Question

Compute $E^0$for the process$\mathrm{Ni}(\mathrm{CN}{)}_4^{2-}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ni}(s)+4{\mathrm{CN}}^{-}$given that the formation constant for the complex is$1.0\times {10}^{22}$.

Short answer

The$E^0$for the processlocalid="1646119167284" $\mathrm{Ni}(\mathrm{CN}{)}_4^{2-}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ni}(s)+4{\mathrm{CN}}^{-}$found to be$-0.90\mathrm{V}$.

Step-by-step solution

Given information

Compute $E^o$for the processdata-custom-editor="chemistry" $\mathrm{Ni}(\mathrm{CN}{)}_4^{2-}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ni}(s)+4{\mathrm{CN}}^{-}$ given that the formation constant for the complex is $1.0\times {10}^{22}$.

Compute E0 for the process

$\mathrm{Ni}(\mathrm{CN}{)}_4^{2-}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ni}(s)+4{\mathrm{CN}}^{-}$

$E_{N{i}^2//ii}^0=-0.250\mathrm{V}$

$E^0$processlocalid="1645933400641" $=E^0\left(N{i}^{2+}/Ni\right)-\frac{0.0592}{n}\log K_f$

$E^0\text{process}=-0.250\mathrm{V}-\frac{0.0592}{2}\log \left(1.0\times {10}^{22}\right)$

$E^0\text{process}=-0.250\mathrm{V}-0.6512\mathrm{V}$

$E^0\text{process}=-0.9012\mathrm{V}$

$E^0\text{process}=-0.90\mathrm{V}$(approximately).