SRI SARVANI EDUCATIONAL SOCIETYSR IPL ASSIGNMENT DT:15-04-2014

SINGLE ANSWER TYPE QUESTIONS:

1. Which of the following best explains why concentration cells must be run under non-standard conditions in order for them to do electrical work?

A) concentration cell cannot do electrical work at 25oC.

B)The change in free energy for a concentration cell is always negative under standard conditions.

C) Concentration cell has a cell potential equal to zero under standard conditions.

D) A concentration cell can only do electrical work when either NH3 or NaOH is added

2. for two reaction are given below :

What will be the for?

A) -1.68V B)1.68V C)-0.20V D)0.20V

3. 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

4. Alkali metals dissolve in liquid NH3 and form blue coloured solution which is conductive due to

ammoniated electrons. The conductivity is

A) Lower than that of completely ionized metal salt in water

B) Higher than that of completely ionized metal salt in water

C) Low than that of fused metal salt

D) Equal to conductivity of liquid metal

5. From the following values of electrode potentials,

i) and

ii) . Calculate for the reaction,

A) -28.95 kJ B) +40.53 kJ C) -75.27 kJ D) -40.53 Kj

6. Given that under acidic conditions. An acidic solution of MnSO4 is mixed with KMnO4 solution. The incorrect statement(s) among the following regarding this system is/are

(I) A stable compound of the stoichiometry Mn(MnO4)2 is formed.

(II) A precipitate of MnO2 is formed.

(III) KMnO4 loses its pink colour.

(IV) A redox reaction with a standard cell potential of +0.47 V occurs.

(A) I (B) II, III, IV (C) I, IV (D) II, III

7. EMF of the following cell is 0.59volts at 298K.

is equilibrium concentration. Coordination no. of is 4. The instability constant of the complex is

(A) (B) (C) (D)

8. 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)279.6 mV

9. 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

10. By how much would the oxidising power of the MnO4− /Mn2+ couple change if the H+ ions

concentration is decreased 100 times?

(a) increases by 189 mV (b) decreases by 189 mV

(c) will increase by 19 mV (d) will decrease by 19 mV

11. The useful work done during the reaction

Ag(s) + Cl2(g) ¾® AgCl(s)

Would be

(a) 110kJ mol–1 (b) 220 kJ mol—1

(c) 55kJ mol–1 (d) 100 kJ mol–1

Given ,

Multiple Answer Type Questions:

12. Which of the following statements are NOT TRUE?

A) Use of salt bridge maximizes liquid junction potential

B) For a spontaneous redox process

C) For a spontaneous redox process

D) A given lead acid storage cell posse’s constant cell potential.

13. Which of the following cell(s) can act as concentration cell?

A)

B)

C)

D)

Integer Type Questions :

14. The voltaic cell using and half cells is set up at standard conditions, and each compartment has a volume of 240 ml. The cell delivers 0.15 A for 52.3 h then the

concentration in Cu chamber is so value of x is : [ Given: Cu (M. W = 63.5)]

15. The standard reduction potential for the half cell.(Give your answer as numerical)

NO3–(aq) + 2H+

(aq) + e– ¾® NO2(g) + H2O(l) is 0.78 V

a) Calculate the reduction potential in 8M H+

b) What will be the reduction potential of the half-cell in a neutral solution? Assume all other species to be at unit concentration

16. At 25°C, the emf of the cell (Give your answer as numerical)

Pb | PbCl2.HCl (0.5 M) || HCl (0.5 M) | AgCl(s) | Ag is 0.49 volts and its temperature coefficient

. Calculate

a) the entropy change when 1 gm mol of silver is deposited and

b) the heat of formation of AgCl, if the heat of formation of lead chloride is – 86000 cal.

