Zn(S) + Cl_2 (1 atm) → Zn^2+ + 2Cl^– . The E° of the cell is 2.12 V.…

The correct answer is option 2. Zn²⁺ concentration should be decreased.

Let us go through the detailed explanation step-by-step to understand why decreasing the concentration of \( \text{Zn}^{2+} \) increases the cell potential (\( E \)).

Given Reaction and Nernst Equation

The given reaction is:

\(\text{Zn}(\text{s}) + \text{Cl}_2(\text{g}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2 \text{Cl}^-(\text{aq})\)

The standard cell potential (\( E^\circ \)) for this reaction is 2.12 V.

The Nernst equation for this reaction is:

\(E = E^\circ - \frac{0.0592}{2} \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}}\)

Breakdown of the Nernst Equation

\(E = E^\circ - \frac{0.0592}{2} \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \)

Here:

\( E \) is the cell potential.

\( E^\circ \) is the standard cell potential (2.12 V).

\( n \) is the number of electrons transferred in the reaction (2 electrons).

\( [\text{Zn}^{2+}] \) is the concentration of zinc ions in solution.

\( [\text{Cl}^-] \) is the concentration of chloride ions in solution.

\( P_{\text{Cl}_2} \) is the partial pressure of chlorine gas.

Understanding the Impact of Each Component

The Nernst equation shows that the cell potential (\( E \)) depends on the ratio of the concentrations of the products to the reactants:

\(E = 2.12 - \frac{0.0592}{2} \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}}\)

Analyzing the Options to Increase \( E \)

To increase \( E \), we need to understand how each variable affects the cell potential:

1. Increasing \( [\text{Zn}^{2+}] \):

Increasing \( [\text{Zn}^{2+}] \) would increase the numerator of the logarithmic term. A larger numerator would result in a larger value inside the logarithm, which would make \( \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \) larger. This would result in a larger value being subtracted from \( E^\circ \), decreasing \( E \). Therefore, increasing \( [\text{Zn}^{2+}] \) decreases \( E \).

2. Decreasing \( [\text{Zn}^{2+}] \):

Decreasing \( [\text{Zn}^{2+}] \) would decrease the numerator of the logarithmic term. A smaller numerator would result in a smaller value inside the logarithm, which would make \( \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \) smaller. This would result in a smaller value being subtracted from \( E^\circ \), increasing \( E \). Therefore, decreasing \( [\text{Zn}^{2+}] \) increases \( E \).

3. Increasing \( [\text{Cl}^-] \):

Increasing \( [\text{Cl}^-] \) would increase the numerator of the logarithmic term significantly since it's squared. A larger numerator would result in a larger value inside the logarithm, which would make \( \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \) larger. This would result in a larger value being subtracted from \( E^\circ \), decreasing \( E \). Therefore, increasing \( [\text{Cl}^-] \) decreases \( E \).

4. Decreasing \( P_{\text{Cl}_2} \):

Decreasing \( P_{\text{Cl}_2} \) would increase the denominator of the logarithmic term.  A smaller denominator would result in a larger value inside the logarithm, which would make \( \log \frac{[\text{Zn}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \) larger. This would result in a larger value being subtracted from \( E^\circ \), decreasing \( E \). Therefore, decreasing \( P_{\text{Cl}_2} \) decreases \( E \).

Conclusion

Given the analysis, to increase the cell potential (\( E \)): Decreasing the concentration of \( \text{Zn}^{2+} \) is the correct approach as it decreases the value inside the logarithm, resulting in a smaller term being subtracted from \( E^\circ \) and thereby increasing \( E \).

Therefore, the correct answer is: \(Zn^{2+}\) concentration should be decreased.