Question

The electrode potentials for
$Cu ^{2+}(a q)+e^{-} \longrightarrow Cu ^{+}(a q)$
and $Cu ^{+}(a q)+e^{-} \longrightarrow Cu (s)$
are $+0.15 V$ and $+0.50 V$ respectively. The value of $E_{ Cu ^{2+} / Cu }^{\circ}$ 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 $Cu ^{2+}( aq )+2 e ^{-} \rightarrow Cu ( s )$, the number of electrons $n =2$.
For the reaction $Cu ^{2+}( aq )+ e ^{-} \rightarrow Cu ^{+}( aq )$, the number of electrons $n =1$.
For the reaction $Cu ^{+}( aq )+ e ^{-} \rightarrow Cu ( s )$, the number of electrons $n=1$.
$n \times E _{ Cu ^{2+} \mid Cu }^{0}= n \times E _{ Cu ^{2+} \mid Cu ^{+}}^{0}+ n \times E _{ Cu ^{+} \mid Cu }^{0}$
$2 \times E _{ Cu ^{2+} \mid Cu }^{0}=1 \times 0.15 V +1 \times 0.50\, V$
$2 \times E _{ Cu ^{2+} \mid Cu }^{0}=0.65\, V$
$E _{ Cu ^{2}+\mid Cu }^{0}=0.325 \,V$