Question

The hydrogen electrode is dipped in a solution of $pH3$at $25^{o}C$. The potential of the cell would be

(the value of 2.303RT/F is 0.059V)

• A. 0.177 V
• B. – 0.177 V
• C. 0.087 V
• D. 0.059 V

Answer 1

Answer: (B)

– 0.177 V

Answer 2

$\mathbf{H}^{\mathbf{+}}\mathbf{+}\mathbf{e}^{\mathbf{-}}\mathbf{\rightarrow}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{H}_{\mathbf{2}}$

$\mathbf{E =}\mathbf{E}^{\mathbf{0}}\mathbf{-}\frac{\mathbf{0.0591}}{\mathbf{n}}\mathbf{\log}\frac{\mathbf{P}_{\mathbf{H}_{\mathbf{2}}}^{\mathbf{1/2}}}{\mathbf{\lbrack}\mathbf{H}^{\mathbf{+}}\mathbf{\rbrack}}$

\mathbf{\&\because}\text{Because pressure is not given } \\ \mathbf{\&}\text{So }\left\lbrack \mathbf{H}_{\mathbf{2}} \right\rbrack\mathbf{= 1}\text{then} \end{array} \right\}\mathbf{=}\mathbf{E}^{\mathbf{0}}\mathbf{-}\mathbf{0.0591}\mathbf{\log}\frac{\mathbf{1}}{\mathbf{\lbrack}\mathbf{H}^{\mathbf{+}}\mathbf{\rbrack}}$$ $$\mathbf{=}\mathbf{E}^{\mathbf{0}}\mathbf{- 0.0591pH}\left( \mathbf{\because pH =}\mathbf{\log}\frac{\mathbf{1}}{\mathbf{\lbrack}\mathbf{H}^{\mathbf{+}}\mathbf{\rbrack}} \right)\mathbf{= 0 - 0.0591 \times 3}\mathbf{=}\mathbf{-}\mathbf{0.177V}$$