NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### 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
2

### 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
3

### 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.
4

### 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

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

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

NEET

Class 12