1

### AIPMT 2014

When 0.1 mol MnO$_4^{2 - }$ is oxidised the quantity of electricity required to completely oxidise MnO$_4^{2 - }$ to MnO$_4^ -$ is
A
96500 C
B
2 $\times$ 96500 C
C
9650 C
D
96.50 C

## Explanation

$\mathop {MnO_4^{2 - }}\limits_{0.1\,mole}^{ - 6}$ $\to$ $\mathop {MnO_4^ - }\limits^{ - 7}$ + e-

Quantity of electricity required = 0.1F

= 0.1 × 96500 = 9650 C
2

### AIPMT 2014

The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be
A
5.4 g
B
10.8 g
C
54.0 g
D
108.0 g

## Explanation

We know, from Faraday’s second law

${{{W_{Ag}}} \over {{E_{Ag}}}} = {{{W_{{O_2}}}} \over {{E_{{O_2}}}}}$

$\Rightarrow$ ${{{W_{Ag}}} \over {108}} = {{{{5600} \over {22400}} \times 32} \over 8}$

$\Rightarrow$ ${{{W_{Ag}}} \over {108}} = {8 \over 8}$

$\Rightarrow$ ${{W_{Ag}} = 108}$ g
3

### NEET 2013

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be
A
0.118 V
B
1.18 V
C
0.059 V
D
0.59 V

## Explanation

Oxidation half reaction is

H2 $\to$ 2H+ + 2e

If pH = 10, [H+] = 10–10

We know, Nernst Equation

Ecell = E°cell - ${{0.0591} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}} \over {{P_{{H_2}}}}}$

For hydrogen electrode E°cell = 0

Ecell = - ${{0.0591} \over 2}\log {{{{\left( {{{10}^{ - 10}}} \right)}^2}} \over 1}$

= 0.0591log1010

Ecell = 0.591 V
4

### NEET 2013 (Karnataka)

Consider the half-cell reduction reaction
Mn2+ + 2e$-$ $\to$ Mn, Eo = $-$1.18 V
Mn2+ $\to$ Mn3+ + e$-$, Eo = $-$ 1.51 V
The $E$o for the reaction 3 Mn2+ $\to$ Mno + 2Mn3+, and possibility of the forward reaction are respectively
A
$-$ 4.18 V and yes
B
+ 0.33 V and yes
C
+ 2.69 V and no
D
$-$ 2.69 V and no

## Explanation

Mn2+ + 2e$-$ $\to$ Mn, Eo = $-$1.18 V

Mn2+ $\to$ Mn3+ + e$-$, Eo = $-$ 1.51 V

3Mn2+ $\to$ Mno + 2Mn3+, $\Delta$Eo = - 1.81 - 1.51 = $-$ 2.69 V

Since $\Delta$E° is negative,

$\therefore$ $\Delta$G = –nFE°, $\Delta$G will have positive value so, forward reaction is not possible.