Question

The cell reaction for the given cell is :
$\operatorname{Pt}\left( H _{2}\right)| pH =2 \| pH =3| Pt \left( H _{2}\right)$ $P_{1}=1$ atm $P _{2}=1 atm$

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

Answer 1

Answer: (B)

$E_{\text {cell }}=E_{O p H}^{\circ}+E_{R p H}^{\circ}+\frac{0.059}{2} \log \frac{\left[H^{+}\right]_{R H S}^{2}}{\left[H^{+}\right]_{L H S}^{2}} \times \frac{P_{1}}{P_{2}}$

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

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

$\Delta G=-n F E > 0$

Thus, cell reaction is non-spontaneous.