Question

Emf of hydrogen electrode in term of $pH$ is (at $1atm$ pressure).

• A. $E_{H_2}=\frac{RT}{F}\times pH$
• B. $E_{H_2}=\frac{RT}{F}\frac{1}{pH}$
• C. $E_{H_2}=\frac{2.303RT}{F}pH$
• D. $E_{H_2}=-0.0591pH$

Answer 1

The correct option is D $E_{H_2}=-0.0591pH$

$E_{el}=E_{el}^{\circ }+\frac{2.303RT}{2F}\left[H^{+}\right]$ [$E_{el}=$ emf of Hydrogen electrode]

$=E_{el}^{\circ }+0.0591\log \left[H^{+}\right]$

$\therefore E_{el}=E_{el}^{\circ }-0.0591pH$