Q. $I_{2}(s)|{\ I}^{-}(0.1\ M)$ half-cell is connected to a $H^{+}(aq)|{\ H}_{2}(1\ bar)|$ Pt half-cell and emf is found to be 0.7714 V. If $E_{I_{2}/I^{-}}^{{^\circ}} = 0.535\ V,$ find the $pH\$of$\ H^{+}/H_{2}$ half-cell

A) 1
B) 2
C) 3
D) 5
Explanation

(c)
The cell reaction is
$H_{2}(g) + I_{2}(s) \rightleftharpoons 2H^{+}(aq) + 2I^{-}(aq)$
$0.7714 = 0.535 - \frac{0.0591}{2}\log\frac{\left\lbrack H^{+} \right\rbrack^{2}\left\lbrack I^{-} \right\rbrack^{2}}{p_{H_{2}}}$
$\therefore pH = 3$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.