+ (aq) + 2Cl – (aq) + 2Hg (l)

Electrode potential and standard electrode potential :

Q.4 For the cell reaction 2Ce4+ + Co 2Ce3+ + Co2+

ocell

E is 1.89 V. If o

Co|Co2E is – 0.28 V, what is the value of o3

Ce|4

CeE

?

Q.5 Determine the standard reduction potential for the half reaction :

Cl2

+ 2e – 2Cl –

Given Pt2+ + 2Cl – Pt + Cl2,

oCellE = – 0.15 V

Pt2+ + 2e – Pt E° = 1.20 V

Q.6 What isoCellE if

2Cr + 3H2O + 3OCl – 2Cr3+ + 3Cl – + 6OH –

2Cr3+ + 3e – Cr, E° = – 0.74 V

OCl – + H2O + 2e – Cl – + 2OH – , E° = 0.94 V

Q.7 Is 1.0 M H+ solution under H2SO

4at 1.0 atm capable of oxidising silver metal in the presence of 1.0 M

Ag+ ion?

o

Ag|AgE = 0.80 V,

o

)Pt(H|H 2

E = 0.0 VV

Q.8 If for the half cell reactions Cu2+ + e – Cu+ E° = 0.15 V

Cu2+ + 2e – Cu E° = 0.34 V



Calculate E° of the half cell reaction

Cu+ + e – Cu

also predict whether Cu+ undergoes disproportionation or not.

Q.9 If o

Fe|2

FeE = – 0.44 V,

o2Fe|3Fe

E = 0.77 V. Calculateo

Fe|3

FeE .

Q.10 Calculate the EMF of a Daniel cell when the concentration of ZnSO4 and CuSO4 are 0.001 M and0.1M resp ectively. The standard potential of the cell is 1.1V.

Q.11 For a cell Mg(s) | Mg2+(aq) || Ag+ (aq) | Ag, Calculate the equilibrium constant at 25°C. Also find the

maximum work per mole Mg that can be obtained by operating the cell.

E0 (Mg2+ /Mg) = 2.37V, E0 (Ag+ /Ag) = 0.8 V.

Q.12 The EMF of the cell M | Mn+ (0.02M) || H+ (1M) | H2(g) (1 atm), Pt at 25°C is 0.81V. Calculate the

valency of the metal if the standard oxidation of the metal is 0.76V.

Q.13 Equinormal Solutions of two weak acids, HA (pKa= 3) and HB (pK

a= 5) are each placed in contact

with equal pressure of hydrogen electrode at 25°C. When a cell is constructed by interconnecting themthrough a salt bridge, find the emf of the cell.

Q.14 In two vessels each containing 500ml water, 0.5m mol of aniline (Kb= 109) and 25mmol of HCl are

added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of

cell made by connecting them appropriately.

Q.15 Calculate E0 and E for the cell Sn | Sn2+ (1M) || Pb2+(103M) | Pb, E0 (Sn2+| Sn) = 0.14V,

E0 (Pb2+| Pb) = 0.13V. Is cell representation is correct?

Q.16 At what concentration of Cu

2+

in a solution of CuSO4 will the electrode potential be zero at 25°C?Given : E0 (Cu | Cu2+) = 0.34 V.

Q.17 A zinc electrode is p laced in a 0.1M solution at 25°C. Assuming that the salt is 20% dissociated at this

dilutions calculate the electrode potential. E0 (Zn2+| Zn) = 0.76V.

Q.18 From the standard potentials shown in the following diagram, calculate the potentials 1E and

2E .

CONCENTRATION CELLS :

Q.19 Calculate the EMF of the following cell

Zn | Zn2+ (0.01M) || Zn2+ (0.1 M) | Zn

at 298 K.



Q.20 Calculate the EMF of the cell,

Zn – Hg(c1M) | Zn2+ (aq)| Hg – Zn(c

2M)

at 25°C, if the concentrations of the zinc amalgam are: c1

= 10g per 100g of mercury and

c2

= 1g per100 g of mercury.

Q.21 Calculate pH using the following cell :

Pt (H2) | H

+

( x M) | | H

+

(1 M) | Pt (H2) if Ecell = 0.2364 V.1 atm 1 atm

Q.22 Calculate the EMF of following cells at 25°C.

(i) Fe | Fe2+ (a1

= 0.3) || Sn2+ (a2

= 0.1) | Sn E0 (Fe2+/Fe) = –0.44 V

(ii) Pt, H2

(2atm) | HCl |H2

(10 atm), Pt. E0 (Sn2+ /Sn) = 0.14 V

Q.23 EMF of the cell Zn | ZnSO4

(a1= 0.2) || ZnSO

4(a

2) | Zn is 0.0088V at 25°C. Calculate the value of a

2.

Q.24 Calculate the equilibrium constant for the reaction

Fe2+ +Ce4+l Fe3+ + Ce3+ , [given : V68.0E;V44.1E 2334 Fe / Fe

0Ce / Ce

0 ]

Q.25 Calculate the equilibrium constant for the reaction Fe + CuSO4l FeSO

4+ Cu at 25°C.

Given E0 (Fe/Fe2+) = 0.44V, E0 (Cu/Cu2+) = 0.337V.

Q.26 The standard reduction potential at 25°C for the reduction of water

2H2O + 2e

l H2

+ 2OH is 0.8277 volt. Calculate the equilibrium constant for the reaction

2H2O l H

3O+ + OH at 25°C.

Q.27 At 250C the value of K for the equilibrium Fe3+ + Ag lFe2+ + Ag+ is 0.531mol/litre. The standard

electrode p otential for Ag+ + e – l Ag is 0.799V. What is the standard potential for

Fe3+ + e – l Fe2+ ?

Q.28 For the reaction, 4Al(s) + 3O2(g) + 6H

2O + 4 OH –

l 4 [Al(OH)4 – ] ;

cellE = 2.73 V. If

)OH(G f = –157 kJ mol –1 and )OH(G 2f

= –237.2 kJ mol –1, determine f G [Al (OH)4 – ].

