Question

The following standard reduction potentials have been determined for the aqueous chemistry of indium:

$\begin{array}{l}{\text{In}}^{\text{3+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to {\text{In}}^{\text{+}}\left(\text{aq}\right){\text{E}}^{\text{o}}\text{=-0.40V}\\ {\text{In}}^{\text{+}}\left(\text{aq}\right){\text{+e}}^{\text{-}}\to \text{In}\left(\text{s}\right){\text{E}}^{\text{o}}\text{=-0.21V}\end{array}$

Calculate the equilibrium constant (K) for the disproportionation of In+(aq) at 25°C.

${\text{3In}}^{\text{+}}\left(\text{aq}\right)\rightleftarrows \text{2In}\left(\text{s}\right){\text{+In}}^{\text{3+}}\left(\text{aq}\right)$

Short answer

The equilibrium constant (K) for disproportionation of at ${\text{25}}^{\text{o}}\text{C}$is${\text{K=2.69×10}}^{\text{6}}$

Step-by-step solution

Equilibrium constant.

The equilibrium constant of the chemical reaction shows the relationship between thereactants and product compounds when the reaction reaches equilibrium. The equilibrium constant is denoted by K.

Balance the reaction

The given half-cell reactions are

${\text{In}}^{\text{3+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to {\text{In}}^{\text{+}}\left(\text{aq}\right){\text{E}}^{\text{o}}\text{=-0.40V}$ --- (a)

${\text{In}}^{\text{+}}\left(\text{aq}\right){\text{+e}}^{\text{-}}\to \text{In}\left(\text{s}\right){\text{E}}^{\text{o}}\text{=-0.21V}$ --- (b)

Now, balance the reaction

First reversing reaction (a), then multiply reaction (a) by three (3) and reaction (b) by two (2), we get

${\text{3In}}^{\text{+}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}\to {\text{3In}}^{\text{+}}\left(\text{aq}\right){\text{E}}^{\text{o}}\text{=-0.21V}$

${\text{2In}}^{\text{+}}\left(\text{aq}\right)\to \text{2In}\left(\text{s}\right){\text{+2e}}^{\text{-}}{\text{-E}}^{\text{o}}\text{=0.40V}$

Calculate the equilibrium constant.

The cell reaction is

${\text{3In}}^{\text{+}}\left(\text{aq}\right){\text{+e}}^{\text{-}}\to \text{2In}\left(\text{s}\right){\text{+In}}^{\text{3+}}\left(\text{aq}\right)$

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}\text{=-0.21V+0.40V}\\ \text{=0.19V}\end{array}$

According to the Nernst equation,

${\text{E}}_{\text{cell}}{\text{=E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{0.0591V}}{\text{2}}\text{log}\left(\frac{\left[{\text{In}}^{\text{3+}}\right]}{{\left[{\text{In}}^{\text{+1}}\right]}^{\text{3}}}\right)$

Since equilibrium exists in this cell. Then,

${\text{E}}_{\text{cell}}\text{=0}$

Here

${\text{E}}_{\text{cell}}^{\text{o}}\text{=}\frac{\text{0.0591V}}{\text{2}}\text{log}\left(\frac{\left[{\text{In}}^{\text{3+}}\right]}{{\left[{\text{In}}^{\text{+1}}\right]}^{\text{3}}}\right)$

$\begin{array}{c}\text{0.19=}\frac{\text{0.0591V}}{\text{2}}\text{logK}\\ \text{logK=}\frac{\text{0.19×2}}{\text{0.0591}}\\ \text{=}\frac{\text{0.38}}{\text{0.0591}}\end{array}$

Therefore,

${\text{K=10}}^{\text{6.429}}$

${\text{K=2.69×10}}^{\text{6}}$