1
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
2
MCQ (Single Correct Answer)

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
3
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.
4
MCQ (Single Correct Answer)

NEET 2013

At 25oC molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm$$-$$1 cm2 mol$$-$$1 and at infinite dilution its molar conductance is 238 ohm$$-$$1 cm2 mol$$-$$1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
A
4.008%
B
40.800%
C
2.080%
D
20.800%

Explanation

Degree of dissociation($$\alpha $$) = $${{9.54} \over {238}}$$ = 0.04008 = 4.008%

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