Explanation

(c)
Reduction hydrogen half-cell is
$\ \ \ \ \ \ \ \ \ \ \ H^{+}\ |\ (xM)\ |\ Pt\ (H_{2})$
+
Pressure$\ pH_{2}$
Half – cell reaction is
$\ \ \ \ \ {2H}^{+}\ (aq)\ + \ {2e}^{-}\ \ \longrightarrow \ \ H_{2}\ (g)$
Reaction quotient
$\mathcal{= \ Q\ = \ }P_{H_{2}}\ /\lbrack H^{+}\rbrack^{2},\ n\ = 2$
$E_{red} = \ E_{red}^{{^\circ}} - \ \frac{0.0591}{n}\log\mathcal{Q} = 0 - \ \frac{0.0591}{2}\log\mathcal{Q}$
[table] | $\mathbf{P}_{\mathbf{H}_{\mathbf{2}}}$ | [$\mathbf{H}^{\mathbf{+}}$] | $\mathcal{Q}$ | $\mathbf{E}_{\mathbf{red}}$ ; (a) | 1atm | 2.0 M | 0.25 | +ve ; (b) | 1atm | 1.0 M | 1.0 | 0 ; (c) | 2atm | 1.0 M | 2.0 | -ve ; (d) | 2atm | 2.0 M | 0.50 | +ve [/table]
$E_{red}^{{^\circ}}$ = 0.00 V for standard hydrogen electrode
If $Q\ > \ 1$, then $E_{red}$ = -ve.
Thus, correct answer is (c).