Question

The electrode potentials for $C{u^{2 + }}\left( {aq} \right) + {e^ - } \to C{u^ + }\left( {aq} \right)$and $C{u^ + }\left( {aq} \right) + {e^ - } \to Cu\left( s \right)$ are $+ 0.15\,V$and $+ 0.50\,V$respectively. The value of ${E^ \circ }_{C{u^{2 + }}/Cu}$ will be:
(i) $0.500\,V$
(ii) $0.325\,V$
(iii) $0.650\,V$
(iv) $0.150\,V$

Answer 1

Standard electrode potential $\left( {{E^ \circ }} \right)$ is defined as the measure of the individual potential of a reversible electrode at standard state with ions at an effective concentration of $1\,mol\,d{m^{ - 3}}$ and at the pressure of $1\,atm$. The standard Gibbs Free energy change for a reaction can be given by the formula $\Delta {G^ \circ } = - nF{E^ \circ }$. Add the given reactions in suitable order to get the final reaction and hence calculate ${E^ \circ }$.

Complete step-by-step answer: Standard electrode potential $\left( {{E^ \circ }} \right)$ is defined as the measure of the individual potential of a reversible electrode at standard state with ions at an effective concentration of $1\,mol\,d{m^{ - 3}}$ and at the pressure of $1\,atm$. It can be of two types standard oxidation potential and standard reduction potential. Generally electrode potentials are represented in terms of their reduction potentials. Hence ${E^ \circ }_{C{u^{2 + }}/Cu}$ corresponds to the reaction, $C{u^{2 + }}\left( {aq} \right) + 2{e^ - } \to Cu\left( s \right)$.
We must add the given reactions: $C{u^{2 + }}\left( {aq} \right) + {e^ - } \to C{u^ + }\left( {aq} \right)$and $C{u^ + }\left( {aq} \right) + {e^ - } \to Cu\left( s \right)$suitably to get the final equation.
Now, the standard Gibbs Free energy change for a reaction can be given by the formula $\Delta {G^ \circ } = - nF{E^ \circ }$where $n$is the number of change in electrons, $F$is Faraday and ${E^ \circ }$ is the standard electrode potential.
So, for the reaction $C{u^{2 + }}\left( {aq} \right) + {e^ - } \to C{u^ + }\left( {aq} \right).......\left( 1 \right),\,\Delta {G^ \circ }_1 = - {n_1}F{E^ \circ }_1$. Since the change in number of electrons is $1$ and the ${E^ \circ }$ value corresponding to the reaction is $+ 0.15\,V$. Therefore, $\Delta {G^ \circ }_1 = - 1 \times F \times 0.15\,V$.
For the reaction $C{u^ + }\left( {aq} \right) + {e^ - } \to Cu\left( s \right).......\left( 2 \right),\,\Delta {G^ \circ }_2 = - {n_2}F{E^ \circ }_2$. Since the change in number of electrons is $1$ and the ${E^ \circ }$ value corresponding to the reaction is $+ 0.50\,V$. Therefore, $\Delta {G^ \circ }_2 = - 1 \times F \times 0.50\,V$.
Now, adding equations $\left( 1 \right)$and $\left( 2 \right)$ we get, $C{u^{2 + }}\left( {aq} \right) + 2{e^ - } \to Cu\left( s \right).........\left( 3 \right),\,\Delta {G^ \circ }_3 = - {n_3}F{E^ \circ }_3$. Since the change in number of electrons is $2$. Therefore, $\Delta {G^ \circ }_3 = - 2 \times F \times {E^ \circ }_3\,$.
Since Gibbs Free Energy is an additive property and we obtain equation $\left( 3 \right)$by adding equations $\left( 1 \right)$and $\left( 2 \right)$. Therefore, $\Delta {G^ \circ }_3 = \Delta {G^ \circ }_1 + \Delta {G^ \circ }_2$.
Substituting the values we get,
- 2 \times F \times {E^ \circ }_3 = \left\\{ {\left( { - 1 \times F \times 0.15} \right) + \left( { - 1 \times F \times 0.5} \right)} \right\\}\,V
$\Rightarrow 2 \times {E^ \circ }_3 = \left( {0.15 + 0.5} \right)\,V\, = \,0.65\,V$
$\Rightarrow {E^ \circ }_3 = \,\dfrac{{0.65}}{2}\,V\, = \,0.325\,V$.
Hence${E^ \circ }_{C{u^{2 + }}/Cu} = \,0.325\,V$.

So the correct answer is (ii) $0.325\,V$.

Note: Always remember electrode potentials are generally represented in terms of the reduction process. You can do the question considering the oxidation potential but then the magnitude of the electrode potential will remain the same but the sign will be reversed.