Incorrect Statement Regarding General Redox Reaction and Cell Work

To determine which statement is incorrect regarding the general redox reaction and electrochemical cell work, we analyze each option:

1. Overall Reaction (Option A):
   Multiplying the anode half-reaction by $n_2$ and the cathode half-reaction by $n_1$: $\text{Anode: } n_2 \text{Red}_1 \rightarrow n_2 \text{Ox}_1^{n_1^+} + n_1 n_2 e^-$ $\text{Cathode: } n_1 \text{Ox}_2 + n_1 n_2 e^- \rightarrow n_1 \text{Red}_2^{n_2^-}$ Adding both half-reactions yields the balanced overall cell reaction: $n_2 \text{Red}_1 + n_1 \text{Ox}_2 \rightleftharpoons n_2 \text{Ox}_1^{n_1^+} + n_1 \text{Red}_2^{n_2^-}$ Therefore, statement A is correct.

2. Electron Balance (Option B):
   The electrons generated at the anode are completely consumed at the cathode to maintain charge conservation, meaning no free electrons appear in the overall balanced reaction. Therefore, statement B is correct.

3. Number of Electrons $n$ (Option C):
   In the Nernst equation for the cell potential, $n = n_1 n_2$ represents the total number of moles of electrons transferred in the balanced redox equation. Therefore, statement C is correct.

4. Electrical Work Done (Option D):
   The maximum electrical work done ($W_{\text{elec}}$) by a reversible electrochemical cell is given by the product of the total charge ($q = nF$) and the potential difference ($E_{\text{cell}}$): $W_{\text{elec}} = q \cdot E_{\text{cell}}$ The statement claims that electrical work is equal to the ratio of charge and potential difference, which is mathematically and physically incorrect ($W \neq \frac{q}{E_{\text{cell}}}$).

Thus, statement D is the incorrect statement.

Correct Answer: D