Question

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^+$ ($M \to M^+ + e^-$). The cation $M^+$ is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?

• A. If $c_1$ is present at anode, then $c_1>c_2$.
• B. If $c_1$ is present at cathode, then $c_1>c_2$.
• C. If $c_1$ is present at anode, then $c_1 = c_2$.
• D. If $c_1$ is present at cathode, then $c_1<c_2$.

Hint:

For a concentration cell to produce positive potential, the concentration at the cathode must be higher than the concentration at the anode.

Answer 1

The Correct Option is D

To solve this problem, we need to understand the Nernst equation and how the concentration of ions affects the cell potential of an electrochemical cell. The Nernst equation is given by:

\(E = E^0 - \frac{RT}{nF} \ln \left(\frac{[M^+_{\text{cathode}}]}{[M^+_{\text{anode}}]}\right)\)

Where:

• \(E\) is the cell potential.
• \(E^0\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred.
• \(F\) is Faraday’s constant.
• \([M^+_{\text{cathode}}]\) and \([M^+_{\text{anode}}]\) are the concentrations of \(M^+\) ions at the cathode and anode, respectively.

For a positive cell potential, i.e., \(E > 0\), the concentration of ions at the cathode should be lower than the concentration of ions at the anode. This promotes the forward reaction (reduction at the cathode) to be more favorable.

Let's analyze the options:

• If \(c_1\) is present at anode, then \(c_1 > c_2\): This would not generate a positive cell potential because the concentration at the anode should ideally be higher.
• If \(c_1\) is present at cathode, then \(c_1 > c_2\): This will generate a negative cell potential because the concentration at the cathode should be lower than the anode, not higher.
• If \(c_1\) is present at anode, then \(c_1 = c_2\): This does not generate any potential difference (potential will be zero) and does not favor cell activity.
• If \(c_1\) is present at cathode, then \(c_1 < c_2\): This scenario supports the generation of a positive cell potential as it promotes a favorable forward reaction due to the concentration gradient. 

Therefore, the correct option is: If \(c_1\) is present at cathode, then \(c_1 < c_2\).