Question

Given the reaction: \[ \text{Cu(s)} + 2\text{Ag}^+_{(aq)} \rightleftharpoons \text{Cu}^{2+}_{(aq)} + 2\text{Ag(s)} \quad (\text{E}_{\text{cell}}^\circ = 0.295 \, \text{V}, \, 2.303 \, \frac{RT}{F} = 0.059 \, \text{V}) \] What is the value of the equilibrium constant \( K \)?

• A. \( 10^{20} \)
• B. \( 10^{15} \)
• C. \( 10^{10} \)
• D. \( 10^{-1} \)
• E. \( 10^{-2} \)

Hint:

The equilibrium constant \( K \) can be calculated using the Nernst equation, which relates the cell potential to the equilibrium constant for a given reaction.

Answer 1

The Correct Option is C

To find the equilibrium constant \( K \), we use the Nernst equation in its logarithmic form: \[ \text{E}_{\text{cell}}^\circ = \frac{0.059}{n} \log K \] Here: - \( \text{E}_{\text{cell}}^\circ = 0.295 \, \text{V} \) - \( n = 2 \) (because there are 2 electrons involved in the reaction) Substitute the given values into the equation: \[ 0.295 = \frac{0.059}{2} \log K \] Solve for \( \log K \): \[ \log K = \frac{0.295 \times 2}{0.059} = 10 \] Thus, \( K = 10^{10} \). So, the correct answer is (C) \( 10^{10} \).