ELECTROLYTIC CELL :

Q.29 Calculate the no. of electrons lost or gained during electrolysis of

(a) 3.55 gm of Cl – ions (b) 1 gm Cu2+ ions (c) 2.7 gm of Al3+ ions

Q.30 How many faradays of electricity are involved in each of the case

(a) 0.25 mole Al3+ is converted to Al.

(b) 27.6 gm of SO3

is convered to 23

SO

(c) The Cu2+ in 1100 ml of 0.5 M Cu2+ is converted to Cu.

Q.31 0.5 mole of electron is passed through two electrolytic cells in series. One contains silver ions, and the

other zinc ions. Assume that only cathode reaction in each cell is the reduction of the ion to the metal.

How many gm of each metals will be deposited.



Q.32 The electrosynthesis of MnO2

is carried out from a solution of MnSO4

in H2SO

4(aq). If a current of

25.5 ampere is used with a current efficiency of 85%, how long would it take to produce 1 kg of MnO2?

Q . 3 3 A c o n s t a n t c u r r e n t o f 3 0 A i s p a s s e d t h r o u g h a n a q u e o u s s o l u t i o n o f N a C l f o r a t i m e o f 1 . 0 h r . H o w m a n y

g r a m s o f N a O H a r e p r o d u c e d ? W h a t i s v o l u m e o f C l

2gas at S.T.P. p roduced?

Q.34 If 0.224 litre of H2 gas is formed at the cathode, how much O2 gas is formed at the anode under identicalconditions?

Q.35 If 0.224 litre of H2

gas is formed at the cathode of one cell at S.T.P., how much of Mg is formed at the

cathode of the other electrolytic cell.

Q.36 Assume 96500 C as one unit of electricity. If cost of electricity of producing x gm Al is Rs x, what is the

cost of electricity of producing x gm Mg?

Q . 3 7 C h r o m i u m m e t a l c a n b e p l a t e d o u t f r o m a n a c i d i c s o l u t i o n c o n t a i n i n g C r O

3according to following equation:

CrO3(aq) + 6H+ (aq) + 6e – Cr(s) + 3H

2O

Calculate :

(i) How many grams of chromium will be plated out by 24000 coulombs and(ii) How long will it take to plate out 1.5 gm of chromium by using 12.5 ampere current

Q.38 Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if

the current efficiency for the process is 50 percent. If the potential drop across the cell is 3.0 volts, how

much energy will be consumed?

Q.39 How long a current of 2A has to be passed through a solution of AgNO3

to coat a metal surface of

80cm2 with 5m thick layer? Density of silver = 10.8g/cm3.

Q.40 3A current was passed through an aqueous solution of an unknown salt of Pd for 1Hr. 2.977g of Pd+n

was deposited at cathode. Find n.

Q.41 50mL of 0.1M CuSO4

solution is electrolyzed with a current of 0.965 A for a p eriod of

200 sec. The reactions at electrodes are:

Cathode : Cu2+ + 2e Cu(s) Anode : 2H2O O

2+ 4H+ + 4e.

Assuming no change in volume during electrolysis, calculate the molar concentration of Cu2+, H+ and

SO42at the end of electrolysis.

Q.42 A metal is known to form fluoride MF2. When 10A of electricity is passed through a molten salt for 330

sec., 1.95g of metal is deposited. Find the atomic weight of M. What will be the quantity of electricity

required to deposit the same mass of Cu from CuSO4?

Q.43 10g solution of CuSO4

is electrolyzed using 0.01F of electricity. Calculate:

(a)The weight of resulting solution (b)Equivalents of acid or alkali in the solution.

Q.44 Cadmium amalgam is p repared by electrolysis of a solution of CdCl2

using a mercury cathode. How

long should a current of 5A be passed in order to prepare 12% CdHg amalgam on a cathode of 2gm

Hg (Cd=112.4)



Q.45 After electrolysis of NaCl solution with inert electrodes for a certain period of time. 600 mL of the

solution was left. Which was found to be 1N in NaOH. During the same time, 31.75 g of Cu was

deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage yield of

NaOH obtained.

Q.46 Three electrolytic cells A, B, C containing solution of ZnSO4, AgNO

3and CuSO

4, respectively are

connected in series. A steady current of 2 ampere was passed through them until 1.08 g of silverdeposited at the cathode of cell B. How long did the current flow? What mass of copper and of zinc

were deposited?

Q.47 A solution of Ni(NO3)2

is electrolysed between p latinum electrodes using a current of 5 ampere for

20 mintue. What mass of Ni is deposited at the cathode?

Q.48 A current of 3.7A is p assed for 6hrs. between Ni electrodes in 0.5L of 2M solution of Ni(NO3)2.

What will be the molarity of solution at the end of electrolysis?

CONDUCTANCE

Conductivities and cell constant:

Q.49 The resistance of a conductivity cell filled with 0.01N solution of NaCl is 210 ohm at18oC.Calculatethe equivalent conductivity of the solution. The cell constant of the conductivity cell is 0.88 cm1.

Q.50 The molar conductivity of 0.1 M CH3COOH solution is 4.6 S cm2 mole1 . What is the specific

conductivity and resistivity of the solution ?

Q.51 The conductivity of pure water in a conductivity cell with electrodes of cross sectional area 4 cm2

and 2 cm ap art is 8 × 107 S cm1.

(i) What is resistance of conductivity cell ?

(ii) What current would flow through the cell under an app lied potential difference of 1 volt?

Q.52 Resistivity of 0.1M KCl solution is 213 ohm cm in a conductivity cell. Calculate the cell constant

if its resistance is 330 ohm.

Q.53 Resistance of a 0.1M KCl solution in a conductance cell is 300 ohm and sp ecific conductance of

0.1M KCl is 1.29 × 10-2 ohm-1 cm-1. The resistance of 0.1M NaCl solution in the same cell is 380

ohm. Calculate the equivalent conductance of the 0.1M NaCl solution.

Q.54 For 0.01N KCl, the resistivity 709.22 mho cm. Calculate the conductivity and equivalent

conductance.

