Explanation

(d)
$Pt(s),\ H_{2}\ (g)\ |\ H^{+}\ (1\ M)\ ||\ {Ag}^{+}\ (aq)\ |\ Ag\ (s)\ \$
EMF of cell = 0.62 V, $E_{cell}^{{^\circ}}\ = \ 0.80\ V$
$H_{2} \rightarrow \ {2H}^{+} + \ {2e}^{-}\ \ \ \ \$ (at anode)
${2Ag}^{+} + \ {2e}^{-}\ \rightarrow 2Ag\ \ (at\ cathode)$
[table] $H_{2} + {2Ag}^{+} \rightarrow 2\ Ag + \ {2H}^{+}\ (overall\ reaction)$ [/table]
$E_{cell}\ = \ E^{{^\circ}}\ - \ \frac{2.303RT}{2F}\log\frac{\lbrack H^{+}\rbrack^{2}}{\lbrack{Ag}^{+}\rbrack^{2}\ \lbrack H_{2}\rbrack}$
$E_{cell}\ = \ E^{{^\circ}}\ - \ \frac{2.303RT}{2F}\log\frac{1}{\lbrack{Ag}^{+}\rbrack^{2}\ }$
$0.62\ = \ 0.80\ + \ \frac{2.303\ \times 0.06}{2}\log\frac{1}{\lbrack{Ag}^{+}\rbrack^{2}}$
$0.62\ = \ 0.80\ + \ \frac{2\ \times 2.303\ \times 0.06}{2}\log{\lbrack{Ag}^{+}\rbrack}$
$- 0.18\ = \ 0.1382\ log\ \lbrack{Ag}^{+}\rbrack\$
$\lbrack{Ag}^{+}\rbrack\ = \ 0.05\ m$
∴ $Mole\ of\ {Ag}^{+}\ \ in\ 100\ mL\ = \ 0.05\ \times \ \frac{100}{1000}$
$Wt.\ of\ {Ag}^{+}\ in\ 100\ mL\ = \ 0.05\ \times \ \frac{100}{1000}\ \times 108$
$\%\ of\ Ag\ in\ 1.08\ g\ alloy = \ \frac{0.05\ \times 100\ \times 108}{1000\ \times 1.08}\ \times 100$
= 50 %