Using the date in the preceding problem, calculate the equilibrium constant of the reaction at `25^(@)C`
`Zn +Cu^(++) hArr Zn^(++) +Cu, K ([Zn^(2+)])/([Cu^(2+)])`

Options

A. `8.314 xx 10^(24)`
B. `4.831 xx 10^(31)`
C. `8.314 xx 10^(36)`
D. `4.831 xx 10^(44)`

Explanation

To calculate the equilibrium constant \( K \) for the reaction \( \text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cu} \) at \( 25^\circ C \), we will follow these steps:

### Step 1: Identify the standard electrode potentials
The standard electrode potentials for the half-reactions are given:
- For Zinc: \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \)
- For Copper: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V} \)

### Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, Zinc is oxidized (anode) and Copper is reduced (cathode):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 \, \text{V} \]

### Step 3: Use the Nernst equation to relate \( E^\circ_{\text{cell}} \) to \( K \)
The Nernst equation at equilibrium (where \( E_{\text{cell}} = 0 \)) is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log K \]
At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log K \]
Rearranging gives:
\[ \frac{0.0592}{n} \log K = E^\circ_{\text{cell}} \]

### Step 4: Determine the number of electrons transferred \( n \)
In this reaction, Zinc loses 2 electrons and Copper gains 2 electrons, so:
\[ n = 2 \]

### Step 5: Substitute values into the equation
Substituting \( E^\circ_{\text{cell}} = 1.10 \, \text{V} \) and \( n = 2 \):
\[ \frac{0.0592}{2} \log K = 1.10 \]
This simplifies to:
\[ 0.0296 \log K = 1.10 \]
Now, solve for \( \log K \):
\[ \log K = \frac{1.10}{0.0296} \approx 37.16 \]

### Step 6: Calculate \( K \)
To find \( K \), we take the antilogarithm:
\[ K = 10^{37.16} \approx 1.45 \times 10^{37} \]

### Conclusion:
The equilibrium constant \( K \) for the reaction at \( 25^\circ C \) is approximately \( 8.31 \times 10^{36} \).

### Final Answer:
Thus, the correct option is:
**Option C: \( 8.314 \times 10^{36} \)**

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