1

### AIPMT 2009

Given :
(i)   Cu2+ + 2e$-$ $\to$ Cu,  Eo = 0.337 V
(ii)  Cu2+ + e$-$ $\to$ Cu+,  Eo = 0.153 V
Electrode potential, Eo for the reaction,
Cu+ + e$-$ $\to$ Cu,  will be
A
0.90 V
B
0.30 V
C
0.38 V
D
0.52 V

## Explanation

For the reaction,

Cu2+ + 2e$-$ $\to$ Cu,  Eo = 0.337 V

$\Delta$Go = - nFEo

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu2+ + e$-$ $\to$ Cu+,  Eo = 0.153 V

$\Delta$Go = - nFEo

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu2+ + e$-$ $\to$ Cu+

$\Delta$Go = –0.521 F = –nFE°

$\Rightarrow$ E° = 0.52 V
2

### AIPMT 2008

Standard free energies of formation (in kJ/mol) at 298 K are $-$237.2, $-$ 394.4 and $-$8.2 for H2O(l), CO2(g) and pentane (g) respectively. The value of Eocell for the pentane-oxygen fuel cell is
A
1.0968 V
B
0.0968 V
C
1.968 V
D
2.0968 V

## Explanation

At Anode:

C5H12 + 10H2O $\to$ 5CO2 + 32H+ + 32e-

At Cathode:

8O2 + 32H+ + 32e- $\to$ 16H2O
-------------------------------------------------
C5H12(g) + 8O2(g) $\to$ 5CO2(g) + 6H2O(l)

$\Delta$G = 5×$\Delta$GCO2 + 6 $\Delta$G(H2O) – [$\Delta$G(C5H12) +8 × $\Delta$GO2 ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol–1

= – 3387 × 103 J mol–1

$\Delta$G = - nFEocell

From the overall equation we find n = 32

– 3387 × 103 = -32 $\times$ 96500 $\times$ Eocell

Eocell = ${{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}$ = 1.0968 V
3

### AIPMT 2008

Kohlrausch's law states that at
A
Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
B
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C
Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D
Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.

## Explanation

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions.

$\lambda$$\infty$ = $\lambda$a + $\lambda$c

where, $\lambda$a = equivalent conductance of the anion

$\lambda$c = equivalent conductance of the cation.
4

### AIPMT 2008

On the basis of the following Eo values, the strongest oxidizing agent is
[Fe(CN)6]4$-$ $\to$ [Fe(CN)6]3$-$ + e$-$;  Eo = $-$0.35 V

Fe2+ $\to$ Fe3+ + e$-$;  Eo = $-$0.77 V
A
Fe3+
B
[Fe(CN)6]3$-$
C
[Fe(CN)6]4$-$
D
Fe2+

## Explanation

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)6]4$-$ $\to$ [Fe(CN)6]3$-$ + e$-$;  Eo = $-$0.35 V

Fe2+ $\to$ Fe3+ + e$-$;  Eo = $-$0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.