Due to the passage of current, the cell reaction is :

$$Ag(s) + {1 \over 2}C{u^{2 + }}(aq) \to A{g^ + }(aq) + {1 \over 2}Cu(s)$$ ...... [1]

From Faraday's first law, $$W = Z \times I \times t$$

or, $$W = {{equivalent\,weight\,(E)} \over {96500}} \times I \times t$$

$$\therefore$$ $${W \over E}$$ (gram equivalent) of $$C{u^{2 + }} = {1 \over {96500}} \times I \times t$$

$$ = {1 \over {96500}} \times 9.65 \times 1 \times 60 \times 60 = 0.36$$

$$\therefore$$ Decrease in concentration of $$C{u^{2 + }} = {{0.36} \over 2}(M) = 0.18\,(M)$$

$$\therefore$$ Remaining concentration of copper = 1 $$-$$ 0.18 = 0.82 (M)

$$\therefore$$ Increase in the concentration of silver ion = 0.36 (M)

$$\therefore$$ Concentration of silver ion = (1 + 0.36) = 1.36 (M)

For the reaction (1), $${E_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{[A{g^ + }]} \over {{{[C{u^{2 + }}]}^{1/2}}}}$$

Before passing current, $${E_{cell}} = E_{cell}^0 - {{0.059} \over 4}\log {1 \over {{1^2}}} = E_{cell}^0\,V$$

When passage of current is stopped,

$$E{'_{cell}} = E_{cell}^0 - {{0.059} \over n}\log {{[A{g^ + }]} \over {{{[C{u^{2 + }}]}^{1/2}}}}$$

or, $$E{'_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{1.36} \over {{{(0.81)}^{1/2}}}} = (E_{cell}^0 - 0.010)\,V$$

$$\therefore$$ $$\Delta$$E = change in cell potential $$ = E{'_{cell}} - {E_{cell}} = - 0.010\,V$$.