Question

Find the equilibrium constant for the reaction
$C{u}^2+I{n}^{2+}\rightleftharpoons C{u}^{+}+I{n}^{3+}$
Given that $E_{C{u}^{2+}/C{u}^{+}}^{\circ }=0.15V,E_{I{n}^{2+}I{n}^{+}}^{\circ }$
$=-0.4V,$
$E_{I{n}^{3+}I{n}^{+}}^{\circ }=-0.42V$

Answer 1

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