Question

What will be $pH$ of the following half-cell $Pt,H_2{H}_{2}S{O}_4$ ? The oxidation potential is $+0.3V.$

• A. 3.085
• B. 4.085
• C. 5.085
• D. 6.085

Answer 1

Answer: (C)

$H_2\to 2H^{+}+2e^{-}$
$\because E_{\text{cell }}=E_{\text{cell }}^o-\frac{0.0591}{n}{\log }_{10}\frac{{\left[H^{+}\right]}^2}{\left[H_2\right]}$
$\therefore 0.3=0-\frac{0.0591}{2}{\log }_{10}{\left[H^{+}\right]}^2$
or $-\log \left[H^{+}\right]=\frac{0.3}{0.0591}=5.085$
$\therefore pH-\log \left[H^{+}\right]$
$=5.085$