Question

In the given reaction,
$2C{u}^{+}(aq)\rightleftharpoons C{u}^{2+}(aq)+Cu(s)$
$E_{C{u}^{+}/Cu}^{\circ }=0.6V\text{ and }{E}_{C{u}^{2+}/Cu}^{\circ }=0.41V$
Find out the equilibrium constant.

• A. $2.76\times 10^2$
• B. $2.76\times 10^4$
• C. $2.76\times 10^6$
• D. $2.76\times 10^8$

Answer 1

Answer: (C)

Right hand cell reaction,
$C{u}^{+}+e^{-}⟶Cu$
Left hand cell reaction,
$C{u}^{+}⟶C{u}^{2+}+e^{-}$
Cell reactions, $2C{u}^{+}⟶Cu+C{u}^{2+}$
Cell potential $E_{\text{cell }}^{\circ }=E_{C{u}^{+}/Cu}^{\circ }-E_{C{u}^{2+}/C{u}^{+}}^{\circ }$
$=0.60-0.41=0.19V$
$-nF{E}^{\circ }=-RT\ln K_{eq}$
$\log K_{eq}=\frac{n{E}^{\circ }}{(2.303RT/F)}=\frac{2\times 0.19V}{0.059V}=6.44$
$K_{eq}=10^{6.44}=2.76\times 10^6$