For the half-cell
$+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \quad \mathrm{E}^{0}=1.30 \mathrm{~V}$
At $\mathrm{pH}=2$, electrode potential is :

Question

For the half-cell $+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \quad \mathrm{E}^{0}=1.30 \mathrm{~V}$ At $\mathrm{pH}=2$, electrode potential is :

Aicharya · Accepted Answer

In this electrode
$[\mathrm{A}]=[\mathrm{B}]$ in quinhydrone electrode

Hence,
$\mathrm{Q}=\left[\mathrm{H}^{+}ight]^{2}$
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.0591}{2} \log \left[\mathrm{H}^{+}ight]^{2} =\mathrm{E}^{0}-0.0591 \log \left[\mathrm{H}^{+}ight] =\mathrm{E}^{\circ}+0.0591 \mathrm{pH} =1.30+0.0591 \times 2 \approx 1.42 \mathrm{~V}$ ,

For the half-cell $+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \quad \mathrm{E}^{0}=1.30 \mathrm{~V}$ At…

Solution