Question

For the disproportionation of copper ${\text{2C}}{{\text{u}}^{\text{ + }}} \to {\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{ + Cu}}$, ${{\text{E}}^{\text{0}}}$ for ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/Cu }}$ is $0.34{\text{V}}$ and ${{\text{E}}^{\text{0}}}$ for ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/C}}{{\text{u}}^{\text{ + }}}$ is $0.15{\text{V}}$
A. ${\text{0}}{\text{.49 V}}$
B. ${\text{ - 0}}{\text{.19V}}$
C. $0.38{\text{V}}$
D. $- 0.38{\text{V}}$

Answer 1

To solve this question, knowledge on electrochemical cells is required. A cell reaction will take place only when it is spontaneous in nature and releases Gibbs free energy in the process. We shall put the appropriate values in the equation given below to find the change in potential of the cell.

Formula used: ${{\Delta G = }} - {{nF\Delta E}}$
Where $\Delta {\text{G}}$ is the Gibbs free energy $\Delta E$ is the change in cell potential, F is the faraday constant, n is the number of electrons transferred.

Complete step by step answer:
Given, ${{\text{E}}^{\text{0}}}$ for ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/Cu }}$ is $0.34{\text{V}}$ while ${{\text{E}}^{\text{0}}}$ for ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/C}}{{\text{u}}^{\text{ + }}}$ is $0.15{\text{V}}$
Net cell reaction is:
${\text{C}}{{\text{u}}^{\text{ + }}}{\text{ + }}{{\text{e}}^ - } \to {\text{Cu}}$
To calculate the value of ${{\text{E}}^{\text{0}}}$for ${\text{C}}{{\text{u}}^ + }{\text{/Cu}}$,
$- {\text{nF}}{{\text{E}}_{{\text{C}}{{\text{u}}^ + }{\text{/Cu}}}}{\text{ = }}\left[ {\left( { - {\text{nF}}{{\text{E}}_{{\text{C}}{{\text{u}}^{2 + }}/{\text{Cu}}}}} \right){\text{ + }}\left( { - {\text{nF}}{{\text{E}}_{{\text{C}}{{\text{u}}^ + }/{\text{C}}{{\text{u}}^{2 + }}}}} \right)} \right]$
Substituting the values we get:
$\Rightarrow - 1 \times {\text{F}} \times {{\text{E}}_{{\text{C}}{{\text{u}}^ + }{\text{/Cu}}}} = \left[ {\left( { - 2 \times 0.34} \right) + \left( { - 1} \right) \times \left( { - 0.15} \right)} \right]$
Therefore, ${{\text{E}}^{\text{0}}}$for ${\text{C}}{{\text{u}}^ + }{\text{/Cu}}$ = $0.53{\text{V}}$
The reactions occurring at the cell are:
${\text{C}}{{\text{u}}^{\text{ + }}} \to {\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{ + e}}$ ${{\text{E}}^{\text{0}}}$for ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{/C}}{{\text{u}}^{\text{ + }}}$is $- 0.15{\text{V}}$
${\text{C}}{{\text{u}}^{\text{ + }}}{\text{ + e}} \to {\text{Cu}}$ ${{\text{E}}^{\text{0}}}$for ${\text{C}}{{\text{u}}^ + }{\text{/Cu}}$= $0.53{\text{V}}$
Total reaction is: $2{\text{C}}{{\text{u}}^{\text{ + }}} \to {\text{C}}{{\text{u}}^{{\text{ + 2}}}}{\text{ + Cu}}$
${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}_{{\text{C}}{{\text{u}}^ + }{\text{/C}}{{\text{u}}^{2 + }}}}{\text{ + }}{{\text{E}}_{{\text{C}}{{\text{u}}^ + }/{\text{Cu}}}}$ = $- 0.15{\text{V}} + 0.53{\text{V}} = 0.38{\text{V}}$
Hence, the potential change for the comproportionation reaction of copper is $0.38{\text{V}}$.

Hence, option C is the correct answer.

Note:
A comproportionation reaction is defined as that reaction in which two reactants each containing the same element but in different oxidation states, form a product in which the elements involved reach the same oxidation number. This is just the opposite of a disproportionation reaction.
In the disproportionation reaction, one compound of intermediate oxidation state converts to two compounds, one with higher oxidation state and the other with lower oxidation state.
Examples of comproportionation reaction include:
Reactions occurring in the lead batteries and the laboratory preparation of manganese oxide.