Q.55 A solution containing 2.08 g of anhydrous barium chloride is 400 CC of water has a sp ecificconductivity 0.0058 ohm –1cm –1. What are molar and equivalent conductivities of this solution.

Application of Kohlrausch's law:

Q.56 Equivalent conductance of 0.01 N Na2SO

4solution is 112.4 ohm –1 cm2 eq –1. The equivalent

conductance at infinite dilution is 129.9 ohm –1 cm2eq –1. What is the degree of dissociation in 0.01

N Na2SO

4solution?



Q.57 Specific conductance of a saturated solution of AgBr is 8.486×10 –7 ohm –1cm –1 at 250C. Specific

conductance of p ure water at 25°C is 0.75 ×10 –6 ohm –1 cm –2. m for KBr , AgNO3

and KNO3

are 137.4 , 133 , 131 ( S cm2 mol –1) resp ectively. Calculate the solubility of AgBr in gm/litre.

Q.58 Saturated solution of AgCl at 25°C has specific conductance of 1.12×10 –6 ohm –1 cm –1. The

Ag+ and Cl – are 54.3 and 65.5 ohm –1 cm2 / equi. resp ectively. Calculate the solubility product

of AgCl at 25°C.

Q.59 Hydrofluoric acid is weak acid. At 25°C, the molar conductivity of 0.002M HF is

176.2 ohm –1 cm2 mole –1. If its m= 405.2 ohm –1 cm2 mole –1, calculate its degree of dissociation

and equilibrium constant at the given concentration.

Q.60 The value of m for HCl, NaCl and CH3CO

2Na are 426.1, 126.5 and 91 S cm2 mol –1 respectively..

Calculate the value of m for acetic acid. If the equivalent conductivity of the given acetic acid is 48.15

at 25° C, calculate its degree of dissociation.

Q.61 Calculate the specific conductance of a 0.1 M aqueous solution of NaCl at room temperature,

given that the mobilities of Na+ and Cl – ions at this temp erature are 4.26×10 –8 and

6.80×10 –8 m2 v –1 s –1, resp ectively.

Q.62 For the strong electroytes NaOH, NaCl and BaCl2

the molar ionic conductivities at infinite dilution

are 248.1×10 –4, 126.5 ×10 –4 and 280.0 ×10-4 mho cm2 mol –1 resp ectively. Calculate the molar

conductivity of Ba(OH)2

at infinite dilution.

Q.63 At 25°C, (H+) = 3.4982 ×10 –2 S m2 mol –1 and (OH – ) = 1.98 ×10 –2 S m2mol –1.

Given: Sp . conductnace = 5.7 ×10 –6 S m –1 for H2O, determine pH and K

w.



APPLICATION

Q.1 The standard reduction potential values, E0(Bi3+|Bi) and E0(Cu2+|Cu) are 0.226V and 0.344V

respectively. A mixture of salts of bismuth and copper at unit concentration each is electrolysed at 25°C.

to what value can [Cu2+] be brought down before bismuth starts to deposit, in electrolysis.

Q.2 The cell Pt, H2(1 atm) | H+(pH=x) || Normal calomel Electrode has an EMF of 0.67V at 25°C. Calculate

the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is 0.28 V.

Q.3 Voltage of the cell Pt, H2(1 atm)|HOCN (1.3 × 103 M)||Ag+ (0.8 M)|Ag(s) is 0.982 V . Calculate

the Kafor HOCN . Neglect [H+] because of oxidation of H

2(g) .

Ag+ + e Ag(s) = 0.8 V.

Q.4 Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally

contained 0.1M MnO4

and 0.8M H+ and which was treated with 90% of the Fe2+ necessary to reduce

all the MnO4

to Mn+2.

MnO4

+ 8H+ + 5e Mn2+ + 4H2O, E0 = 1.51V

Q.5 Calculate the emf of the cell

Pt, H2(1.0 atm) | CH

3COOH (0.1M) || NH

3(aq, 0.01M) | H

2(1.0 atm),

Pt Ka(CH

3COOH) = 1.8 × 105, K

b(NH

3) = 1.8 × 105.

Q.6 The Edison storage cell is rep resented as Fe(s) | FeO(s) | KOH(aq) | Ni2O

3(s) | Ni(s) The half cell

reaction are

Ni2O

3(s) + H

2O(i) + 2e

l 2NiO(s) + 2OH E0 = + 0.40V

FeO(s) + H2O(l) + 2el Fe(s) + 2OH E0 = 0.87V

(i) What is the cell reaction?

(ii) What is the cell e.m.f.? How does it depend on the concentration of KOH?

(iii) What is the maximum amount of electrical energy that can be obtained from one mole of Ni2O

3?

Q.7 The standard reduction potential for Cu2+ / Cu is 0.34 V. Calculate the reduction p otential at

pH = 14 for the above couple. Ksp

of Cu(OH)2

is 1 × 1019.

Q.8 Determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in M/32 solution of

salt at 298 K from the following cell data at 298 K.

Pt | H2

(1 atm) | H+(1M) || M/32 C6H

5NH

3Cl | H

2(1 atm) | Pt ; E

cell= – 0.188 V.

Q.9 The emf of the cell, Pt | H2

(1 atm), | H+ (0.1 M, 30 ml) || Ag+ (0.8 M) | Ag is 0.9 V. Calculate the emf

when 40 ml of 0.05 M NaOH is added.

Q.10 The emf of the cell Ag|AgI|KI(0.05M) || AgNO3(0.05M) |Ag is 0.788V. Calculate the solubility product

of AgI.

Q.11 Consider the cell Ag|AgBr(s)|Br ||AgCl(s), Ag | Cl at 25º C . The solubility product constants of AgBr

& AgCl are respectively 5 × 1013 & 1 × 1010 . For what ratio of the concentrations of Br & Cl ions

would the emf of the cell be zero ?



Q.12 The p Ksp

of Agl is 16.07 . If the Eº value for Ag+Ag is 0.7991 V . Find the Eº for the half cell reaction

AgI(s) + e Ag + I.

Q.13 For the galvanic cell : Ag|AgCl(s)| KCl (0.2M) || K Br (0.001 M)| AgBr(s) | Ag,

Calculate the EMF generated and assign correct polarity to each electrode for a sp ontaneous process

after taking into account the cell reaction at 250C.

