Question

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

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

Answer 1

Answer: (B)

$E_{C{u}^{2+}/C{u}^{+}}^o=0.15V$
$E_{C{u}^{2+}/C{u}^{+}}^o=0.34V$
$C{u}^{2+}+e^{-}\to C{u}^{+};E^o=+0.15V$
Or $C{u}^{+}\to C{u}^{2+}+e^{-};E_1^o=-0.15V$
$\Delta G_1^o=-nF{E}_1^o=-1\times F\times (-0.15)=+0.15F$
$C{u}^{2+}+2e^{-}\to Cu;E_2^o=0.34V$
$C{u}^{+}+e\to Cu$
$E_3^o=?$
$\Delta G_2^o=-2\times 0.34\times F$
$\Delta G_3^o=-1\times E_3^o\times F$
$\Delta G_3^o=\Delta G_1^o+\Delta G_2^o$
$-E_3^{o}F=+0.15F-0.68F$
$E_3^o=+0.53V$
For disproportionation reaction of $C{u}^{+}$for half-cell reaction:
$C{u}^{+}+e^{-}\to Cu;E^o=+0.53V$
$C{u}^{+}\to C{u}^{2+}+\bar{e};E^o=-0.15V$
$\therefore 2C{u}^{2+}\to C{u}^{2+}+Cu;E_{\text{cell }}^o=0.38V$