Question

The equilibrium constant for the reaction:
$I{n}^{2+}+C{u}^{2+}\rightleftharpoons I{n}^{3+}+C{u}^{+}$ at 298 K
Given $E_{C{u}^{2+}/Cu}^o=0.15V;E_{I{n}^{3+}/I{n}^{+}}^o=0.42V$ and $E_{I{n}^{2+}/I{n}^{+}}^o=-0.40V$

• A. ${10}^{10}$
• B. ${10}^9$
• C. ${10}^{11}$
• D. None of these

Answer 1

The correct option is A ${10}^{10}$

$C{u}^{2+}+e\to C{u}^{+};E_1^o=0.15V$ ...... (i)
$I{n}^{3+}+2e\to I{n}^{+};E_2^o=-0.42V$ ...... (ii)
$I{n}^{2+}+e\to I{n}^{+};E_3^o=-0.40V$ ...... (iii)
By subtracting Equation (iii) from equation (ii) and a half reaction is obtained as
$I{n}^{3+}+e\to I{n}^{2+};E_4^o$ ......... (iv)
$\therefore -\Delta G^o=-\Delta G_2^o+\Delta G_3^o$
$+1\times E_4^o\times F=+2\times (-0.42)\times F-1\times (-0.40)\times F$
$\therefore E_4^o=0.44V$
Now by two half reactions Equation (i) and (iv) redox change is
$C{u}^{2+}+I{n}^{2+}\to C{u}^{+}+I{n}^{3+}$
$\therefore E_{cell}^o=\frac{0.059}{1}log{K}_c$
$E_{R{P}_{Cu}}^o+E_{O{P}_{In}}^o=\frac{0.059}{1}log{K}_c$
$0.15+0.44=\frac{0.059}{1}log{K}_c$
$\therefore K_c={10}^{10}$