Question

Given the following two standard reduction potentials,

$$\begin{gathered} {\mathit{M}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{3}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{M}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{V} \\ {\mathit{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{M}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{50}\mathit{V} \end{gathered}$$

determine the standard reduction potential of the half-reaction

${\mathit{M}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }{\mathit{M}}^{\mathbf{2}\mathbf{+}}$

(Hint:You must use the extensive property ΔG⁰ to determine the standard reduction potential.)

Short answer

The standard reduction potential of the half reaction is 0.70 V

Step-by-step solution

analyzing the given data

$$\begin{gathered} M^{3+}+3e^{-}\to M \\ \left(E°=-0.10V\right)and[G_1=-\left(3\right)F\left(-0.10\right)] \\ M\to M^{2+}+2e^{-} \\ \left(E°=-0.50V\right)and[G_2=-\left(2\right)F\left(0.50\right)] \end{gathered}$$

Determining the standard reduction potential of the half-reaction

Using Gibbs free energy method, we get,

$$\begin{gathered} (\Delta G=\Delta G_1+\Delta G_2) \\ =0.30F-F \\ =-0.70F \end{gathered}$$

Equating this with$-nFE,$

$$\begin{gathered} -\mathrm{nFE}=-0.7\mathrm{F} \\ \mathrm{Here},\mathrm{n}=1 \\ \mathrm{E}=0.70 \mathrm{V} \end{gathered}$$

Hence, the standard reduction potential of the half-reaction will be 0.70V.