Question

Calculate the standard potential for the half-reaction$\mathrm{BiOCl}\left(s\right)+2{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Bi}\left(s\right)+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$

given that $K_{\text{sp}}$for localid="1646120146262" $\mathrm{BiOCl}$has a value of localid="1646120154469" $8.1\times {10}^{-19}$.

Short answer

The standard potential for the half-reaction $\mathrm{BiOCl}\left(s\right)+2{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Bi}\left(s\right)+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$is $E^0\text{process}=-0.037\mathrm{V}$approximately

Step-by-step solution

Given information

Calculate the standard potential for the half-reaction$\mathrm{BiOCl}\left(s\right)+2{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Bi}\left(s\right)+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$

given that $K_{sp}$for data-custom-editor="chemistry" $\mathrm{BiOCl}$has a value of $8.1\times {10}^{-19}$.

Calculation

The standard potential is

$\mathrm{BiOCl}\left(s\right)+2{\mathrm{H}}^{+}+3e^{-}\rightleftharpoons \mathrm{Bi}\left(s\right)+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O},E^0\left({\mathrm{BiO}}^{+}/\mathrm{Bi}\right)=+0.320\mathrm{V}$

$E^0\text{process}=E^0\left({\mathrm{BiO}}^{+}/\mathrm{Bi}\right)-\frac{0.0592}{n}\log \frac{1}{Ksp}$

$E^0\text{process}=+0.320\mathrm{V}-\frac{0.0592}{3}\log \frac{1}{8.1\times {10}^{-19}}$

$E^0\text{process}=+0.320\mathrm{V}-0.3569\mathrm{V}$

$E^0\text{process}=-0.03697\mathrm{V}$

$E^0\text{process}=-0.037\mathrm{V}$