Question

For a certain redox reaction in galvanic cell \(\text{X}_{(s)}+\text{Y}_{(aq)}^{2+}⟶\text{X}_{(aq)}^{2+}+\text{Y}_{(s)}\), \(\text{E}^{\circ}\) cell is \(0.0296 \text{V}\) at \(298 \text{K}\). What is equilibrium constant of reaction?

• A. \(1\)
• B. \(10\)
• C. \(100\)
• D. \(1000\)

Hint:

Use E0 = (0.0592/n) log K with n = 2 for this reaction.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
At equilibrium the cell potential is zero, which links the standard cell potential to the equilibrium constant: \(E^{\circ}_{cell} = \frac{0.0592}{n}\log K\) at \(298\) K.

Step 2: Key Formula or Approach:
The reaction \(\text{X} + \text{Y}^{2+} \to \text{X}^{2+} + \text{Y}\) transfers \(n = 2\) electrons.

Step 3: Detailed Explanation:
\[ \log K = \frac{n E^{\circ}}{0.0592} = \frac{2 \times 0.0296}{0.0592} = 1 \]
\[ K = 10^{1} = 10 \]
A value of \(K = 1\) would need \(E^{\circ} = 0\), and \(K = 100\) or \(1000\) would need \(E^{\circ}\) of \(0.0592\) V or \(0.0888\) V.

Final Answer:
\(K = 10\), option (B). \[ \boxed{10} \]