Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B ... · = antilog 37.28 = 1.94 × 10 37. Dr. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important

$Page 1: Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B ... · = antilog 37.28 = 1.94 × 10 37. Dr. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important$
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+2\Physical\Grand Test\+2 Grand Test - 3 11.6.2017\+2 Grand Test-3-Redox & Electrochemistry.doc

Test Venue:

Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Dr. Sangeeta Khanna Ph.D


READ THE INSTRUCTIONS CAREFULLY

1. The test duration is of 2 hour.

2. The maximum marks are 221. 3. There are total 51 questions.

4. Fill your particulars correctly and completely on the OMR ANSWERSHEET.

SECTION – A (Single Correct Choice Type) Negative Marking [-1] This Section contains 30 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 30 × 4 = 120 Marks

1. The cell reaction for the given cell is spontaneous if

Pt, C2 | C– (1 M) || C– (1 M) | Pt, Cl2

(P1) (P2) a. P1 > P2 b. P1 < P2 c. P1 = P2

d. P2 = 1 atm. B

Sol. 1

20red

0oxicell

P

P log

2

059.0EEE Pt ,Cl|Cl||Cl|Cl ,Pt

21 P

2

P

2

= 1

20cell P

Plog

2

059.0E

Now Ecell will be + ve if P2 > P1 2. The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The potential of the cell would be

(the value of 2.303 RT/F is 0.059 V)

a. 0.177 V b. 0.087 V c. -0.177 V d. 0.059 V C

Sol. H+ + e– 2H2

1

E = E° - ]H[

1log

n

059.0

= 0 - pH1

059.0 = -0.059 × 3 = -0.177 V

3. If the pressure of hydrogen gas is increased from 1 atm. to 100 atm., keeping the hydrogen ion concentration constant at 1 M, the voltage of the hydrogen half cell at 25°C will be

a. 0.059 V b. -0.059 V c. 0.295 V d. 0.118 V B

Sol. H+ + e–

2

1H2,

E = E° - ]H[

plog

n

059.02/1

1H

= 0 -

1

)100(log

1

06.0 2/1 = -0.059 V

4. E° for the cell Zn | Zn2+(aq) || Cu2+ (aq) | Cu is 1.10 V at 25°C. The equilibrium constant for the cell reaction Zn + Cu2+(aq) Cu + Zn2+ (aq) is of the order of a. 10–37 b. 1037 c. 10–17 d. 1017 B

Sol. eq0cell

K logn

059.0E or log Keq =

059.0

210.1 = 37.28

Keq = antilog 37.28 = 1.94 × 1037



5. An aqueous solution of conc. H2SO4 is electrolysed using platinum electrodes. The products at the anode and cathode are

a. O2, H2 b. S2O 28

, Na c. O2, Na d. S2O 28

, H2

D

6. Choose the incorrect statement:

a. A reaction is spontaneous from left to right if Keq> q in which case Ecell >0

b. A reaction occurs from right to left if Keq < q, in which case Ecell < 0 c. If the system is at equilibrium no net reaction occurs

d. Ecell is temperature independent D 7. Given E° values for some reactions as

(i) I2 + 2e– 2I– ; E° = 0.54 V

(ii) MnO 4

+ 8H+ + 5e– Mn2+ + 4H2O; E° = 1.52 V

(iii) Fe3+ + e– Fe2+ ; E° = 0.77 V

(iv) Sn4+ + 2e– Sn2+ ; E° = 0.1 V The strongest reductant and oxidant respectively are

a. Sn2+, MnO 4

b. MnO 4

, Sn4+ c. I2, Fe3+ d. I2, Sn2+

A

Sol. Larger the value of E 0red

larger is tendency for reduction and consequently stronger will be the oxidant.

