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

1. Identify the half-cell reactions and the number of electrons transferred
The given cell reaction is:

H_2(g) + 2AgCl_(s) ⇌ 2Ag_(s) + 2HCl_(aq)

We can break this into two half-cell reactions:

Oxidation (Anode): H_2(g) → 2H⁺_(aq) + 2e^-
Reduction (Cathode): 2AgCl_(s) + 2e^- → 2Ag_(s) + 2Cl^-_(aq)

From these half-cell reactions, it's clear that the number of electrons transferred (n) is 2.

2. Recall the relationship between $E^0_{cell}$ and the equilibrium constant (K)
The relationship between the standard cell potential ($E^0_{cell}$) and the equilibrium constant (K) is given by:

$E^0_{cell} = \frac{RT}{nF} \ln K$

where:
- $E^0_{cell}$ is the standard cell potential
- $R$ is the ideal gas constant (8.314 J/(mol·K))
- $T$ is the temperature in Kelvin
- $n$ is the number of moles of electrons transferred in the cell reaction
- $F$ is Faraday's constant (96485 C/mol)
- $K$ is the equilibrium constant

At 298 K, this equation can be simplified to:

$E^0_{cell} = \frac{0.0592}{n} \log K$

3. Substitute the given values and solve for K
We are given:

- $E^0_{cell} = 0.22$ V
- $n = 2$
- $T = 25^\circ C = 298 K$

Substitute these values into the simplified equation:

$0.22 = \frac{0.0592}{2} \log K$

Solve for $\log K$:

$\log K = \frac{0.22 \times 2}{0.0592} = \frac{0.44}{0.0592} \approx 7.4324$

Solve for K:

$K = 10^{7.4324} \approx 2.706 \times 10^7 \approx 2.8 \times 10^7$

Final Answer:
(A) 2.8 x 10⁷

Answer 2

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. \(2.8 \times 10^7\)