]103.3K;108.2K[13

)AgBr(sp10

)AgCl(sp

Q.14 Given, E° = –0.268 V for the Cl – | PbCl2

| Pb couple and – 0.126 V for the Pb2+ | Pb couple, determine

Ksp

for PbCl2

at 25°C?

Q.15 Calculate the voltage, , of the cell at 25° C

Mn(s) | Mn(OH)2(s) | Mn2+(x M), OH – (1.00 x 10 –4M) || Cu2+(0.675M) | Cu(s)

given that Ksp

= 1.9 × 10 –13 for Mn(OH)2(s) E0 (Mn2+/Mn) = –1.18 V

Q.16 Calculate the voltage, E, of the cell

Ag(s) | AgIO3

(s) | Ag+(x M), HIO3

(0.300M) || Zn2+ (0.175M) | Zn(s)

if Ksp

= 3.02 × 10 –8 for AgIO3(s) and K

a= 0.162 for HIO

3.

Q.17 The voltage of the cell

Pb(s) | PbSO4(s) | NaHSO

4(0.600M) || Pb2+(2.50 x 10 –5M) | Pb(s)

is E = +0.061 V. Calculate K2

= [H+] [SO42– ] / [HSO

4 – ], the dissociation constant for

4HSO .

Given : Pb (s) + SO42– (aq) l PbSO

4(s) + 2e – (E0 = 0.356) E0(Pb2+/Pb) = –0.126 V

Q.18 The voltage of the cell

Zn(s) | Zn(CN)42– (0.450M), CN – (2.65 × 10 –3M) || Zn2+(3.84 × 10 –4M) | Zn(s)

is E = +0.099 V. Calculate the constant Kf

for Zn2+ + 4CN – l Zn(CN)

4

2– , the only

Zn2+ + CN – complexation reaction of importance.

Q.19 Given the standard reduction p otentials Tl+ + e TI, E0 = 0.34V and

TI3+ + 2e TI+, E0 = 1.25V. Examine the spontaneity of the reaction, 3TI+ 2TI + TI3+. Also find

E0 for this disproportionation.

Q.20 Estimate the cell potential of a Daniel cell having 1 M Zn++ & originally having 1M Cu++after sufficient

NH3

has been added to the cathode compartment to make NH3

concentration 2 M.

K f

for [Cu(NH3)4]2+ = 1 × 1012, E0 for the reaction,

Zn + Cu2+ Zn2+ + Cu is 1.1 V.

Q.21 The normal oxidation potential of Zn referred to SHE is 0.76V and that of Cu is0.34V at 25°C. When

excess of Zn is added to CuSO4, Zn displaces Cu2+ till equilibrium is reached. What is the ratio of Zn2+

to Cu2+ ions at equilibrium?

Q.22 Kd

for complete dissociation of [Ag(NH3)2]+ into Ag+ and 2NH

3is 6 × 108. Calculate E0 for the

following half reaction; Ag(NH3)2+ + e Ag + 2NH

3

Ag+ + e Ag, E0 = 0.799 V



Q.23 The overall formation constant for the reaction of 6 mol of CN with cobalt (II) is

1 × 1019. The standard reduction potential for the reaction

[Co(CN)6]3 + e Co(CN)

64 is 0.83 V. Calculate the formation constant of [Co(CN)

6]3

Given Co3+ + e Co2+ ; E0 = 1.82 V

Q.24 Calculate E° for the following reactions at 298 K,

23)NH(Ag + e – l Ag + 2NH3

2)CN(Ag + e –

l Ag + 2CN –

Given: ])NH(Ag[K,V7991.0E 23InsAg|Ag

= 6.02 × 10 –8 and ])CN(Ag[K 2Ins = 1.995 ×10 –19

Q.25 Calculate the equilibrium constant for the reaction:

3Sn(s) + 2Cr2O

72– + 28H+ 3Sn4+ + 4Cr3+ + 14H

2O

E0 for Sn/Sn2+ = 0.136 V E0 for Sn2+ /Sn4+ = – 0.154 V

E0 for Cr2O

72– /Cr3+ = 1.33 V

Q.26 Calculate the equlibrium concentrations of all ions in an ideal solution prepared by mixing 25.00 mL of 0.100M Tl+ with 25.00mL of 0.200M Co3+.

E0 ( Tl+ /Tl3+ )= –1.25 V ; E0 (Co3+ /Co2+ ) = 1.84 V

Q.27 Same quantity of electricity is being used to liberate iodine (at anode) and a metal x (at cathode). The

mass of x dep osited is 0.617g and the iodine is comp letely reduced by 46.3 cc of 0.124M sodium

thiosulphate. Find the equivalent mass of x.

Q.28 In a fuel cell, H2

& O2

react to produce electricity. In the process, H2

gas is oxidized at the anode & O2

at the cathode . If 67.2 litre of H2

at STP react in 15 minutes, what is the average current produced ? If

the entire current is used for electrode deposition of Cu from Cu (II) solution, how many grams of Cuwill be deposited?

Anode : H2

+ 2OH 2H2O + 2 e – Cathode : O

2+ 2 H

2O + 4e 4 OH –

Q.29 One of the methods of p reparation of per disulphuric acid, H2S

2O

8, involve electrolytic oxidation of

H2SO

4at anode (2H

2SO

4 H

2S

2O

8+ 2H+ + 2e) with oxygen and hydrogen as

byp roducts. In such an electrolysis, 9.722 L of H2

and 2.35 L of O2

were generated at STP. What is

the weight of H2S

2O

8formed?

Q.30 During the discharge of a lead storage battery the density of sulphuric acid fell from 1.294 to

1.139 g.ml1. H2SO

4of density 1.294 g mL1 is 39% and that of density 1.139 g mL1 is 20% by

weight. The battery holds 3.5L of acid and the volume practically remains constant during the discharge.Calculate the number of amp ere hours for which the battery must have been used. The discharging

reactions are:

Pb + SO42 PbSO

4+ 2e (anode)

PbO2

+ 4H+ + SO42 + 2e PbSO

4+ 2H

2O (cathode)

Q.31 A current of 3 amp was passed for 2 hour through a solution of CuSO4

,3 g of Cu2+ ions were deposited

as Cu at cathode. Calculate percentage current efficiency of the p rocess.



