Question

Calculate the standard potential for the half-reaction

$\mathrm{Al}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2^{-}+3{\mathrm{e}}^{-}\to \mathrm{Al}\left(s\right)+2{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}$

if the formation constant for the complex is$1.3\times {10}^{13}$.

Short answer

The standard potential for the half-reaction $\mathrm{Al}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2^{-}+3{\mathrm{e}}^{-}\to \mathrm{Al}\left(s\right)+2{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}$is localid="1646120589558" $E^0\text{process}=-1.92\mathrm{V}$.

Step-by-step solution

Introduction

The standard reduction potential is determined close to a standard hydrogen electrode (SHE) connection electrode, which is arbitrarily shown a possibility of $0.00$volts.

Since the reduction potential calculates the ingrained preference for a species to experience drop, comparing standard reduction potential for two methods can be helpful for resolving how a response will proceed.

Calculation

The calculation for the standard potential is

$\mathrm{Al}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2^{-}+3{\mathrm{e}}^{-}\stackrel{}{\leftrightarrows }\mathrm{Al}\left(\mathrm{s}\right)+2{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}{E}^0\left({\mathrm{Al}}^{3+}/\mathrm{Al}\right)=-1.662\mathrm{V}$

$E^0\text{process}=E^0\left({\mathrm{Al}}^{3+}/\mathrm{Al}\right)-\frac{0.0592}{n}\log K_f$

$E^0\text{process}=-1.662\mathrm{V}-\frac{0.0592}{3}\log \left(1.3\times {10}^{13}\right)$

$E^0\text{process}=-1.662\mathrm{V}-0.2587\mathrm{V}$

$E^0\text{process}=-1.9207\mathrm{V}$

$E^0\text{process}=-1.92\mathrm{V}$