Q. The hydrogen electrode is dipped in a solution of $pH = 3\ at\ 25℃.$ The potential of the cell would be $\left( the\ value\ of\ \ \frac{2.303RT}{T}\ is\ 0.059\ V \right)$
A) 0.059 V
B) 0.088 V
C) 0.178 V
D) $- 0.177$ V
Explanation
(d)
Reduction potential of hydrogen electrode,
$E_{H} = - \frac{2.303RT}{F}\log\frac{1}{\left\lbrack H^{+} \right\rbrack}$
$= - 0.059\ pH = - 0.059 \times 3 = - 0.177\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.