Calculate equilibrium constant for the reaction :
`Zn+Cd^(2+) hArr Zn^(2+)+Cd`,
(Given `E_(cell)^(@)=0.36 "V "`)

Explanation

To calculate the equilibrium constant (K) for the reaction:

\[ \text{Zn} + \text{Cd}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cd} \]

given that the standard cell potential \( E^\circ_{\text{cell}} = 0.36 \, \text{V} \), we can use the relationship between the standard cell potential and the equilibrium constant:

### Step 1: Use the Nernst Equation
The Nernst equation relating the standard cell potential to the equilibrium constant is given by:

\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K \]

where:
- \( E^\circ_{\text{cell}} \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the balanced equation,
- \( K \) is the equilibrium constant.

### Step 2: Identify the Number of Electrons Transferred
In the given reaction, zinc (Zn) is oxidized to zinc ions (\( \text{Zn}^{2+} \)), and cadmium ions (\( \text{Cd}^{2+} \)) are reduced to cadmium (Cd). Each zinc atom loses 2 electrons, and since one cadmium ion gains 2 electrons, the total number of electrons transferred \( n \) is 2.

### Step 3: Rearrange the Equation to Solve for \( \log K \)
Now, rearranging the Nernst equation to solve for \( \log K \):

\[ \log K = \frac{E^\circ_{\text{cell}} \cdot n}{0.0591} \]

### Step 4: Substitute the Values
Substituting the values into the equation:

\[ \log K = \frac{0.36 \, \text{V} \cdot 2}{0.0591} \]

### Step 5: Calculate \( \log K \)
Calculating \( \log K \):

\[ \log K = \frac{0.72}{0.0591} \approx 12.182 \]

### Step 6: Calculate \( K \)
To find \( K \), we take the antilog of \( \log K \):

\[ K = 10^{12.182} \approx 1.596 \times 10^{12} \]

### Final Answer
Thus, the equilibrium constant \( K \) for the reaction is approximately:

\[ K \approx 1.596 \times 10^{12} \]

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