Question

For the given cell reaction \( E_{\text{cell}} = 1.1 \, \text{V} \), what is the equilibrium constant \( K_c \) for the reaction?

• A. \( K_c = 10^{-0.0591} \)
• B. \( K_c = 10^{0.0591} \)
• C. \( K_c = 10^{2.0591} \)
• D. \( K_c = 10^{2.059} \)

Hint:

To calculate the equilibrium constant from the cell potential, use the Nernst equation, and remember to account for the number of electrons involved.

Answer 1

The Correct Option is D

Step 1: Understanding the Relationship.
The relationship between the cell potential and the equilibrium constant \( K_c \) is given by the Nernst equation: \[ E_{\text{cell}} = \frac{0.0591}{n} \log K_c \] Where \( n \) is the number of electrons involved in the reaction.
Step 2: Calculation.
Given that \( E_{\text{cell}} = 1.1 \, \text{V} \) and \( n = 2 \), we can solve for \( K_c \): \[ 1.1 = \frac{0.0591}{2} \log K_c \] \[ \log K_c = \frac{2 \times 1.1}{0.0591} = 2.059 \] \[ K_c = 10^{2.059} \]
Step 3: Conclusion.
The correct answer is (D) \( K_c = 10^{2.059} \).