1

### 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

### 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

### 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

### 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