Similarly, smaller the value of E 0red

larger the tendency for oxidation and consequently stronger will be

the reductant. 8. In the electrolysis of copper chloride solution using copper electrodes, the mass of cathode increases

by 3.25 g. In the anode,

a. 0.05 mol of Cu will go into solution as Cu2+ ions b. 560 mL of O2 at S.T.P. will be liberated c. 112 mL of Cl2 will liberate d. 3.2 mol of copper metal will dissolve A

Sol. CuCl2 Cu+2 + 2Cl–

Copper deposited = 5.63

25.3 = 0.05 mol

The Cu anode will give same amt. of Cu+2 ions in solution. 9. Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of

water. The total volume of the two gases (dry and at STP) produced will be approximately (in litres)

a. 22.4 b. 44.8 c. 67.2 d. 89.4 C

Sol. H2O H2 + 2O2

1

or 2H+ + 2e– H2

i.e., 2 moles of e– produce 1 mole of H2 i.e., 22.4 L. Hence 4 moles of e– will give H2 = 44.8 L.

O2 produced = 2

1of H2 = 22.4 L

Total volume = 44.8 + 22.4 = 67.2 L 10. Red hot carbon will remove oxygen from the oxide AO and BO but not from MO, while B will remove

oxygen from AO. The activity of metals A, B and M in decreasing order is:

a. A > B > M b. B > A > M c. M > B > A d. M > A > B C



Sol. C removes oxygen from AO and BO, C is better reducing agent than A and B; B remove oxygen from

AO, B is better reducing agent than A. Better reducing means more active. Hence the correct order is

M > C > B > A.

11. Point out the correct statement:

1. Electrolysis of aqueous solution of LiCl shows pH > 7

2. Oxidation number of C in carbon suboxide is +2/3

3. Oxidation number of P in sodium hypophosphite is +1

4. The oxidation number and covalency of oxygen in O2 molecule are 0 and 2

a. 1, 3 b. 1, 2, 4 c. 1, 2, 3, 4 d. 1, 3, 4

D

Sol. Carbon suboxide is 2

2

3

x

OC

3x – 4 = 0

3x = +4 or x = +4/3

12. Conductometric titration curve of a equimolar mixture of a HCl and HCN with NaOH (aq) is:

a. b.

c. d.

D

Sol. Molar conductivity of H+ and OH– are very high as compare to other ions. Initially conductance of solution sharply decreases due to consumption of free H+ then increases due to formation of salt

(NaCN) and After complete neutralization further sharply increases due to presence of OH– . 13. Consider the cell given below:

Cu/Cu+2 || Ag+/Ag

Ag+ + e Ag; E0 = x

Cu2+ + 2e Cu; E0 = y E0 of the cell is: a. x – y b. 2x – y c. y – x d. y – 2x A 14. For a cell reaction involving two-electron change, the standard emf of the cell is found to be 0.295 V at

250C. The equilibrium constant of the reaction at 250C will be: a. 1.0 x 10–10 b. 29.5 x 10–2 c. 10.0 d. 1.0 x 1010 D

Sol. K log 2

059.0E

10059.0

20.295K log

Vol of NaOH

Cond

ucta

nce

C

ond

ucta

nce

Vol of NaOH

Cond

ucta

nce

Vol of NaOH

Vol of NaOH

Cond

ucta

nce



15. Equal quantities of electricity are passed through three voltameters containing FeSO4, Fe2(SO4)3 and Fe(NO3)3. Consider the following statements in this regard:

1. The amount of iron deposited in FeSO4 and Fe2(SO4)3 is equal 2. The amount of iron deposited in Fe(NO3)3 is two-third of the amount of iron deposited in FeSO4 3. The amount of iron deposited in Fe2(SO4)3 and Fe(NO3)3 is equal a. 1 alone is correct b. 1 and 2 are correct c. 2 and 3 are correct d. 3 alone is correct C 16. Consider following electrolytes: 1. CuSO4 2. Fe2(SO4)3 3. AlCl3 4. AgNO3

