1
MCQ (Single Correct Answer)

AIPMT 2009

Given :
(i)   Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V
(ii)  Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V
Electrode potential, Eo for the reaction,
Cu+ + e$$-$$ $$ \to $$ Cu,  will be
A
0.90 V
B
0.30 V
C
0.38 V
D
0.52 V

Explanation

For the reaction,

Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V

$$\Delta $$Go = - nFEo

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V

$$\Delta $$Go = - nFEo

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu2+ + e$$-$$ $$ \to $$ Cu+

$$\Delta $$Go = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V
2
MCQ (Single Correct Answer)

AIPMT 2008

Standard free energies of formation (in kJ/mol) at 298 K are $$-$$237.2, $$-$$ 394.4 and $$-$$8.2 for H2O(l), CO2(g) and pentane (g) respectively. The value of Eocell for the pentane-oxygen fuel cell is
A
1.0968 V
B
0.0968 V
C
1.968 V
D
2.0968 V

Explanation

At Anode:

C5H12 + 10H2O $$ \to $$ 5CO2 + 32H+ + 32e-

At Cathode:

8O2 + 32H+ + 32e- $$ \to $$ 16H2O
-------------------------------------------------
C5H12(g) + 8O2(g) $$ \to $$ 5CO2(g) + 6H2O(l)

$$\Delta $$G = 5×$$\Delta $$GCO2 + 6 $$\Delta $$G(H2O) – [$$\Delta $$G(C5H12) +8 × $$\Delta $$GO2 ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol–1

= – 3387 × 103 J mol–1

$$\Delta $$G = - nFEocell

From the overall equation we find n = 32

– 3387 × 103 = -32 $$ \times $$ 96500 $$ \times $$ Eocell

Eocell = $${{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}$$ = 1.0968 V
3
MCQ (Single Correct Answer)

AIPMT 2008

Kohlrausch's law states that at
A
Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
B
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C
Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D
Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.

Explanation

Kohlrausch’s law states that the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductance of the component ions.

$$\lambda $$$$\infty $$ = $$\lambda $$a + $$\lambda $$c

where, $$\lambda $$a = equivalent conductance of the anion

$$\lambda $$c = equivalent conductance of the cation.
4
MCQ (Single Correct Answer)

AIPMT 2008

On the basis of the following Eo values, the strongest oxidizing agent is
[Fe(CN)6]4$$-$$ $$ \to $$ [Fe(CN)6]3$$-$$ + e$$-$$;  Eo = $$-$$0.35 V

Fe2+ $$ \to $$ Fe3+ + e$$-$$;  Eo = $$-$$0.77 V
A
Fe3+
B
[Fe(CN)6]3$$-$$
C
[Fe(CN)6]4$$-$$
D
Fe2+

Explanation

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)6]4$$-$$ $$ \to $$ [Fe(CN)6]3$$-$$ + e$$-$$;  Eo = $$-$$0.35 V

Fe2+ $$ \to $$ Fe3+ + e$$-$$;  Eo = $$-$$0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.

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