Question

What will be the emf for the given cell :

$Pt|H_2\left(P_1\right)|H^{+}\left(aq\right)|H_2\left(P_2\right)|Pt$

• A. $\frac{RT}{F}log\frac{P_1}{P_2}$
• B. $\frac{RT}{2F}log\frac{P_1}{P_2}$
• C. $\frac{RT}{F}log\frac{P_2}{P_1}$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

The Nernst equation for the emf of the cell is $E_{cell}=E_{cell}^0-\frac{2.303RT}{nF}log{\left\{\frac{P_{H_2}}{{\lfloor H^{+}\rfloor }^2}\right\}}_2\times {\left\{\frac{{\lfloor H^{+}\rfloor }^2}{P_{H_2}}\right\}}_1$.
Since both the electrodes dip in same, solution, the hydrogen ion concentration is same. Hence, $E_{cell}=-\frac{2.303RT}{nF}log\frac{{\left\{P_{H_2}\right\}}_2}{{\left\{P_{H_2}\right\}}_1}.$