The quantity of electricity needed to electrolyse completely 1 M, 1 lit. solution of these electrolytes will be: a. 2F, 6F, 3F and 1 F respectively b. 6F, 2F, 3F and 1F respectively c. 2F, 6F, 1F and 3F respectively d. 6F, 2F, 1F and 3F respectively A 17. Zn| Zn2+ (a = 0.1 M)II Fe2+ (a= 0.01 M) | Fe The emf of the above cell is 0.2905 V. The equilibrium constant for the cell reaction is: a. 1. 0 x 100.32/0.0591 b. 1.0 x 100.32/0.0295 c. 1.0 x 100.26/0.0295 d. e0.32/0.295 B

Sol. 0.2905 = E° - 01.0

1.0log

2

059.0

0.2905 + 0.0295 = E° 0.32 = E°

E° = K log2

059.0

0295.0

32.0

059.0

2EK log

18. For which of these electrodes will the reduction potential vary with pH

,Am/ AmO(II) , AmO/AmO )I( 4222

22

24 Am/Am )III(

a. (I) only b. (II) only c. (I) and (II) only d. (I), (II) and (III) B

Sol.

52

6

22

AmO AmOe

[No H+ is used]

OH2AmOAmOe2H4 24

4

62

[E° will depend on H+]

2e– + Am+4 Am+2

19. In the diagram given below the value of X is:

a. 0.325 V b. 0.65 V c. -0.35 V d. -0.65 V A

Sol. 02

01

03

GGG

= -0.15 F – 0.5 F = -0.65 F

03

03

nFEG

-0.65 F = 03

FE2

03

E = 0.325 V

Cu2+

Cu+ Cu

+ 0.15 V + 0.50 V

E0 = X volt



20. The ionization constant of a weak electrolyte is 25 x 10–6 while the equivalent conductance of its 0.01 M solution is 19.6 s cm2 eq-1. The equivalent conductance of the electrolyte at infinite dilution (in s cm2 eq–1) will be:

a. 250 b. 196 c. 392 d. 384 C

Sol. K = 25 x 10–6 ; eq = 19.6 S cm2 eq–1 ; = C2

05.0C

K

6.1905.0C

392

21. The ionic conductance of Al3+ and SO42– ions at infinite dilution are x and y ohm-1 cm2 mol-1

respectively. If Kohlrausch’s Law is valid, the molar conductance of aluminium sulphate at infinite dilution will be:

a. 3x + 2y b. 3y + 2x c. 2x + 2y d. 3x + 3y B

Sol. y

24

x

3342 SO3Al2)SO(Al

x2y3m

22. Electrolyte KCl KNO3 HCl NaOAc NaCl (S cm2 mol–1 ) 149.9 145.0 426.2 91.0 126.5

Calculate HOAc using appropriate molar conductance of the electrolytes listed above at infinite

dilution in H2O at 250C. a. 517.2 b. 552.7 c. 390.7 d. 217.5 C

Sol. NaClHClAcONaAcOH

= 91.0 + 426.2 – 126.5 = 390.7 Scm2 mol – 1 23. The molar conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 M acetic acid solution is 5.2

mho cm2 mol– 1.[H+] in solution will be

a. 0.00133 M b. 0.133 M c. 0133 M d. 1.33 M A

Sol. The degree of dissociation ‘’ is given by

v

0133.076.390

2.5 or 1.33%

CH3COOH CH3COO– + H+ Before dissociation 1 0 0

After dissociation (1 - )

[H+] = C = 0.1 x 0.0133 = 0.00133 M 24. A 0.05 N solution of a salt occupying a volume between two platinum electrodes separated by a

distance of 1.72 cm and having an area of 4.5 cm2 has a resistance of 250 ohm. Calculate the equivalent conductance of the solution.

