Question

The electrode potentials for
$C{u}^{2+}(aq)+e^{-}⟶C{u}^{+}(aq)$
and $C{u}^{+}(aq)+e^{-}⟶Cu(s)$
are $+0.15V$ and $+0.50V$ respectively. The value of $E_{C{u}^{2+}/Cu}^{\circ }$ will be

• A. 0.325 V
• B. 0.650 V
• C. 0.150 V
• D. 0.500 V

Answer 1

Answer: (A)

For the reaction $C{u}^{2+}(aq)+2e^{-}\to Cu(s)$, the number of electrons $n=2$.
For the reaction $C{u}^{2+}(aq)+e^{-}\to C{u}^{+}(aq)$, the number of electrons $n=1$.
For the reaction $C{u}^{+}(aq)+e^{-}\to Cu(s)$, the number of electrons $n=1$.
$n\times E_{C{u}^{2+}\mid Cu}^0=n\times E_{C{u}^{2+}\mid C{u}^{+}}^0+n\times E_{C{u}^{+}\mid Cu}^0$
$2\times E_{C{u}^{2+}\mid Cu}^0=1\times 0.15V+1\times 0.50V$
$2\times E_{C{u}^{2+}\mid Cu}^0=0.65V$
$E_{C{u}^2+\mid Cu}^0=0.325V$