Q.32 An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolyzed until all the copper is deposited.

The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the

current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.

Q.33 In the refining of silver by electrolytic method what will be the weight of 100 gm Ag anode if

5 amp ere current is passed for 2 hours? Purity of silver is 95% by weight.

Q.34 Dal lake has water 8.2 ×1012 litre approximately. A power reactor produces electricity at the rate of

1.5×106coulomb per second at an appropriate voltage.How manyyears would it take to electrolyse the

lake?

Q.35 A lead storage cell is discharged which causes the H2SO

4electrolyte to change from a concentration of

34.6 % by weight (density 1.261g ml –1 at 25°C) to 27 % by weight. The original volume of electrolyte

is one litre. Calculate the total charge released at anode of the battery. Note that the water is p roduced

by the cell reaction as H2SO

4is used up . Over all reaction is

Pb(s) + PbO2(s) + 2H

2SO

4(l) 2PbSO

4(s) + 2H

2O(l)

Q.36 100ml CuSO4(aq) was electrolyzed using inert electrodes by passing 0.965 A till the pH of the resulting

solution was 1. The solution after electrolysis was neutralized, treated with excess KI and titrated with

0.04M Na2S

2O

3. Volume of Na

2S

2O

3required was 35 ml. Assuming no volume change during

electrolysis, calculate:

(a) duration of electrolysis if current efficiency is 80% (b) initial concentration (M) of CuSO4.

Q.37 An external current source giving a current of 5.0 A was joined with Daniel cell and removed after

10 hrs. Before passing the current the LHE and RHE contained 1L each of 1M Zn2+ and Cu2+respectively.

Find the EMF supplied by the Daniel cell after removal of the external current source. E0 of Zn2+ /Zn and

Cu2+ /Cu at 25°C is 0.76 and +0.34V respectively.

Q.38 Determine at 298 for cell

Pt | Q, QH2, H+ || 1M KCl | Hg

2Cl

2(s) | Hg(l) | Pt

(a) it's emf when pH = 5 .0

(b) the p H when Ecell

= 0

(c) the positive electrode when pH = 7.5

given E0RP(RHS)

= 0.28, E0RP(LHS)

= 0.699

Q.39 At 25°C, Hf (H

2O,l) = –56700 cal / mol and energy of ionization of H

2O (l) = 19050 cal/mol. What

will be the reversible EMF at 25°C of the cell,

Pt | H2(g) (1 atm) | H+ || OH – | O

2(g) (1 atm) | Pt, if at 26°C the emf increas by 0.001158 V.

Q.40 Calculate the cell potential of a cell having reaction: Ag2S + 2e –

l2Ag + S2– in a solution buffered at

pH = 3 and which is also saturated with 0.1 M H2S.

For H2S : K

1= 10 –8 and K

2= 1.1 × 10 –13, K

sp(Ag

2S) = 2 × 10 –49, .8.0E

Ag / Ag

Q.41 The equivalent conductance of 0.10 N solution of MgCl2

is 97.1 mho cm2 equi –1 at 25°C. a cell

with electrode that are 1.5 cm2 in surface area and 0.5 cm ap art is filled with 0.1 N MgCl2

solution.

How much current will flow when p otential difference between the electrodes is 5 volt.



Q.42 A dilute aqueous solution of KCl was p laced between two electrodes 10 cm apart, across which a

p otential of 6 volt was app lied. How far would the K+ ion move in 2 hours at 25°C? Ionic

conductance of K+ ion at infinite dilution at 25°C is 73.52 ohm –1 cm2 mole –1?

Q.43 When a solution of sp ecific conductance 1.342 ohm –1 metre –1 was p laced in a conductivity cell

with p arallel electrodes, the resistance was found to be 170.5 ohm. Area of electrodes is1.86×10 –4 m2. Calculate separation of electrodes.

Q.44 The specific conductance at 25°C of a saturated solution of SrSO4is 1.482×10 –4 ohm –1 cm –1while

that of water used is 1.5×10 –6 mho cm –1. Determine at 25°C the solubility in gm per litre of SrSO4

in water. Molar ionic conductance of Sr2+ and SO4

2– ions at infinite dilution are 59.46 and

79.8 ohm –1 cm2 mole –1 respectively. [ Sr = 87.6 , S = 32 , O = 16 ]

Q.45 Calculate the solubility and solubility product of Co2

[Fe(CN)6] in water at 250C from the following

data:

Conductivity of a saturated solution of Co2[Fe(CN)

6] is 2.06 × 10 –6 –1 cm –1 and that of water used

4.1 × 10 –7 –1 cm –1 . The ionic molar conductivities of Co2+ and Fe(CN)64– are 86.0 –1 cm2 mol –1

and 444.0 –1 cm –1mol –1.

Q.46 A sample of water from a large swimming pool has a resistance of 9200 at 25°C when placed in a

certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 85 at

25°C. 500 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution

gave a resistance of 7600 . Calculate the volume of water in the pool.

Given : Molar conductance of NaCl at that concentration is 126.5 –1 cm2 mol –1 and molar conductivity

of KCl at 0.02 M is 138 –1 cm2 mol –1.