a. 30.56 ohm-1 cm2 eq-1 b. 305.6 ohm-1 cm2 eq-1

c. 3056 ohm-1 cm2 eq-1 d. 3.056 ohm-1 cm2 eq-1

A



Sol. Specific conductance = conductance × cell constant

K = C × A

= AR

1

= 5.4

72.1

250

1 = 1.5288 × 10-3 ohm-1 cm-1

e = K × N

1000

= 1.5288 × 10-3 × 05.0

1000

= 30.56 ohm-1 cm2 eq-1 25. The normal oxidation potential of zinc referred to the standard hydrogen electrode is 0.76 volt and that

of copper is –0.34 volt at 250C. When excess of zinc is added to a solution of copper sulphate, the zinc displaces copper till equilibrium is reached. What is the ratio of concentration of Zn2+ to Cu2+ ions at equilibrium?

a. 1.679 × 1036 : 2 b. 1.679 × 1037 : 1 c. 1.679 × 10 : 1 d. 1679 × 1037 : 1 B

Sol. Zn + CuSO4 Cu + ZnSO4

Zn + Cu2+ Cu + Zn2+

Ecell = E°cell - ]Cu[

]Zn[log

2

0591.02

2

At equilibrium Ecell = 0

E°cell = ]Cu[

]Zn[log

2

0591.02

2

or log 0591.0

E2

]Cu[

]Zn[0cell

2

2

[ 0cell

E = 0.76 + 0.34 = 1.10 volt]

log2

2

Cu

Zn

=

0591.0

10.12 = 37.225

]Cu[

]Zn[2

2

= 1.679 × 1037 : 1

26. NaOH is added in R.H.S. of the given cell to consume all the H+ present in R.H.S. of the cell of emf + 0.701 volt at 250C before its use. What is the emf of the cell after addition of NaOH.

Zn | Zn2+

(0.1 M) || HCl (1 litre) | Pt(H2g) V 760.0E0

2Zn/Zn

1 atm a. 0.3759 volt b. 3759 volt c. 3.759 volt d. 37.59 volt

A

Sol. Zn + 2H+ Zn2+ + H2

Applying Nernst eq. Ecell = E° - 2

2

]H[

]Zn[log

2

0591.0

After addition of NaOH, the solution becomes neutral i – e, the concentration of H+ ions in cathodic solution becomes 10-7, Applying again Nernst equation

Ecell = E 0cell - 2

2

]H[

]Zn[log

2

0591.0

= 0.760 - 27 )10(

1.0log

2

0591.0

= 0.3759 volt

27. Consider the following reaction: 5H2O2 + xClO2 +2OH– xCl– + yO2 + 6H2O a. x= 5, y = 2 b. x= 2, y = 5 c. x= 4, y = 10 d. x= 5, y = 5 B



28. Balance the following redox reaction by using following correct coefficients:

C12H22O11 + HNO3 H2C2O4 + H2O+NO2

a. 1 : 18 : 6 : 18 : 19 b. 1 : 36 : 6 : 36 : 23 c. 1 : 36 : 6 : 18 : 36 d. 1 : 36 : 6 : 23 : 36 D 29. The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3 and H4P2O6 is

a. H3PO4 > H4P2O6 > H3PO3 > H3PO2 b. H3PO3 > H3PO2 > H3PO4 > H4P2O6 c. H3PO2 > H3PO3 > H4P2O6 > H3PO4 d. H3PO4 > H3PO2 > H3PO3 > H4P2O6 A

Sol.

30. For the following cell:

Zn(s) | ZnSO4 (aq.) || CuSO4 (aq.) | Cu(s)

when the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for G (in J mol-1) is [F is Faraday constant, R is gas constant, T is temperature, E°(cell) = 1.1V] a. 2.303 RT + 1.1F b. 2.303 RT – 2.2F c. 1.1 F d. – 2.2 F B

