Question

The cell reaction for the given cell is :
$\mathrm{Pt}\left(H_2\right)|pH=2\Vert pH=3|Pt\left(H_2\right)$ $P_1=1$ atm $P_2=1atm$

• A. Spontaneous
• B. Non-spontaneous
• C. In equilibrium
• D. Either of these

Answer 1

Answer: (B)

$E_{\text{cell }}=E_{OpH}^{\circ }+E_{RpH}^{\circ }+\frac{0.059}{2}\log \frac{{\left[H^{+}\right]}_{RHS}^2}{{\left[H^{+}\right]}_{LHS}^2}\times \frac{P_1}{P_2}$

$=\frac{0.059}{2}\log \frac{{\left(10^{-3}\right)}^2}{{\left(10^{-2}\right)}^2}=-0.059V$

$\because \left(E_{OP}^{\circ }=-E_{RP}^{\circ }\right)$

$\Delta G=-nFE>0$

Thus, cell reaction is non-spontaneous.