FEEL THE HEAT

Q.1 Calculate the equilibrium constant for the reaction, 2Fe3+ + 3Il 2Fe2+ + I

3. The standard

reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for Fe3+ / Fe2+ and I3

/ I

couples. [JEE 1998]

Q.2 Find the solubility p roduct of a saturated solution of Ag2CrO

4in water at 298 K if the emf of the cell

Ag|Ag+

(satd.Ag2 CrO4 soln.) || Ag+

(0.1 M) | Ag is 0.164 V at 298K. [JEE 1998]

Q.3 A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y and 1 M Z at 25°C. If the

reduction potential of Z > Y > X, then

(A) Y will oxidise X and not Z (B) Y will oxidise Z and X

(C) Y will oxidise both X and Z (D) Y will reduce both X and Z. [JEE 1999]

Q.4 For the electrochemical cell, M | M+ || X – | X, E° (M+ /M) = 0.44 V and E° (X/X – ) = 0.33V. From this

data , one can deduce that

(A) M + X M+ + X – is the sp ontaneous reaction

(B) M+ + X – M + X is the sp ontaneous reaction

(C) Ecell= 0.77 V(D) E

cell= –0.77 V [JEE 2000]

Q.5 Copp er sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A

constant current of 2 mA was passed for 16 mintue. It was found that after electrolysis, the absorbance

(concentration) of the solution was reduced to 50% of its original value. Calculate the concentration of

copper sulphate in the solution to begin with. [JEE 2000]

Q.6 The following electrochemical cell has been set up

Pt(I)

| Fe3+, Fe2+(a =1) || Ce4+ , Ce3+ (a = 1) | Pt(II)

23 Fe / FeE = 0.77 V and 34 Ce / Ce

E = 1.61 V

If an ammetter is connected between the two platinum electrodes. predict the direction of flow of current.

Will the current increase or decrease with time? [JEE 2000]

Q.7 The reaction, [JEE 2001]

3ClO – (aq) 3

ClO (aq) + 2Cl – (aq)

is an example of

(A) Oxidation reaction

(B) Reduction reaction

(C) Disproportionation reaction

(D) Decomposition reaction

Q.8 The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is

(A) LiCl > NaCl > KCl [JEE 2001]

(B) KCl > NaCl > LiCl

(C) NaCl > KCl > LiCl

(D) LiCl > KCl > NaCl



Q.9 Saturated solution of KNO3

is used to make salt bridge because

(A) velocity of K+ is greater than that of 3

NO

(B) velocity of 3

NO is greater than that of K+

(C) velocities of both K+ and 3

NO are nearly the same

(D) KNO3

is highly soluble in water [JEE 2001]

Q.10 The standard potential of the following cell is 0.23 V at 15° C & 0.21 V at 35° C

Pt | H2(g) | HCl (aq) | AgCl(s) | Ag(s)

(i) Write the cell reaction.

(ii) Calculate H0 ,S0 for the cell reaction by assuming that these quantities remain unchanged in the range

15°C to 35°C.

(iii) Calculate the solubility of AgCl in water at 25°C. Given standard reduction potential of the

Ag+ /Ag couple is 0.80 V at 25°C. [JEE 2001]

Q.11 Standard electrode potential data are useful for understanding the suitablilty of an oxidant in a redox

titration. Some half cell reactions and their standard potentials are given below:

4

MnO (aq) + 8H+(aq) + 5e – Mn2+ (aq) + 4H2O (l); E° = 1.51 V

272

OCr (aq) + 14 H+ (aq) + 6e – 2Cr3+ (aq) +7H2O (l); E° = 1.38 V

Fe3+ (aq) + e – Fe2+ (aq); E° = 0.77 V

Cl2

(g) + 2e – 2Cl – (aq); E° = 1.40 V

Identify the only incorrect statement regarding quantitative estimation of aqueous Fe(NO3)2

(A)

4MnO can be used in aqueous HCl (B)

2

72OCr can be used in aqueous HCl

(C) 4

MnO can be used in aqueous H2SO

4(D) 2

72OCr can be used in aqueous H

2SO

4

[JEE 2002]

Q.12 In the electrolytic cell, flow of electrons is from:

(A) Cathode to anode in solution (B) Cathode to anode through external supp ly

(C) Cathode to anode through internal supp ly (D) Anode to cathode through internal supp ly.

[JEE 2003]

Q.13 Two students use same stock solution of ZnSO4

and a solution of CuSO4. The e.m.f of one cell is

0.03 V higher than the other. The conc. of CuSO4

in the cell with higher e.m.f value is 0.5 M. Find out

the conc. of CuSO4

in the other cell

06.0

F

RT303.2. [JEE 2003]

Q.14 Zn | Zn2+ (a = 0.1M) || Fe2+ (a = 0.01M)|Fe. The emf of the above cell is 0.2905 V. Equilibrium

constant for the cell reaction is

(A) 100.32/0.0591 (B) 100.32/0.0295

(C) 100.26/0.0295 (D) e0.32/0.295 [JEE 2004]



Q.15 Find the equilibrium constant at 298 K for the reaction,

Cu2+(aq) + In2+(aq) lCu+(aq) + In3+(aq)

Given that V15.0ECu|Cu2

, V42.0EIn|In3

, V40.0EIn|In2

[JEE 2004]

Q.16 The half cell reactions for rusting of iron are: [JEE 2005]

2H+ +21 O

2+ 2e – H

2O; E0 = + 1.23 V, Fe2+ + 2e – Fe; E0 = –0.44 VV

G0 (in kJ) for the reaction is:

(A) –76 (B) –322

(C) –122 (D) –176

Q.17

(a) Calculate0G f

of the following reaction

)aq(Ag + )aq(Cl AgCl(s)

Given :0G f

(AgCl) = –109 kJ/mole,0G f

(Cl – ) = –129 kJ/mole,0G f

(Ag+) = 77 kJ/mole

Represent the above reaction in form of a cell

Calculate E0 of the cell. Find log10

KSP

of AgCl

(b) 6.539 × 10 –2 g of metallic Zn (amu = 65.39) was added to 100 ml of saturated solution of AgCl.

Calculate log10

2

2

]Ag[

]Zn[

, given that

Ag+ + e – Ag E0 = 0.80 V ; Zn2+ + 2e – Zn E0 = –0.76V

Also find how many moles of Ag will be formed? [JEE 2005]

Question No. 18 to 20 (3 questions)

Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3is added to glucose with

NH4OH then gluconic acid is formed

Ag+ + e – Ag ; 0red

E = 0.8 V

C6H

12O

6+ H

2O C

6H

12O

7(Gluconic acid) + 2H+ + 2e – ; 0

redE = – 0.05 V

23

)NH(Ag + e – Ag(s) + 2NH3

;0E = – 0.337 V

[Use 2.303 ×

F

RT= 0.0592 and

RT

F= 38.92 at 298 K] [JEE 2006]

Q.18 2Ag+ + C6H

12O

6+ H

2O 2Ag(s) + C

6H

12O

7+ 2H+

Find ln K of this reaction

(A) 66.13 (B) 58.38 (C) 28.30 (D) 46.29

Q.19 When ammonia is added to the solution, p H is raised to 11. Which half-cell reaction is affected by pH

and by how much?