Sol. G = G° + 2.303 RT log Q

G = -nFE° + 2.303RT log Q Given: E° = 1.1 V and n = 2

G = (-2 × 1.1 × F) + 2.303RT log ]Cu[

]Zn[2

2

G = -2.2 F + 2.303 RT

H3PO4 HO

P

OH

OH

O

; oxidation state of P = +5

H2P2O6 HO

P

OH

OH

O


P

OH

O

H3PO4 HO

P

H

OH

O


H3PO2 HO

P

H

O

; oxidation state of P = +1 H



SECTION – B (Assertion and Reason) Negative Marking [-1]

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)

out of which ONLY ONE is correct. 5 × 4 = 20 Marks

(a) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the

assertion. (b) If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the

assertion. (c) If assertion is CORRECT, but reason is INCORRECT. (d) If both assertion & reason are INCORRECT.

1. Assertion:- At the end of electrolysis using platinum electrodes, an aqueous solution of copper

sulphate turns colourless. Reason: Copper in copper sulphate is converted to copper hydroxide during the electrolysis. a. (a) b. (b) c. (c) (d). d C 2. Assertion: The conductivity of an electrolytic solution decreases with dilution whereas equivalent

conductance increases with dilution. Reason: Conductivity is for 1 cc of the solution whereas equivalent conductance is for any volume of

the solution containing one gram equivalent of the electrolyte. a. (a) b. (b) c. (c) (d). d B 3. Assertion: In electrolysis, the quantity of electricity needed for depositing 1 mole of silver is different

from that required for 1 mole of copper. Reason: The molecular weights of silver and copper are different. a. (a) b. (b) c. (c) (d). d B 4. Assertion: When aqueous solution of NaNO3 is electolysed, sodium is liberated at cathode: (using Hg

cathode) Reason: Na+ is neutralized at cathode. a. (a) b. (b) c. (c) (d). d B 5. Assertion: Electrolysis of an aqueous solution of a bromide or iodide gives Br2 or I2 at the anode

whereas that of NaF gives O2 and not F2. Reason: F2 is highly reactive whereas Br2 and I2 are less reactive gases. a. (a) b. (b) c. (c) (d). d B



SECTION – C (Statement Based) Negative Marking [-1]

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 5 × 4 = 20 Marks

In each of the following questions (Q.1 – Q.5), three statements I, II, III are given. Mark

(a) if all the statements are correct (b) if II and III are correct (c) if I and III are correct (d) if only II is correct 1. I: Conductance of electrolytic solution increases with temperature.

II: Resistivity is reciprocal of molar conductivity of electrolyte. III: Cell constant has a unit cm–1. a. (a) b. (b) c. (c) d. (d) C

Sol. Conductance increases with temp. due to increase in mobility of ions

Cell constant = a

= cm–1

2. I: Oxidation number is same as the formal charge. II: The numerical value of O.N. and valency may differ. III: Absolute value of electrode potential cannot be determined a. (a) b. (b) c. (c) d. (d) B

Sol. 0.N is residual charge and is different for same element in different compounds. The valency of an element remains the same in all the compounds

3. In CrO5, I: Oxidation number of Cr is +6 II: Four O atoms of molecule are involved in peroxide linkage III: Only one O atom has O.N. = -2 a. (a) b. (b) c. (c) d. (d) A

Sol. The structure of CrO5 is

4. I: The conductivity of molten NaCl is due to movement of Na+ and Cl– ions II: Solid NaCl is also good conductor of electricity III: Sodium is a good conductor because of mobile electrons a. (a) b. (b) c. (c) d. (d) C

Sol. Molten NaCl and molten Na are good conductors. 5. I: Cathode is –ve terminal both electrochemical and electrolytic cell.

