11 05 13 Chemistry Electrochemistry Assignment 2

Sri Chaitanya Narayana IIT Academy Chemistry_Electrochemistry

Sri Chaitanya Narayana IIT AcademySri Sarvani Educational Society

COMMON CENTRAL OFFICE_MADHAPUR_HYDCHEMISTRY : ELECTROCHEMISTRY

Assignment-2

1)Match the following :-

Column-I Column-II(A) For a galvanic cell, the Ecell =

0(p) 19300 coulombs

(B) The charge carried by

electrons

(q) The cell reactions are at equilibrium

(C) During discharge of lead storage battery

(r) 2F

(D) (s) 2 moles of H2SO4 are consumed

(t) Cell reactions are reversed2)

The chlorate ion can disproportionate in basic solution according to the reaction

. The equilibrium concentration of the ions resulting from a solution initially at 0.1 M in chlorate ion is then a is _______

( and ).PARAGRAPH FOR QUESTION NO 28 - 30

. Following technique is adopted to check suspected drunk drivers. Read the passage and answer the questions at the end of it.When suspected drunk drivers are tested with a Breathalyzer, the alcohol (ethanol) in the exhaled breath is oxidized to ethanoic acid with an acidic solution of potassium dichromate:

The Breathalyzer measures the colour change and produces a meter reading calibrated in terms of blood alcohol content.3) Colour of the testing solution changes from:4) If the standard half-cell potential for reduction of ethanoic acid to ethanol is 0.06 V and that of reduction

of to is 1.33 V, then of the reaction taking place in alcohol meter is:1) 1.39 V 2) 1.27 V 3) -1.39 V 4) -1.51 V

5)

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What is the value of E for the reaction when the concentrations of ethanol, ethanoic acid, ,and are 1.0 M and the pH is 4.00 ?

1) 0.31 V 2) -1.01 V 3) 1.64 V 4) 0.95 VPARAGRAPH FOR QUESTION NO 6 – 8

One of the nicest application of reduction potential is the prediction of the products of reaction involving elements having several oxidation states. Consider the reaction of iodine with permanganate in acid solution. The pertinent diagram is shown below

6) What are species which are unstable towards disproportion7)

When the reaction between and carried out with iodide in excess, then the products of the reaction are8) To the acidic permanganate solution, iodide solution is added drop wise the products of reaction are1) 2) 3) 4)

PARAGRAPH FOR QUESTION NO 9 – 1185.5ml of 0.672 Normal solution of Sodium bromate was prepared. Half cell reaction isThe driving force diminishes to zero on the way to equilibrium, just as in any other spontaneous

process. Both and the corresponding cell potential are zero when the redox reaction comes to equilibrium.The Nernst equation for the redox process of the cell may be given as

The key to the relationship is the standard cell potential E0, derived from the

standard free energy change as :

9) The equilibrium constant Kc for the reaction :

Will be :10)

The value of the reaction quotient Q, for the cell is

11)

The standard emf of the cell, in which the cell reaction is 0.6195 V

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Sri Chaitanya Narayana IIT Academy Chemistry_Electrochemistryat 00C and 0.6753 V at 250C. The value of of the reaction at 250C is.

12. A hydrogen electrode placed in a buffer solution of NaCN and HCN in the mole ratio of X : Y and Y : X has an oxidation electrode potential values E1 and E2 volts respectively at 25°C. The pKa value of HCN is

(A) (B)

(C) (D)

13. The e.m.f. of the cell Cd (s) + Hg2+ (aq) ® Cd+2 (aq) + Hg is given by E = 0.6708 – 1.02 × 10–4 (T – 25 V) where T is the temperature in °C and E in volts. The entropy change for the reaction is (A) – 19.69 J deg–1 (B) – 129.3 kJ

(C) 19.69 kJ deg–1 (D) – 9.85 J deg–1 14. If and , then is

(A) +0.28 V (B) – 0.28 V (C) + 0.74 V (D) + 1.69 V

15. A graph is plotted between Ecell of quinhydrone half cell and pH of solution. Intercept is 0.699 V. At what pH, Ecell will become 0.492 V? (A) 3.5 (B) 4.2 (C) 7 (D) 10

16. Calculate the cell EMF in mV for at 298 K

Given : (AgCl) = – 109.56 kJmol–1 and (H+ + Cl–)(aq) = kJ mol–1

(A) 456 mV (B) 654 mV (C) 546 mV (D) none of these

17. At what does the following cell have its reaction at equilibrium?

Given : Ksp = 8 × 10–12 for Ag2CO3 and Ksp = 4 × 10–13 for AgBr (A) (B)(C) (D)

18. Calculate the EMF of the cell at 298 K Pt | H2 (1 atm) | NaOH (xM), NaCl (xM) | AgCl (s) | Ag (Given )(A) 1.048 V (B) – 0.04 V (C) – 0.604 V (D) emf depends on x and cannot be determined unless value of x is given

19. Consider the cell Ag Ag. Given that C2 = 10C1 and that x and y are degrees

of dissociation in AgNO3 solution of concentrations C1 and C2 in the two half cells, then log in

terms of the emf E of the cell at 298 Kelvin will be

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(A) (B)

(C) (D) .