(A) Eoxd

will increase by a factor of 0.65 from 0oxd

E



(B) Eoxd

will decrease by a factor of 0.65 from 0oxd

E

(C) Ered

will increase by a factor of 0.65 from 0red

E

(D) Ered

will decrease by a factor of 0.65 from 0red

E

Q.20 Ammonia is always is added in this reaction. Which of the following must be incorrect?

(A) NH3 combines with Ag+

to form a complex.

(B) 23

)NH(Ag is a weaker oxidising reagent than Ag+.

(C) In absence of NH3

silver salt of gluconic acid is formed.

(D) NH3

has affected the standard reduction potential of glucose/gluconic acid electrode.

Q.21 We have taken a saturated solution of AgBr.Ksp

of AgBr is 12 × 10 –14. If 10 –7 mole of AgNO3

are

added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of 10 –7

S m –1 mol –1.

Given :0

)Ag( = 6 ×10 –3 S m2 mol –1 ;0

)Br( = 8 ×10 –3 S m2 mol –1 ;0

)NO( 3 = 7×10 –3 S m2 mol –1

[JEE 2006]

Paragraph for Question Nos. 22 to 24 (3 questions)

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules

(approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their

atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.

This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry

and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical

reaction, which requires a clear understanding of the mole concept.

A 4.0 molar aqueous solution of NaCl is p repared and 500 mL of this solution is electrolysed. This leads

to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23,

Hg = 200; 1 Faraday = 96500 coulombs) [JEE 2007]Q.22 The total number of moles of chlorine gas evolved is

(A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0

Q.23 If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is

(A) 200 (B) 225 (C) 400 (D) 446

Q.24 The total charge (coulombs) required for complete electrolysis is

(A) 24125 (B) 48250 (C) 96500 (D) 193000

Paragraph for Question Nos. 25 & 26 (2 questions)

Redox reactions p lay a p ivoted role in chemistry and biology. The values of standard redox potential(E°) of two half-cell reactions decide which way the reaction is expected to proceed. A simp le example

is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of

half-cell reactions (acidic medium) along with their E° (V with respect to normal hydrogen electrode)

values. Using this data obtain the correct exp lanations to Questions 14-16.

I2

+ 2e – 2I – E° = 0.54

Cl2

+ 2e – 2Cl – E° = 1.36

Mn3+ + e – Mn2+ E° = 1.50



Fe3+ + e – Fe2+ E° = 0.77

O2

+ 4H+ + 4e – 2H2O E° = 1.23 [JEE 2007]

Q.25 Among the following, identify the correct statement.

(A) Chloride ion is oxidised by O2

(B) Fe2+ is oxidised by iodine

(C) Iodine ion is oxidised by chlorine (D) Mn2+ is oxidised by chlorine

Q.26 While Fe

3+

is stable, Mn

3+

is not stable in acid solution because(A) O2

oxidises Mn2+ to Mn3+

(B) O2

oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+

(C) Fe3+ oxidises H2O to O

2

(D) Mn3+ oxidises H2O to O

2

Q.27 Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time

required to liberate 0.01 mol of H2

gas at the cathode is (1 Faraday = 96500 C mol –1)

(A) 9.65 × 104 sec (B) 19.3 × 104 sec

(C) 28.95 × 104 sec (D) 38.6 × 104 sec [JEE 2008]

Q.28 For the reaction of NO3¯ ion in an aqueous solution, E° is +0.96 V. Values of E° for some metal ions aregiven below

V2+ (aq) + 2e¯ V E° = – 1.19 V

Fe3+ (aq) + 3e¯ Fe E° = – 0.04 V

Au3+ (aq) + 3e¯ Au E° = + 1.40 V

Hg2+ (aq) + 2e¯ Hg E° = + 0.86 V

The pair(s) of metal that is(are) oxidised by NO3¯ in aqueous solution is(are) [JEE 2009]

(A) V and Hg (B) Hg and Fe

(C) Fe and Au (D) Fe and VQ.29.. AgNO

3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was

measured. The measured p lot of conductance() versus the volume of AgNO3

[JEE 2011]

(A) (P) (B) (Q) (C) (R) (D) (S)



ANSWER KEY

“Boost your basics”1.A 2.A 3.B 4.B 5.C 6.B 7.A 8.D 9.A 10.B 11.A 12.B 13.B

14B. 15C. 16.B 17.B 18.C 19.C 20.C 21.D 22.B 23.A 24.B 25.D 26.A

27D. 28A. 29 B 30 A 31 A 32 A 33 B

Multiple Choice Questions

42. A,B Q.43 C,D Q.44 B,C,D Q.45 A,B Q.46 B,C

FUNDAMENTALS

Q.1 (a) 2Ag + Cu2+ 2Ag+ + Cu, (b) 4

MnO + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

(c) 2Cl – + 2Ag+ 2Ag + Cl2, (d) H

2+ Cd2+ Cd + 2H+

Q.2 (a) Zn | Zn2+ | | Cd2+ | Cd, (b) Pt, H2

| H+ | | Ag+ | Ag ,

(c) Pt | Fe2+, Fe3+ | | 272

OCr , H+ , Cr3+ | Pt

Q.3 Anode Cathode

(a) Zn | Zn2+

H+

, H2 | Pt(b) Pt | Sn2+, Sn4+ Fe3+, Fe2+ | Pt

(c) Pt | Fe2+, Fe3+ 4

MnO , H+, Mn2+ | Pt

(d) Pb | Pb2+ Br2, Br – | Pt

(e) Cu | Cu2+ | | Cl – | Hg2Cl

2| Hg]

