Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions increases which in turn increases equivalent conductance of the solution.

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

The degree of dissociation ($$\alpha $$)

$$\alpha $$ = $${8 \over {400}}$$ = 2 $$ \times $$ 10^-2

From Ostwald’s dilution law for weak monobasic acid, we have

k_c = C$$\alpha $$²

= $${1 \over {32}} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2}$$

= 1.25 $$ \times $$ 10^-5

For the reaction,

Cu²⁺ + 2e^$$-$$ $$ \to $$ Cu,  E^o = 0.337 V

$$\Delta $$G^o = - nFE^o

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺,  E^o = 0.153 V

$$\Delta $$G^o = - nFE^o

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺

$$\Delta $$G^o = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V