Question

For the half cell

At $pH=2$, the electrode potential is

• A. 1.36 V
• B. 1.30 V
• C. 1.42 V
• D. 1.20 V

Answer 1

Answer: (C)

In this electrode,
$[A]=[B]$ in quihydrone electrode
Hence, $Q={\left[H^{\oplus }\right]}^2$
Hence, $Q={\left[H^{\oplus }\right]}^2$

$E=E^{\ominus }-\frac{0.0591}{2}\log {\left[H^{\oplus }\right]}^2$
$=E^{\ominus }-0.0591\log \left[H^{\oplus }\right]$
$=E^{{\Theta }_{+}}0.0591pH$
$=1.30+0.059\times 2$
$=1.42V$