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1

### 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
2

### 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.
3

### NEET 2013 (Karnataka)

How many gram of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1 Faraday = 96,500 C; Atomic mass of Co = 59 u)
A
4.0
B
20.0
C
40.0
D
0.66

## Explanation

W = $${{IEt} \over {96500}}$$

= $${{10 \times 109 \times 60 \times 59} \over {96500}}$$

= 20
4

### NEET 2013

A button cell used in watches function as following.
Zn(s) + Ag2O(s) + H2O(l) $$\rightleftharpoons$$ 2Ag(s) + Zn2+(aq) + 2OH$$-$$(aq)

If half cell potentials are
Zn2+(aq) + 2e$$-$$ $$\to$$ Zn(s);  Eo = $$-$$0.76 V
Ag2O(s) + H2O(l) + 2e$$-$$ $$\to$$ 2Ag(s) + 2OH$$-$$(aq), Eo = 0.34 V

The cell potential will be
A
0.84 V
B
1.34 V
C
1.10 V
D
0.42 V

## Explanation

At anode :Zn2+(aq) + 2e$$-$$ $$\to$$ Zn(s);  Eo = $$-$$0.76 V

At cathode : Ag2O(s) + H2O(l) + 2e$$-$$ $$\to$$ 2Ag(s) + 2OH$$-$$(aq), Eo = 0.34 V

cell = E°cathode – E°anode

= 0.34 – (–0.76) = 1.10 V

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Class 12