II: Reduction occurs at cathode both in galvanic as well as electrolytic cell III: Chemical change in electrolytic cell is non-spontaneous. a. (a) b. (b) c. (c) d. (d) B

Cr

O

O

O O

O

–1 –1

–1 –1

–2



SECTION – D (Multiple Correct Choice Type) No Negative Marking

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 5 × 5 = 25 Marks

1. Which of the following facts is true? a. If E0 (Mn+/M) is negative, H+ will be reduced to H2 by the metal M b. If E0 (Mn+/M) is positive, Mn+ will be reduced to M by H2 c. In a cell Mn+/M assembly is attached to hydrogen – half cell. To produce spontaneous cell reaction,

metal M will act as negative electrode, if the potential Mn+/M is negative. It will serve as positive electrode, If Mn+/M has a positive potential

d. Compounds of active metals (Zn, Na, Mg) are reducible by H2 whereas those of noble metals (Cu, Ag, Au) are not reducible

A, B, C,

2. For a strong electrolyte, increased slowly with dilution and can be represented by the equation:

= 0 – AC1/2 Which electrolyte (s) have the same value of the constant A? a. NaCl b. CaCl2 c. ZnCl2 d. AlCl3 B,C Sol. Both are 2 – 1 type electrolyte 3. Using the standard potential values given below, decide which of the statements I, II, III, IV are correct.

Choose the right answer from (a) , (b), (c) and (d) Fe2+ + 2e– = Fe, E0 = -0.44 V Cu2+ + 2e– = Cu, E0 = + 0.34 V Ag+ + e– = Ag, E0 = + 0.80 V I. Copper can displace iron from FeSO4 solution II. Iron can displace copper from CuSO4 solution III. Cu can displace Ag from AgNO3 solution IV. Iron can displace silver from AgNO3 solution a. I and II b. II and III c. II and IV d. I and IV B, C

4. Given V,37.2E V,80.0E 0

/Mg2Mg

0

Ag/Ag

V79.0E V,34.0E 0

/Hg2Hg

0

Cu/2Cu

. Which of the following statement is/are correct?

a. AgNO3 can be stored in copper vessel b. Cu(NO3)2 can be stored in silver vessel c. CuCl2 can be stored in silver vessel d. HgCl2 can be stored in copper vessel B, C 5. The values of E0 of some of the reactions are given below:

I2 +2e–2I– ; E0 = +0.54 volt

Cl2 + 2e– 2Cl– ; E0 = +1.36 volt

Fe3+ + e– Fe2+; E0 = +0.76 volt

Ce4+ + e–Ce3+; E0 = +1.60 volt

Sn4+ + 2e–Sn2+; E0 = +0.15 volt

Which of the following statement is correct? a. Fe+3 will oxidise Ce+3 b. Fe+2 will be oxidised by Ce+4 c. I2 does not displace Cl2 from KCl d. FeCl3 can oxidise SnCl2 B, C, D



SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 Marks Each

1. Match the Column – I with Column – II. [Single Match]

Column – I Column – II

(A) No. of Faradays of electricity produced by combustion of 1 mol benzene (p) 10

(B) No. of electrons gained by 2 mol of KMnO4 in acidic medium (q) 20

(C) Eq. Wt. of NH4NO3 in given reaction

OHONNOHN 2234

(r) 30

(D) No. of electron exchanged in the given reaction:

3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O

(s) 5

Sol. A R; B P; C Q; D S

A. C6H6 + 2

15O2 6CO2 + 3H2O x = 30

B. 2KMnO4 + 16H+ + 10e– 2Mn+2 + 8H2O + 2K+ x = 10

C. eq. NH 4

= 5.44

18 ; eq. mass of NO

3 = 5.154

62

eq. mass NH4NO3 = 4.5 + 15.5 = 20

D. 2e– + Cl2 2Cl–) × 5

Cl2 2ClO3– + 10e–

6Cl2 10Cl– + 2ClO3–

3Cl2 5Cl– + ClO3– (5e–)

2. The standard reduction potential data at 25°C is given below. E°(Fe3+, Fe2+) = +0.77 V; E°(Fe2+, Fe) = -0.44 V E°(Cu2+, Cu) = +0.34 V; E°(Cu+, Cu) = +0.52 V

