Question

$C{u}^{+}$ ion is not stable in aqueous solution because of disproportionation reaction. $E^{\circ }$ value for disproportionation of $C{u}^{+}$ (given $E^{\circ }C{u}^2/Cu=0.34V)$

• A. $-0.19V$
• B. $0.49V$
• C. $-0.38V$
• D. $0.38V$

Answer 1

The correct option is D $0.38V$

$\begin{array}{c}\text{Cut ion is unstable in aqueous solution and readily}\\ \text{disproportionate to}{Cu}^{+2}\text{and}Cu\text{.}\end{array}$

$2c{u}^{+}⟶c{u}^{+2}+cu{E}_{cell}^0=?$

$\text{Given,}{E}_{{Cu}^{+2}/Cu}^{\circ }=0.34V\text{and}{E}_{{Cu}^{+2}/{Cu}^{+}}^{\circ }=0.15V\text{.}$

$\begin{array}{cc}{Cu}^{+2}+2e^{-}⟶Cu-\dots \left(i\right)& [E_{{Cu}^{+2}/Cu}^{\circ }=0.34v]\\ {Cu}^{+2}+e^{-}⟶C{u}^{2+}....\left(ii\right)& [E_{{Cu}^{+2}/{cu}^{+}}^0=+0\cdot 15]\end{array}$

$\text{Reverse reaction (ii) and add with (i):}$

$\begin{array}{cc}& {Cu}^{+2}+2e^{-}\to Cu{E}_{{Cu}^{+2}/Cu}^0=0.34V\\ & {Cu}^{+}⟶{cu}^{+2}+e^{-}{E}_{C{u}^{+}}^{\circ }/{Cu}^{+2}=-0.15v\\ & \therefore C{u}^{+}+e^{-}⟶E_{C{u}^{+}/C}^0=?\end{array}$

$\text{Here,}\Delta G_{\text{net}}=\Delta G_1+\Delta G_2(\because \Delta G=-nF{E}^{\circ })$

$\Rightarrow -1\times E_{C{u}^{2+}/Cu}^0=-2\times 0.34+1\times 0.15$

$\therefore E_{C{u}^{+}/Cu}=0.53V$

$\begin{array}{c}\text{Now,}{E}_{Cu}^0=E_{oxidation}^0+E_{Reduction}^0=E_{C{u}^{+}/C{u}^{2+}}^0\\ =-0.15V+0.53V\\ =0.38V\end{array}$