For a general redox reaction

Anode: $$\text{Red}_1 \to \text{Ox}_1^{n_1+} + n_1 e^-$$

Cathode: $$\text{Ox}_2 + n_2 e^- \to \text{Red}_2^{n_2-}$$

Which of the following statement is **incorrect** ?

(A)

Anode reaction: $$\mathrm{Red_1 \rightarrow Ox_1^{n_1+} + n_1e^-}$$

Cathode reaction: $$\mathrm{Ox_2 + n_2e^- \rightarrow Red_2^{n_2-}}$$

Multiplying the anode reaction by $$n_2$$ and the cathode reaction by $$n_1$$ and adding,

$$n_2\mathrm{Red_1}+n_1\mathrm{Ox_2}\rightarrow n_2\mathrm{Ox_1^{n_1+}}+n_1\mathrm{Red_2^{n_2-}}$$

Hence, statement (A) is correct.

(B)

In an overall balanced redox reaction, electrons released during oxidation are completely consumed during reduction. Therefore, free electrons do not appear in the overall cell reaction.

Hence, statement (B) is correct.

(C)

The Nernst equation is $$E=E^\circ-\frac{2.303RT}{nF}\log Q$$

Rearranging,

$$\frac{E-E^\circ}{RT/F}=-\frac{2.303}{n}\log Q$$

This represents a straight line passing through the origin with slope $$-\frac{2.303}{n}$$

Since the slope is inversely proportional to $$n$$, the graph shown is correct.

Hence, statement (C) is correct.

(D)

The maximum electrical work obtained from a cell is $$W_{\max}=qE=nFE_{\text{cell}}$$

Electrical work is equal to the product of charge and potential difference, not their ratio.

Hence, statement (D) is incorrect.

Therefore, the incorrect statement is option (D).