Question
Consider the following standard electrode potentials and calculate the equilibrium constant
at $${25^ \circ }C$$ for the indicated disproportion nation reaction :
$$\eqalign{
& 3M{n^{2 + }}\left( {aq} \right) \to Mn\left( s \right) + 2M{n^{3 + }}\left( {aq} \right) \cr
& M{n^{3 + }}\left( {aq} \right) + {e^ - } \to M{n^{2 + }}\left( {aq} \right);{E^ \circ } = 1.51\,V \cr
& M{n^{2 + }}\left( {aq} \right) + 2{e^ - } \to Mn\left( s \right);{E^ \circ } = - 1.185\,V \cr} $$
A.
$$1.2 \times {10^{ - 43}}$$
B.
$$2.4 \times {10^{ - 73}}$$
C.
$$6.3 \times {10^{ - 92}}$$
D.
$$1.5 \times {10^{ - 62}}$$
Answer :
$$6.3 \times {10^{ - 92}}$$
Solution :
$$\eqalign{
& 2M{n^{2 + }} \to 2M{n^{3 + }} + 2{e^ - },\Delta G_1^ \circ \cr
& M{n^{2 + }} + 2{e^ - } \to Mn,\Delta G_2^ \circ \cr
& \overline {3M{n^{2 + }}\left( {aq} \right) \to Mn\left( s \right) + 2M{n^{3 + }}\left( {aq} \right)} \cr} $$
$$ - 2 \times F \times E_3^ \circ = - 2 \times F \times \left[ { - 1.51} \right]$$ $$ - 2 \times F \times \left( { - 1.185} \right)$$
$$\eqalign{
& E_3^ \circ = - 2.695 \cr
& E_3^ \circ = + \frac{{0.0591}}{2}\log {K_{eq}};{K_{eq}} \simeq 6.3 \times {10^{ - 92}}. \cr} $$