Question

Find the equilibrium constant for the reaction
$ Cu^2+In^{2+} \rightleftharpoons Cu^+ + In^{3+} $
Given that $ E^\circ_{Cu^{2+}/Cu^+}=0.15V, E^\circ_{In^{2+}\, In^+} $
$ = - 0.4\,V,$
$ E^\circ_{In^{3+}\, In^+} =-0.42\, V$

Answer 1

Given,
$ In^{2+} + e^- \longrightarrow In^+ \, \, E^\circ=-0.40 \, \, $ ...(i)
$\Rightarrow \Delta G^\circ = 0.40\, F$
$ In^{3+} + 2e^- \longrightarrow In^+ \, \, E^\circ=-0.42 \, \,$ ...(ii)
$\Rightarrow \Delta G^\circ = 0.84\, F$
Subtracting (i) from (ii)
$ In^{3+} + e^- \longrightarrow In^{2+} \, \, \, \Delta G^\circ=-0.44\, F=- E^\circ F $
$\Rightarrow E^\circ=-0.44\, V$
Now, for : $ Cu^{2+}+In^{2+} + \longrightarrow Cu^+ + In^{3+}$
$ E^\circ=E^\circ (Cu^{2+}/Cu^+)- E^\circ (In^{3+} /In^{2+})$
$ = 0.15-(-0.44) = 0.59\, V $
Also $ E^\circ = 0.0590\, \log\, K $
$\Rightarrow \log\, K= \frac{E^\circ}{0.059}= 10 $
$ \Rightarrow \, K=10^{10}$