E°[O2(g) + 4H+ + 4e– 2H2O] = +1.23 V; E°[O2(g) + 2H2O + 4e– 4OH–] = +0.40 V E° (Cr3+, Cr) = -0.74 V; E° (Cr2+, Cr) = - 0.91 V Match E° of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists. (Single Match Only )

Column – I Column – II

(A) E° (Fe3+, Fe) (p) - 0.36 V

(B) E° (4H2O 4H+ + 4OH–) (q) -0.4 V

(C) E° (Cu2+ + Cu 2 Cu+) (r) -0.04 V

(D) E° (Cr3+, Cr2+) (s) -0.83 V

Sol. A r; B s; C p; D q

(A) Fe+3 + 3e– Fe ...(1) (B) 4H2O 4H+ + 4OH– ...(1)

Fe+3 + e– Fe+2 ...(2) O2 + 4H+ + 4e– 2H2O ...(2)

Fe+2 + 2e– Fe ...(3) O2 + 2H2O + 4e– 4OH– ...(3)

G1 = G2 + G3 G1 = G3 - G2

3FE 1 = 32 2FE FE 1 231 EEE

3

E2EE 32

1

231 EEE

= 0.4 – 1.23 = -0.83



(C) Cu+2 + Cu 2Cu+ ...(1) (D) Cr+3 + e– Cr+2

Cu+2 + 2e– Cu ...(2) Cr+3 + 3e– Cr

(Cu+ + e– Cu) ...(3) Cr+2 + 2e– Cr

G1 = G2 – 2 × G3 G1 = G2 - G3

52.0234.02E1

SECTION – E (Integer Type) No Negative Marking

This Section contains 4 questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 4 × 5 = 20 Marks

1. How many of the following have eq lower than m (molar conductance)

(a) KCl (b) Ba(OH)2 (c) NaAlO2 (d) Na3AlO3 (e) K2(HgI4) (f) NaBH4 (g) FeSO4 (h) K2Cr2O7

Sol. (5)

.eqM x ; x = charge

eqM if x > 1

b, d, e, g, h

2. How many of the following Half cell have Reduction electrode potential more than standard Reduction

potential (E°).

(a) )M 1(atm 2

2 H|H|Pt (b) )M 2(atm 1

2 H|H|Pt (c) Cu|Cu 2)M01.0(

(d) Pt|Fe ,Fe||Zn|Zn2M

2

M1

3

M2

2 (e) 2pHatm 1

2 H|H|Pt

(f) )M1(atm 2

2 Cl|Cl|Pt

Sol. (2)

b, f

(a) 2H+ + 2e– H2

E E

22

Hatm 1PH ;

]H[

plog

2

059.0E 2

(b) H+ > 1 M E > E°

(c) Cu+2 + 2e– Cu ; )Culog(2

059.0EE 2

Cu+2 < 1 M ; E < E°

Zn + 2Fe+3 Zn+2 + 2Fe+2

(d) 23

222

]Fe[

]Fe][Zn[log

2

059.0EE

[Zn+2] & [Fe+2] > 1 M

E < E°

(e) atm 1

210

He2H22

E E ;1

]10[log

2

059.0E

2

(f) Cl2 + 2e– Cl–

2

1log

2

059.0EE

E > E°



3. How many Faraday’s will be required to convert nitrobenzene to nitrosobenzene Sol.(2)

4. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm2. The conductance of this solution was found to be 5 × 10–7S. The pH of the solution is 4. The value of limiting molar conductivity

0M

of this weak monobasic acid in aqueous solution is Z × 102S cm-1 mol-1. The value of Z

is. Sol. (6)

mmmm

Cm

1000a

G1000c

1000

ccc][H acid w eakFor

6Z10Z

10001

120105

102

7

4

NO2 +3

N = O

+1

2e– +

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Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B ... · = antilog 37.28 = 1.94 × 10 37. Dr. Sangeeta Khanna Ph.D Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important