Q. EMF of hydrogen electrode in term of pH is (at 1 atm 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}} = \$ - 0.0591 pH
Explanation
(d)
${2H}^{+} + \ {2e}^{-}\ \rightarrow \ H_{2}\$
According to Nernst equation,
$E\ = \ E^{{^\circ}}\ + \ \frac{0.0591}{n}\log\frac{1}{\lbrack H^{+}\rbrack^{2}}$
$E\ = \ 0\ - \ \ \frac{0.0591}{2}\log{\lbrack H^{+}\rbrack^{2}}$
= - 0.0591 pH
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.