Question

The potential of the cell containing two hydrogen electrodes as represented below, is:

$Pt,H_{2\left(g\right)}|H^{+}\left({10}^{-6}M\right)||H^{+}\left({10}^{-4}M\right)|H_{2\left(g\right)},Pt$ at $298K$

• A. $-0.0591V$
• B. $-0.118V$
• C. $0.118V$
• D. $0.0591V$

Answer 1

Answer: (C)

At cathode: $H_c^{+}+e^{-}⟶1/2H_2$ $E^o=0V$

At anode: $1/2H_2⟶H_a^{+}+e^{-}$ $E^o=0V$

_____________________

Cell reaction $H_c^{+}⟶H_a^{+}$ $E_{cell}^o=0$

For concentration $E_{cell}^o$ is $0$, as both the compartment have same half cell.

Using Nernst Equation

$[H^{+}{]}_a=$ concentration of $H^{+}$ at anode; $[H^{+}{]}_c=$ concentration of $H^{+}$ at cathode

$\Rightarrow E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{[H^{+}{]}_a}{[H^{+}{]}_c}$

$\Rightarrow E_{cell}=-0.0591\log \frac{\left[{10}^{-6}\right]}{\left[{10}^{-4}\right]}$

$\Rightarrow E_{cell}=-0.0591\log {10}^{-2}$

$\Rightarrow E_{cell}=2\times 0.0591=0.1182V$