1

### 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.
2

### AIPMT 2010 Mains

Consider the following relations for emf of an electrochemical cell
(i)   EMF of cell = (Oxidation potential of anode) $-$ (Reduction potential of cathode)
(ii)  EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode)

Which of the above relations are correct?
A
(iii) and (i)
B
(i) and (ii)
C
(iii) and (iv)
D
(ii) and (iv)

## Explanation

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode – Oxidation potential of cathode.
3

### 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
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