Explanation

(d)
$E_{cell} = \ E_{cell}^{{^\circ}} - \ \frac{0.05915}{n}\log\mathcal{Q}$
For standard hydrogen electrode,
$E_{cell}^{{^\circ}} = 0.00V$
∴ $E_{cell} = \ - \ \frac{0.05915}{n}\log\mathcal{Q}$
Given, pH = 1.0
∴ $\ \ \lbrack H^{+}\rbrack\ = \ 1\ \times \ 10^{- 1}$
$E_{cell} = \ - \ \frac{0.05915}{n}\log\frac{1}{\lbrack H^{+}\rbrack}$
[∵ The reaction occurring is ${2H}^{+} + {2e}^{-}\ \rightarrow \ H_{2}$ ]
$\ \ \ \ \ \ \ \ \ \ = \ + \ \frac{0.05915}{1}\log{(H^{+})}$
$\ \ \ \ \ \ \ \ \ \ = \ 0.05915\ \log{(10^{- 1})}$
= - 0.05915 V
= - 59.15 mV