Q. The hydrogen electrode is dipped in Asolution of pH 3 at $25℃.$ The potential would be (the value of 2.303 RT /F is 0.059 V)
A) 0.177 V
B) 0.087 V
C) 0.059 V
D) -0.177 V
Explanation
(d)
$pH = 3,\ \left\lbrack H^{+} \right\rbrack = 10^{- 3}$
$E\ = \ E_{red}^{{^\circ}}\ + \ 0.059\ \log\ (ion)$
$E\ = \ 0\ + \ 0.059\ log\ (10^{- 3})$
$E\ = \ + \ 0.059\ ( - 3)\ = \ - \ 0.177\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.