Question

Compute E⁰ for the process

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

given that the formation constant for the complex is 1.0 x 10²².

Short answer

The E⁰ for the given reaction is -0.901 V.

Step-by-step solution

Given Information

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

K_fis 1.0x10²²

Explanation

The expression for the calculation of E⁰ is:

${\mathrm{E}}^0={\mathrm{E}}^0\left({\mathrm{Ni}}^{2+}/\mathrm{Ni}\right)-\frac{0.0592}{\mathrm{n}}\log {\mathrm{K}}_{\mathrm{f}}$ ...... (I)

Here, the electrons involved in the reaction is n, the ratio of activity of product to the activity if reactant is K_f, and the standard electrode potential for the reaction is localid="1650437033744" ${\mathrm{E}}^0\left({\mathrm{Ni}}^{2+}/\mathrm{Ni}\right)$.

Substitute -0.250 V for ${\mathrm{E}}^0\left({\mathrm{Ni}}^{2+}/\mathrm{Ni}\right)$, 2 for n, and 1.0 x 10²² for K_f in Equation (I).

$$\begin{gathered} E^0=-0.250\mathrm{V}-\frac{0.0592}{2}\log \left(1.0\times {10}^{22}\right)\mathrm{V} \\ =-0.250\mathrm{V}-\frac{0.0592}{2}\left(22\right)\mathrm{V} \\ =0.250\mathrm{V}-0.651\mathrm{V} \\ =-0.901\mathrm{V} \end{gathered}$$

Conclusion

-0.901 V is the E⁰ for the given reaction.