Question

\(H_{2(g)}+2AgCl_{(s)}⇌2Ag_{(s)}+2HCl_{(aq)}\)
\(E^{\degree}_{cell}\) at 25° C for the cell is 0.22 V. The equilibrium constant at 25° C is

• A. 2.8 × 10⁷
• B. 5.2 × 10⁸
• C. 2.8 × 10⁵
• D. 5.2 × 10⁴

Answer 1

The Correct Option is A

We are given the following reaction:
\(H_2 (g) + 2 \, \text{AgCl} (s) \rightleftharpoons 2 \, \text{Ag} (s) + 2 \, \text{HCl} (aq) \)

The standard cell potential \(( E^0_{\text{cell}})\) is given as 0.22 V.

We use the relationship between the cell potential and the equilibrium constant:
\(E^0_{\text{cell}} = \frac{0.0592}{n} \log K \)
Where:
\(E^0_{\text{cell}} = 0.22 \, \text{V}\)
\(n = 2\) (since 2 moles of \(\text{Ag}\) are involved),
\(K\) is the equilibrium constant.

Rearranging the equation to solve for \( K \):
\(K = 10^{\frac{n E^0_{\text{cell}}}{0.0592}} \)

Substituting the known values:
\(K = 10^{\frac{2 \times 0.22}{0.0592}} = 10^{7.44} \approx 2.8 \times 10^7 \)

Therefore, the equilibrium constant is:
Answer: A. \(2.8 \times 10^7\)