Q.4 1.61 V Q.5 1.35 V

Q.6 1.68 V Q.7 – 0.80 V, NO

Q.8 0.53 V, disproportionation Q.9 – 0.0367 V

Q.10 E = 1.159V Q.11 Kc= 1.864 × 10107, G0 = – 611.8 kJ

Q.12 n = 2 Q.13 E = 0.059

Q.14 E = 0.395 VQ.15 E0

cell= +0.01V, E

cell= 0.0785V, correct representation is Pb|Pb2+ (103M)||Sn2+(1M)|Sn

Q.16 [Cu2+] = 2.97 × 1012M for E = 0 Q.17 E = 0.81 V

Q.18 0.52 V, 0.61 V Q.19 0.0295 V

Q.20 0.0295 V Q.21 pH = 4

Q.22 (i) E = 0.286V; (ii) E = 0.0206V Q.23 a2

= 0.1006 M

Q.24 Kc

= 7.6 × 1012 Q.25 Kc

= 1.96 × 1026

Q.26 Kw

1014 Q.27 E0 = 0.7826 V

Q.28 –1.30 ×103 kJ mol –1

Q.29 (a) 6.02 × 1022 electrons lost, (b) 1.89 × 1022 electrons gained, (c) (b) 1.80 × 1023 electrons gained

Q.30 (a) 0.75 F, (b) 0.69 F, (c)1.1 F Q.31 (i) 54 gm, (ii) 16.35 gm

Q.32 1.023 × 105 sec Q.33 1.12 mol, 12.535 litreQ.34 0.112 litre Q.35 0.24 gms

Q.36 Rs. 0.75 x Q.37 (i) 2.1554 gm ; (ii) 1336. 15 sec

Q.38 115800C, 347.4 kJ Q.39 t = 193 sec

Q.40 n = 4 Q.41 Cu2+ = 0.08M, H+ = 0.04M, SO24

= 0.1M

Q.42 A = 114, Q = 5926.8C Q.43 Final weight = 9.6g, 0.01 Eq of acid

Q.44 t = 93.65 sec. Q.45 60 %

Q.46 (i) 482.5 sec (ii) 0.3175 gm (iii) 0.327 gm Q.47 1.825 g



Q.48 2M Q.49 419 S cm2 equivalent –1

Q.50 0.00046 S cm1 ; 2174 ohm cm Q.51 (i) 6.25 × 105 ohm, (ii) 1.6 × 106 amp

Q.52 1.549 cm –1 Q.53 101.8 ohm –1 cm2 / gm equivalent

Q.54 0.0141 mho g equiv –1 m2, 0.141 mho m –1

Q.55 (i) 232 Mho cm2 mol –1 , (ii) 116 Mho cm2 equivalent –1

Q.56 0.865 Q.57 1.33 ×10 –4 gm/litre

Q.58 8.74 × 10

–11

mole

2

/litre

2

Q.59 = 0.435 , k = 6.7×10

–4

Q.60 (i) 390.6 S cm2 mol –1 (ii) 12.32% Q.61 1.067 S m –1

Q.62 523.2 ×10 –4 mho cm2 mol –1 Q.63 (i) 6.98 (ii) 1.08 × 10 –14

APPLICATION

Q.1 [Cu2+] = 104M Q.2 pH = 6.61

Q.3 Ka

= 6.74 × 104 Q.4 1.39V

Q.5 – 0.46 V Q.6 (ii). 1.27 V, (iii) 245.1 kJ

Q.7 E0 = 0.22 V Q.8 h = 2.12 ×10 –2, Kh= 1.43 × 10 –5 M

Q.9 0.95 V Q.10 Ksp

= 1.1 × 1016

Q.11 [Br ] : [Cl] = 1 : 200 Q.12 E0 = 0.1511V

Q.13 – 0.037 V Q.14 1.536 ×10 –5 M3

Q.15 1.66V Q.16 –1.188V

Q.17 10 –2 Q.18 5.24 × 1016

Q.19 E° = 1.59V, non-spontaneous Q.20 E0 = 0.71V

Q.21 [Zn2+]/[Cu2+] = 1.941 × 1037 Q.22 0.373V

Q.23 Kf

= 8.227 × 1063 Q.24 0.372 V , – 0.307 V

Q.25 K = 10268 Q.26 Tl+ = 10 –8; Co3+ = 2 × 10 –8

Q.27 Eq. wt. = 107.3 Q.28 643.33amp ,190.5g

Q.29 43.456g Q.30 265 Amp. hr.

Q.31 42.2 %

Q.32 V(O2) = 99.68 mL, V(H2) = 58.46 mL, Total vol. = 158.1 mLQ.33 57.5894 gm Q.34 1.9 million year

Q.35 1.21 × 105 coulomb Q.36 1250 s, 0.064 M

Q.37 1.143V Q.38 (a) –0.124 V, (b) 7.1, (c) calomel electrode

Q.39 0.4414 V Q.40 – 0.167 V

Q.41 0.1456 amp ere Q.42 3.29 cm

Q.43 4.25×10 –2 metre Q.44 0.1934 gm/litre

Q.45 KSP

= 7.682 × 10 –17 Q.46 2 × 105 dm3

FEEL THE HEAT

Q.1 KC = 6.26 × 107

Q.2 Ksp = 2.287 × 1012

Q.3 A Q.4 B

Q.5 7.95 × 10 –5M Q.6 decrease with time

Q.7 C Q.8 B

Q.9 C Q.10 H0 = – 49987 Jmol –1 , S0 = – 96.5 J mol –1 K –1 , s = 1.47 ×10 –5 M

Q.11 A Q.12 C Q.13 0.05 M Q.14 B Q.15 KC

= 1010 Q.16 B

Q.17 (a) E0 = 0.59 V, log10

KSP

= –10 (b) 52.8, 10 –6 moles

Q.18 A Q.19 A Q.20 D Q.21 55 S m –1 Q.22 B Q.23 D

Q.24 D Q.25 C Q.26 D Q.27 B Q.28 A,B,D Q.29. D



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Electro Chemistry