Question

If for the cell reaction,

\[ \mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{A}_{(\mathrm{aq})}^{+}+\mathrm{B}_{(\mathrm{s})} \]

if \(\mathrm{E}_{\text {cell }}^{\circ}=0.0592 \mathrm{~V}\). What is equilibrium constant of reaction at 298 K ?

• A. 1
• B. 10
• C. 100
• D. 1000

Hint:

Use the relation between standard cell potential and the equilibrium constant at 298 K with n equal to 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At 298 K, \(E^{\circ}_{cell} = \dfrac{0.0592}{n}\log K\).
In the reaction \(\text{A} + \text{B}^{+} \rightarrow \text{A}^{+} + \text{B}\), one electron is transferred, so \(n = 1\).

Step 2: Calculation
\(0.0592 = \dfrac{0.0592}{1}\log K\)
\(\log K = 1\).

Step 3: Compare the options
\(K = 10^{1} = 10\).

Final Answer:
The equilibrium constant is 10. \[ \boxed{\text{(B) }10} \]