1
MCQ (Single Correct Answer)

AIPMT 2004

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25oC. The equilibrium constant of the reaction would be
A
2.0 $$ \times $$ 1011
B
4.0 $$ \times $$ 1012
C
1.0 $$ \times $$ 102
D
1.0 $$ \times $$ 1010

Explanation

We know, from Nernst Equation

Ecell = Eocell - $${{2.303RT} \over {nF}}{\log _{10}}K$$

At equilibrium Ecell = 0

$$ \therefore $$ 0 = Eocell - $${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$$

$$ \Rightarrow $$ 0 = 0.295 - $${{2.303 \times 8.314 \times 298} \over {2 \times 96500}}{\log _{10}}K$$

$$ \Rightarrow $$ 0.295 = $${{0.0591} \over 2}{\log _{10}}K$$

$$ \Rightarrow $$ $${\log _{10}}K$$ = 10

$$ \Rightarrow $$ K = 1 $$ \times $$ 1010
2
MCQ (Single Correct Answer)

AIPMT 2003

On the basis of the information available from the reaction,

4/3Al + O2 $$ \to $$ 2/3Al2O3,   $$\Delta $$G = $$-$$ 827 kJ mol$$-$$1 of O2,

the minimum e.m.f. required to carry out an electrolysis of Al2O3 is
(F = 96500 C mol$$-$$1)
A
2.14 V
B
4.28 V
C
6.42 V
D
8.56 V

Explanation

4/3Al + O2 $$ \to $$ 2/3Al2O3,   $$\Delta $$G = $$-$$ 827 kJ mol$$-$$1

For 1 mol of Al, n = 3

$$ \therefore $$ For $${4 \over 3}$$ mol of Al, n = $$3 \times {4 \over 3} = 4$$

As $$\Delta $$G = - nFEo

$$ \Rightarrow $$ – 827 × 103 J = – 4 × E° × 96500

$$ \Rightarrow $$ Eo = $${{827 \times {{10}^3}} \over {4 \times 96500}}$$ = 2.14 V
3
MCQ (Single Correct Answer)

AIPMT 2003

The e.m.f. of a Daniell cell at 298 K is E1.

When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2?
A
E1 > E2
B
E1 < E2
C
E1 = E2
D
E2 = 0 $$ \ne $$ E1

Explanation

Cell reaction is,

Zn + Cu2+ $$ \to $$ Zn2+ + Cu

Ecell = Eocell - $${{RT} \over {nF}}\ln {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}$$

Greater the factor $${{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}$$, less is the EMF.

Hence E1 > E2
4
MCQ (Single Correct Answer)

AIPMT 2002

In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam
A
Hg is more inert than Pt
B
More voltage is required to reduce H+ at Hg than at Pt
C
Na is dissolved in Hg while it does not dissolve in Pt
D
Conc. of H+ ions is larger when Pt electrode is taken.

Explanation

In electrolysis of NaCl when Pt electrode is taken then H2 liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H+ at Hg than Pt.

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