Question

a) Use the data from AppendixE to calculate the half-cell potential ${\mathit{E}}^{\mathbf{o}}$for the half-reaction${\mathbf{\text{Mn}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{e}}}^{\mathbf{-}}\mathbf{\to }{\mathbf{\text{MN}}}_{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$

b) Consider the disproportionation reaction$\mathbf{3}{{\mathbf{\text{Mn}}}^{\mathbf{2}\mathbf{+}}}_{\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}}\mathbf{\rightleftharpoons }{{\mathbf{\text{Mn}}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}}^{\mathbf{+}\mathbf{2}}{{\mathbf{\text{.Mn}}}^{\mathbf{3}\mathbf{+}}}_{\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}}$ Will${\text{Mn}}^{2+}$ disproportionate in aqueous solution?

Short answer

a) Half-cell potential$E^o$ for the half-reaction is$-0.183\text{V}$

b) No, it will not be disproportionate in an aqueous solution.

Step-by-step solution

Half-cell potential.

Half-cell potential refers to the potential at the electrode of each half cell in an electrochemical cell. In an electrochemical cell, the total potential is the total potential calculated from the potentials of two half cells.

Half-cell potential  Eo for the half-reaction 

a)

The given equation is,

${\text{Mn}}^{3+}\left(aq\right)+3{\text{e}}^{-}\to {\text{MN}}_{\left(s\right)}$

Use standard reduction potentials:

1)${\text{Mn}}^{3+}\left(aq\right)+{\text{e}}^{-}\to {{\text{Mn}}^{2+}}_{\left(aq\right)}$

${\text{E}}^{°}1=-1.002\text{V}$

2)${{\text{Mn}}^{2+}}_{\left(aq\right)}+2{\text{e}}^{-}\to {\text{Mn}}_{\left(s\right)}$

${{\text{E}}^{°}}_{-}2=-1.185\text{V}$

3)${\text{Mn}}^{3+}\left(aq\right)+3{\text{e}}^{-}\to {\text{MN}}_{\left(s\right)}$

$\begin{array}{l}{\text{E}}_{-}^{°}3=?\\ {\text{E}}^{°}3=-1.185\text{V}-(-1.002\text{V})\\ {\text{E}}^{°}3=-0.183\text{V}\end{array}$

Check ifMn2+   disproportionate in aqueous solution.

b)

No, it will not disproportionate in aqueous solution.