Q. For the following cell with hydrogen electrodes at two different pressures $p_{1}\ and\ p_{2}$ $Pt\ (H_{2})\ |\ H^{+}\ (aq)\ |\ Pt\ (H_{2})$ $p_{1}$ 1 M $p_{2}$ emf is given by
A) $\frac{RT}{F}\ \log_{e}\ \frac{p_{1}}{p_{2}}$
B) $\frac{RT}{2F}\ \log_{e}\ \frac{p_{1}}{p_{2}}$
C) $\frac{RT}{F}\ \log_{e}\ \frac{p_{2}}{p_{1}}$
D) $\frac{RT}{2F}\ \log_{e}\ \frac{p_{2}}{p_{1}}$
Explanation
(b)
LHS half cell
$H_{2}(g) \rightarrow \ {2H}^{+}\ (1\ M) + \ {2e}^{-}$
$P1$
RHS half cell
${2H}^{+}\ (1\ M) + \ {2e}^{-}\ \ \rightarrow \ H_{2}\ (g)$
$P2$
[table] $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{2}(g) \rightarrow \ H_{2}\ (g)$ [/table]
P1 p2
$E_{cell}^{{^\circ}}\ = \ 0.00\ V,\ K\ = \ \frac{p_{2}}{p_{1}},\ n\ = \ 2$
$E_{cell} = \ E_{cell}^{{^\circ}} - \ \frac{RT}{nF}\ log_{e}\ K$
$\ \ \ \ = \ 0\ - \ \frac{RT}{2F}\ log_{e}\ \frac{P_{2}}{P_{1}}$
$E_{cell} = \ \frac{RT}{2F}\ \log_{e}\ \frac{p_{1}}{p_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.