Question

The equilibrium constant for the reaction is :
$I{n}^{2+}+C{u}^{2+}⟶I{n}^{3+}+C{u}^{+}$at $298K$
Given, $E_{C{u}^{2+}/C{u}^{+}}^{\circ }=0.15V$
$E_{{\text{In}}^{3+}/I{n}^{+}}^{\circ }=-0.42V,E_{{\text{In}}^{2+}/I{n}^{+}}^{\circ }=-0.40V$

• A. $10^5$
• B. $10^3$
• C. $10^{10}$
• D. $10^6$

Answer 1

Answer: (C)

Given, ${\text{In}}^{3+}+2e^{-}⟶{\mathrm{In}}^{+};E_1^{\circ }=-0.42V\dots$ (i)
$I{n}^{2+}+e^{-}⟶{\text{ In }}^{+};E_2^{\circ }=-0.40V\dots \text{ (ii) }$
By subtracting. Eq. (ii) from Eq. (i) a third half-cell reaction can be obtained as:

$I{n}^{3+}+e^{-}⟶{\mathrm{In}}^{2+};E_3^{\circ }=-0.44V$
For the reactions: $C{u}^{2+}+e^{-}⟶C{u}^{+};E_4^{\circ }=0.15$
$I{n}^{2+}⟶{\mathrm{In}}^{3+}+e^{-};E_3^{\circ }=+0.44$
The net redox change: $C{u}^{2+}+I{n}^{2+}⟶C{u}^{+}+I{n}^{3+}$;
$E_{\text{cell }}^{\circ }=E_4^{\circ }+E_3^{\circ }=0.15+0.44=0.59V$
Also. $E_{\text{cell }}^{\circ }=\frac{0.059}{1}\log K_c$
$0.59=\frac{0.059}{1}\log K_c$
$\therefore K_c=10^{10}$