1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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.

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI