1
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25oC. The value of standard Gibb's energy, $$\Delta $$Go will be
(F = 96500 C mol$$-$$1)
A
$$-$$ 89.0 kJ
B
$$-$$ 89.0 J
C
$$-$$ 44.5 kJ
D
$$-$$ 98.0 kJ

Explanation

The cell reaction

Cu + 2Ag+ $$ \to $$ Cu2+ + 2Ag

We know, $$\Delta $$G° = – nFE°cell

= – 2 × 96500 × 0.46 = – 88780 J

= – 88.780 kJ = – 89 kJ
2
MCQ (Single Correct Answer)

AIPMT 2010 Mains

Which of the following expressions correctly represents the equivalent conductance at infinite diluation of Al2(SO4)3. Given that $$\mathop \Lambda \limits^ \circ $$Al3+ and $$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$ are the equivalent conductances at infinite dilution of the respective ions?
A
$$2\mathop \Lambda \limits^ \circ $$Al3+   +   $$3\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$
B
$$\mathop \Lambda \limits^ \circ $$Al3+   +   $$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$
C
($$\mathop \Lambda \limits^ \circ $$Al3+   +   $$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$) $$ \times $$ 6
D
$${1 \over 3}$$$$\mathop \Lambda \limits^ \circ $$Al3+   +   $${1 \over 2}$$$$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$

Explanation

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.
3
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
A
increase in ionic mobility of ions
B
100% ionisation of electrolyte at normal dilution
C
increase in both i.e., number of ions and ionic mobility of ions
D
increase in number of ions.

Explanation

Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions increases which in turn increases equivalent conductance of the solution.
4
MCQ (Single Correct Answer)

AIPMT 2009

Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 $$ \times $$ 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is product? (Assume 100% current efficiency, at mass of Al = 27 g mol$$-$$1).
A
8.1 $$ \times $$ 104 g
B
2.4 $$ \times $$ 105 g
C
1.3 $$ \times $$ 104 g
D
9.0 $$ \times $$ 103 g

Explanation

E = Z × 96500

$$ \Rightarrow $$ $${{27} \over 3}$$ = Z $$ \times $$ 96500

$$ \Rightarrow $$ Z = $${9 \over {96500}}$$

Now applying the formula, W = Z × I × t

W = $${9 \over {96500}}$$ $$ \times $$ 4 $$ \times $$ 104 $$ \times $$ 6 $$ \times $$ 60 $$ \times $$ 60

= 8.1 × 104 g

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