Question

What will be $p H$ of the following half-cell $Pt, H _{2} H _{2} SO _{4}$ ? The oxidation potential is $+ 0.3\,V.$

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

Answer 1

Answer: (C)

5.085

Answer 2

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