For a general redox reaction}
\[
\text{Anode: } \text{Red}_1 \rightarrow \text{Ox}_1^{n_1+} + n_1e^-
\]
\[
\text{Cathode: } \text{Ox}_2 + n_2e^- \rightarrow \text{Red}_2^{n_2-}
\]
Which of the following statement is incorrect?
The overall reaction can be written as \( n_2\text{Red}_1 + n_1\text{Ox}_2 \rightarrow n_2\text{Ox}_1^{n_1+} + n_1\text{Red}_2^{n_2-} \)
The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
In the Nernst equation plot, slope \( \propto \frac{1}{n} \), where \(n\) is number of electrons transferred in the redox reaction.
If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.
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The Correct Option isD
Solution and Explanation
Concept:
Electrochemical cell reactions follow electron transfer balance and thermodynamic relations.
Step 1: {Check option (A)}
To balance electrons:
\[
n_2(\text{Red}_1 \rightarrow \text{Ox}_1^{n_1+}+n_1e^-)
\]
\[
n_1(\text{Ox}_2+n_2e^- \rightarrow \text{Red}_2^{n_2-})
\]
Adding both reactions:
\[
n_2\text{Red}_1+n_1\text{Ox}_2 \rightarrow n_2\text{Ox}_1^{n_1+}+n_1\text{Red}_2^{n_2-}
\]
Thus statement (A) is correct.
Step 2: {Check option (B)}
Electrons cancel when half reactions are added, hence they do not appear in overall reaction.
Thus (B) is correct.
Step 3: {Check option (C)}
Nernst equation:
\[
E = E^\circ - \frac{0.0591}{n}\log Q
\]
Thus slope of the graph between \(E\) and \(\log Q\) is proportional to \( \frac{1}{n} \).
Thus (C) is correct.
Step 4: {Check option (D)}
Electrical work done:
\[
w = qV
\]
Work equals product of charge and potential difference, not their ratio.
Thus statement (D) is incorrect.
