For a general redox reaction Anode : Red_1 arrow Ox_1^ n_1^+ + n_1...

For the given redox reaction, the half-cell reactions are:
Anode: $\text{Red}_1 \rightarrow \text{Ox}_1^{n_1^+} + n_1 e^-$
Cathode: $\text{Ox}_2 + n_2 e^- \rightarrow \text{Red}_2^{n_2^-}$

To obtain the overall balanced chemical equation, we multiply the anode reaction by $n_2$ and the cathode reaction by $n_1$ so that the number of electrons lost equals the number of electrons gained:
$n_2\text{Red}_1 \rightarrow n_2\text{Ox}_1^{n_1^+} + n_1 n_2 e^-$
$n_1\text{Ox}_2 + n_1 n_2 e^- \rightarrow n_1\text{Red}_2^{n_2^-}$

Adding these gives the overall reaction:
$n_2\text{Red}_1 + n_1\text{Ox}_2 \rightleftharpoons n_2\text{Ox}_1^{n_1^+} + n_1\text{Red}_2^{n_2^-}$
Thus, statement (A) is correct.

In the overall reaction, electrons do not appear because the electrons produced at the anode are exactly consumed at the cathode. Thus, statement (B) is correct.

From the Nernst equation:
$E = E^\circ - \dfrac{RT}{nF} \ln Q = E^\circ - \dfrac{2.303 RT}{nF} \log_{10} Q$
$\dfrac{E - E^\circ}{RT/F} = -\dfrac{2.303}{n} \log_{10} Q$
The graph of $\dfrac{E - E^\circ}{RT/F}$ versus $\log_{10} Q$ is a straight line with a slope of $-\dfrac{2.303}{n}$, where $n$ is the number of electrons transferred in the balanced redox reaction. Thus, statement (C) is correct.

The electrical work done by a galvanic cell is given by the product of the total charge transferred and the cell potential (potential difference), not their ratio.
$W_{\text{electrical}} = \text{Charge} \times \text{Potential difference} = nF E_{\text{cell}}$
Thus, statement (D) is incorrect.

Answer: 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.