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1

AIPMT 2008

MCQ (Single Correct Answer)
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.
2

AIPMT 2007

MCQ (Single Correct Answer)
The efficiency of a fuel cell is given by
A
$$\Delta $$G/$$\Delta $$S
B
$$\Delta $$G/$$\Delta $$H
C
$$\Delta $$S/$$\Delta $$G
D
$$\Delta $$H/$$\Delta $$G

Explanation

Efficiency of a fuel cell ($$\phi $$) = $${{\Delta G} \over {\Delta H}} \times 100$$

Generally, fuel cells are expected to have an efficiency of 100 percent.
3

AIPMT 2007

MCQ (Single Correct Answer)
The equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) $$ \to $$ Cu2+(aq) + 2Ag(s);
Eo = 0.46 V at 298 K is
A
2.0 $$ \times $$ 1010
B
4.0 $$ \times $$ 1010
C
4.0 $$ \times $$ 1015
D
2.4 $$ \times $$ 1010

Explanation

RT ln K = nFE°

ln K = $${{nFE^\circ } \over {RT}}$$

= $${{2 \times 0.46} \over {0.0591}}$$

$$ \Rightarrow $$ K = 4 $$ \times $$ 1015
4

AIPMT 2006

MCQ (Single Correct Answer)
A hypothetical electrochemical cell is shown below.

$$A\left| {{A^ + }\left( {xM} \right)} \right|\left| {{B^ + }\left( {yM} \right)} \right|B$$

The emf measured is + 0.20 V. The cell reaction is
A
A + B+ $$ \to $$ A+ + B
B
A+ + B $$ \to $$ A + B+
C
A+ + e$$-$$ $$ \to $$ A;  B+ + e$$-$$ $$ \to $$ B
D
the cell reaction cannot be predicted.

Explanation

From the given expression:

At anode : A $$ \to $$ A+ + e

At cathode : B+ + e $$ \to $$ B

Overall reaction is : A + B+ $$ \to $$ A+ + B

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