Question

$C u^{+}$ion is not stable in aqueous solution because of disproportionation reaction. $E^{\circ}$ value for disproportionation of $C u^{+}$is:
(Given: $E_{C u^{2+} | C u^{+}}^{o}=0.15 V, E_{C u^{2+}|C u}^{o}=0.34 V$ )

• A. −0.19 V
• B. 0.38
• C. 0.94V
• D. −0.38 V

Answer 1

Answer: (B)

$E_{C u^{2+} / C u^{+}}^{o}=0.15\, V$
$E_{C u^{2+} / C u^{+}}^{o}=0.34 \,V$
$C u^{2+}+e^{-} \rightarrow C u^{+} ; E^{o}=+0.15 \,V$
Or $C u^{+} \rightarrow C u^{2+}+e^{-} ; E_{1}^{o}=-0.15\, V$
$\Delta G_{1}^{o}=-n F E_{1}^{o}=-1 \times F \times(-0.15)=+0.15 F$
$C u^{2+}+2 e^{-} \rightarrow C u ; E_{2}^{o}=0.34\, V$
$C u^{+}+e \rightarrow C u$
$E_{3}^{o}=?$
$\Delta G_{2}^{o}=-2 \times 0.34 \times F$
$\Delta G_{3}^{o}=-1 \times E_{3}^{o} \times F$
$\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}$
$-E_{3}^{o} F=+0.15 F-0.68 F$
$E_{3}^{o}=+0.53 \,V$
For disproportionation reaction of $C u^{+}$for half-cell reaction:
$C u^{+}+e^{-} \rightarrow C u ; E^{o}=+0.53\, V$
$C u^{+} \rightarrow C u^{2+}+\bar{e} ; E^{o}=-0.15\, V $
$\therefore 2 C u^{2+} \rightarrow C u^{2+}+C u ; E_{\text {cell }}^{o}=0.38\, V$