20. The standard reduction potentials of some half cell reactions are given below :

E° = 1.51 V

E° = 1.61V

E° = 1.71 V Pick out the correct statement : (A) Ce+4

will oxidize Mn+2 to (B) will oxidize Pb to PbO2

(C) H2O2 will oxidize Mn+2 to (D) PbO2 will oxidize Mn+2 to

21. Identify the incorrect statement (s)(A) In the spontaneous chemical reaction of a galvanic cell, electrons flow from the cathode to the

anode. (B) In the equation the and are the standard reduction potentials. (C) Doubling the coefficients in a chemical reaction will square the value of Keq will double DG° and

will double .(D)Salt Bridge is essential for all types of cells.

22. Consider the cell Ag | AgBr | KBr (0.01 M) | | KCl (1M) | AgCl | Ag with EMF 0.059 V. Conclusion inferred may be (A) The ratio of the simultaneous solubilities of AgCl and AgBr in pure water is 1000(B) It is concentration cell (C) Change in concentration of KCl will not affect EMF. (D) KSP (AgCl) > KSP (AgBr)

PARAGRAPH FOR QUESTION NO 23 - 25Comprehension:The potential of an electrode when each species involved in it exists in the standard state is called its standard potential. The standard reduction potential of a couple is the measure of its tendency to get reduced. A series obtained by arranging the various couples in order of their decreasing standard potential is called electrochemical series. Any of the two couples of this series joined together gives an electrochemical cell in which reduction occurs at the electrode which occupies the higher position. The standard potential of hydrogen electrode is taken to be zero by convention.Given below are the sequence of half–reactions (acidic media) with relevant E0 values in volt at 298K.

Also given some more data are:

23. The E0 for is equal to(A) 1.054V (B) 1.506V(C) 5.27V (D) 7.53V

24. The value of E0 of reaction:

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(A) 0.734V(B) 0.282V(C) 4.498V(D) Can’t be calculated as concentrations of species are not given.

25. For the allFe(s) | Fe2+ (0.1M) || 100 mL 0.3M HA (pKa =5.2) mixed with 50 mL 0.4M NaOH | H2 (1 atm), Pt

the Ecell at 298 ( and log1020 = 1.3) is equal to

(A) –0.15V (B) +0.185V(C) –0.44V (D) +0.145V

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ELECTROCHEMISTRY – 2 KEY & HINTS

1. A- Q ; B-P,R; C- P,R,S ; D- T 2.2 3.3 4.2 5.4 6.3

7.2 8.1 9.1 10.4 11.1 12.A 13.A 14.D 15.A 16.A 17.B

18.A 19.C 20.A,B,C 21.A,B,C,D 22.A,B,D 23.B 24.A 25.D

1)Hint) Conceptual.2)Hint) 23) Hint)

is orange changing to 4) Hint)

5) Hint)

6) Hint)For disproportion of a species its successive reduction potential should be greater than

previous education potential.7) Hint)

As, oxidizes to , it will be absorbed by present in the mixture forms 8) Hint)

The products of the reaction must be stable in presence at is not formed as

oxidize it to and is also not formed only is formed.

9)

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Sri Chaitanya Narayana IIT Academy Chemistry_ElectrochemistryHint)

10) Hint)

11) Hint)

12. E1 =

= 0.059

E2 = + 0.059 pHII = 0.059

pKa = .

13. here V degree–1

= 2 × 96500 (– 1.02 × 10–4) = – 19.69 Jdeg–1.

16. Cell Reaction : H2 + 2AgCl ¾® 2Ag + 2H+ + kJ mol–1

Ecell = +0.220 + = 456 mV.

17. Ag+ (cathod)

Ecell = Ecell + .

18.

Ecell = 0.22 + .

19. Cell reaction : Ag+ (y, C2) ¾® Ag+ (x, C1) COIPL_Chemistry 7


E = .

.

20. For 2Mn2+ + 5PbO2 + 4H+ ¾® 5Pb + + 2H2O E° = 1.455 – 1.51 = negative. Reaction is not fissible.

22. The cell can be represented as

Ag | Ag+

Net reaction :

Ecell = 0.059 log

Let the simultaneous solubilities of AgCl and AgBr and x¢ and y¢ moles / litre respectively.

AgCl (s)

AgBr (s)

Ksp (AgCl) =

Ksp (AgBr) =

Thus, the simultaneous solution of AgCl is 103 times the simultaneous solution of AgBr. 24.

–––––––––––––––––––––––––––––––––––––––––––––––––––

= 1.506V

25.

–––––––––––––––––––––––––––––––––––

= –0.772V

=

= 1.506 – (+0.772)= +0.734V

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11 05 13 Chemistry Electrochemistry Assignment 2