Question

The solubility product for Cul (s) is 1.1 x 10^-12. Calculate the value of E° for the half-reaction.
$\mathit{C}\mathit{u}\mathit{I}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{C}\mathit{u}\mathbf{+}{\mathit{I}}^{\mathbf{-}}$

Short answer

The value of standard reduction electrode potential is -0.19 V.

Step-by-step solution

Calculation for the value of Standard Gibbs free energy change

The expression for standard Gibbs free energy change can be denoted as:

$$\begin{gathered} \Delta G^o=-RT \ln K_{sp} \\ \Delta G^o=-(8.314 J/K mol)(298 K)\ln (1.1\times {10}^{-12}) \\ \Delta G^o=6.84\times {10}^{4}J \end{gathered}$$

Calculation for the standard electrode potential for cell

The formula can be expressed as:

$$\begin{gathered} {E^o}_{cell}=-\frac{\Delta G^o}{nF} \\ {E^o}_{cell}=\frac{(6.84\times {10}^{4}J)}{(1 mole)(96485 C/mol)} \\ {E^o}_{cell}=-0.709 V \end{gathered}$$

Calculation for the standard reduction potential for cell

The value of${E^o}_{cell} \& {E^o}_{red}=-0.709 V \& -0.52 V$

On substituting the values,

$$\begin{gathered} {E^o}_{red}={E^o}_{cell}-{E^o}_{oxd} \\ {E^o}_{red}=(-0.709 V)-(-0.52 V) \\ {E^o}_{red}=-0.19 V \end{gathered}$$