The `E^(@)` at `25^(@)`C for the following reaction is 0.22 V. Calculate the equilibrium constant at `25^(@)`C :
`H_(2)(g)+2AgCl(s)to2Ag(s)+2HCl(aq)`

Options

A. `2.8xx10^(7)`
B. `5.2xx10^(8)`
C. `5.2xx10^(6)`
D. `5.2xx10^(3)`

Answer

To calculate the equilibrium constant (K) for the given reaction at 25°C, we can use the Nernst equation. The reaction is:

\[ H_2(g) + 2AgCl(s) \rightarrow 2Ag(s) + 2HCl(aq) \]

Given that the standard electrode potential (\(E^0\)) for this reaction is 0.22 V, we can follow these steps:

### Step 1: Identify the number of electrons transferred (n)
In the reaction, hydrogen is oxidized from 0 to +1 (2 electrons for 1 molecule of \(H_2\)), and silver ions are reduced from +1 to 0 (2 electrons for 2 \(AgCl\)). Therefore, the total number of electrons transferred in the reaction is:

\[ n = 2 \]

### Step 2: Use the Nernst equation
The Nernst equation relates the standard electrode potential to the equilibrium constant:

\[ E^0 = \frac{0.0591}{n} \log K \]

At equilibrium, \(E\) is 0, so we can rearrange the equation to solve for \(K\):

\[ E^0 = \frac{0.0591}{n} \log K \]

### Step 3: Substitute the known values into the equation
Substituting \(E^0 = 0.22\) V and \(n = 2\):

\[ 0.22 = \frac{0.0591}{2} \log K \]

### Step 4: Solve for \(\log K\)
Rearranging the equation gives:

\[ \log K = \frac{0.22 \times 2}{0.0591} \]

Calculating the right side:

\[ \log K = \frac{0.44}{0.0591} \approx 7.44 \]

### Step 5: Calculate \(K\) using the antilog
To find \(K\), we take the antilogarithm:

\[ K = 10^{7.44} \]

Calculating \(K\):

\[ K \approx 2.8 \times 10^7 \]

### Final Answer
The equilibrium constant \(K\) at 25°C for the reaction is approximately:

\[ K \approx 2.8 \times 10^7 \]

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