Question

\( A_1^{n_1^+} + n_1 e^- \rightarrow A_1 \)
\( A_2 \rightarrow A_2^{n_2^+} + n_2 e^- \)

• A. Overall cell reaction:
  $n_2 A_1^{n_1^+} + n_1 A_2 \rightarrow n_2 A_1 + n_1 A_2^{n_2^+}$
• B. Electrical work done by the cell = charge $\times$ potential difference.
• C. \frac(E - E₀)RT log Q

  

• D. In the overall cell reaction, the number of electrons are not present, as electron liberated at anode, are consumed at cathode.

Hint:

The Nernst equation describes the relationship between the cell potential and the concentrations of the reacting species. The slope of a graph based on the Nernst equation depends on the number of electrons involved and the temperature.

Answer 1

The Correct Option is C

Step 1: Understanding the Cell Reaction.
In the given question, the overall cell reaction is the combination of half-reactions at the anode and cathode. The electrons are transferred between the two reactions, which represent the redox process in a cell.
Step 2: Analyzing the Electrical Work.
The electrical work done by the cell is indeed the charge multiplied by the potential difference, which is a fundamental concept in electrochemistry.
Step 3: Analysis of the Graph.
Option (C) shows a relationship between the cell potential and the reaction quotient, represented as a plot. This is not correctly depicting the Nernst equation, as the slope of such a plot should be the negative of $nF/RT$ and not in the current form.
Step 4: Electron Transfer Clarification.
In option (D), the statement about the electrons not being present in the overall cell reaction is incorrect. Electrons are transferred from anode to cathode, and they are indeed a part of the overall reaction. Final Answer: Option (C) is incorrect as it misrepresents the Nernst equation graph.