17. 25 mL of a solution of HCl (0.1M) is being titrated potentiometrically against 0.1 M NaOH solution using a hydrogen electrode as the indicator electrode and saturated calomel electrode (SCE) as the reference electrode. What would be the EMF of the cell initially and after the addition of 20 mL of alkali at 25°C? Given Reduction potential of SCE = 0.2422V. [log 9 = 0.95]. (Give your answer as numerical)

18. Calculate the potential of an indicator electrode versus the standard hydrogen electrode which originally 0.1M MnO4

– and 0.8M H+ and which has been treated with 90% of the Fe2+ necessary to reduce all the MnO4

– to Mn2+.(Give your answer as numerical)MnO4

– + 8H+ + 5e– ¾® Mn2+ + 4H2O E° = 1.51 V

19. b) What is the maximum value of ratio for which the following will act as electrochemical cell? (Give your answer as numerical)

[log 6 = 0.778] [1 + 4]

20. When a rod of metallic lead was added to a 0.01 M solution of [Co(en) 3]3+, it was found that 68% of the cobalt complex was reduced to [Co(en)3] 2+ by lead.i) Find the value of K for Pb + 2[Co(en)3]3+ Pb2+ + 2[Co(en)3] 2+

ii) What is the value of Eo [Co(en)3]3+|[Co(en)3] 2+

Given: Eo (Pb2+|Pb) = - 0.126 V(Give your answer as numerical)

SOLUTIONS

1. C 2. B 3. A 4. B 5. D 6.A 7.B 8.D 9.A

10. (b) + 5e– + 8H+ ¾® Mn2+ + 4H2O

According to Nernst equation,

Ered = Eredo

– log Let [H+]initial = X

Ered(initial) = [H+]final = X

100= X

102

Ered(final) =

Ered(final) – Ered(initial) = log 1016 = – 0.1891 V

This Ered decreases by 0.189 V. The tendency of the half cell to get reduced is its oxidising power. Hence the oxidising power decreases by 0.189V

11.A

(a)For the cell reaction

Ag(s) + Cl2(g) ¾® AgCl(s)

E0 = – 1.14V

or E = E0 –

Under standard conditions,

Useful work = – Wmax = – nFE

= (–1) × (–1.14) × 96500 × 10–3 kJ = 110 kJ mol–1

12. ABCD

13. ABC

14. 4

15. For the half cell reaction

NO3–(aq) + 2H+

(aq) + e– ¾® NO2(g) + H2O(l)

The Nernst equation is E = E0 –

= 0.78 – = 0.78 + 0.059 log 8

= 0.833V

Substituting this value for case (b)

E = 0.78 –

= 0.78 – 0.059 ´ 7 = 0.367V

16. The net cell reaction

= – 11260 cal [1]

= = – 4.14 cal/degree [1]

DH of the reaction

DH = DG + TDS

= – 11200 + 298 ´ (–4.14)

= – 12494 cal [1]

This heat of reaction is the algebraic sum of the heats of formation of the components.

= – 30506 cal/mole

17. The galvanic cell formed in this case is as follows:

[1]

=

= 0.2422 - 0.0591 log [

= 0.2422 + 0.0591 pH at [1]

Initial pH of the 0.1 HCl:

= 0.2422 + 0.0591 = 0.3013V [1]

pH after addition of 20 mL alkali:

Amount of HCl initially present =

Amount of NaOH added =

Amount of HCl left unreached = = 0.5 millimole

= 0.3574V [1]

18. Let us consider Galvanic cell is H+ (1M) | H2(1atm), Pt || MnO4

– (H+) | Mn+2, Pt Anode half cell : 2H+ (1M) ¾® H2 (1atm) + 2e–

Cathode half cell: MnO4– + 8H+ + 5e– ¾® Mn+2 + 4H2O

Initial Conc.: 0.1 0.8 0 0

Alter Complete reaction with Fe+2

(0.01) (0.08) (0.09) So, electrode potential of indicator electrode

=

= 1.51 –

= 1.51 –

= 1.51 – (5.36 ´ 109)= 1.51 – 0.1149= 1.395 V

Thus, potential of an indicator electrode versus the SHE is 1.395 V because ESHE = 019.

b) The cell reaction is

[1]

to act as electrochemical cell,

[1]

Here

or, [1]

or,

should be less than [1]

20.

i) [Co(en)3]3+ = 0.0032, [Co(en)3]2+ = 0.0068 , [Pb2+] = 0..0034

K =

[Pb2+ ] [Co (en )32+ ]−1

[Co (en )33+ ]−1

On putting the various known values , we get K = 0.0154

ii) DG10 = -nFE°cell = -2.303 RT log K.

From here we get, E°cell = -0.0536 V From which we can calculate E [Co(en)3

3+/[Co(en)32+] = -0.18V