`E^(@)` for the cell
`Zn(s)|Zn^(2+)(aq)|Cu^(2+)(aq)|Cu(s)` is 1.1V at `25^(@)C` the equilibrium constant for the cell reaction is about

Options

A. `10^(-37)`
B. `10^(37)`
C. `10^(-17)`
D. `10^(17)`

Answer

To find the equilibrium constant (K) for the cell reaction represented by the cell notation `Zn(s)|Zn^(2+)(aq)|Cu^(2+)(aq)|Cu(s)` with a standard cell potential (E°) of 1.1 V at 25°C, we can follow these steps:

### Step 1: Write the half-reactions
The half-reactions for the cell are:
- At the anode (oxidation):
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
- At the cathode (reduction):
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)} \]

### Step 2: Write the overall cell reaction
Combining the half-reactions, the overall cell reaction can be written as:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

### Step 3: Use the Nernst equation
The Nernst equation relates the standard cell potential to the reaction quotient (Q) and the equilibrium constant (K):
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where:
- \(E_{cell}\) is the cell potential at non-standard conditions,
- \(E^\circ_{cell}\) is the standard cell potential (1.1 V),
- \(n\) is the number of moles of electrons transferred (which is 2 in this case),
- \(Q\) is the reaction quotient.

### Step 4: Set up the equation at equilibrium
At equilibrium, \(E_{cell} = 0\) and \(Q = K\):
\[ 0 = E^\circ_{cell} - \frac{0.0591}{n} \log K \]

### Step 5: Rearrange the equation to solve for K
Rearranging the equation gives:
\[ \frac{0.0591}{n} \log K = E^\circ_{cell} \]
\[ \log K = \frac{n \cdot E^\circ_{cell}}{0.0591} \]
Substituting \(n = 2\) and \(E^\circ_{cell} = 1.1\) V:
\[ \log K = \frac{2 \cdot 1.1}{0.0591} \]

### Step 6: Calculate K
Calculating the right-hand side:
\[ \log K = \frac{2.2}{0.0591} \approx 37.24 \]
Now, converting from logarithmic form to exponential form:
\[ K = 10^{37.24} \approx 10^{37} \]

### Final Answer
Thus, the equilibrium constant \(K\) for the cell reaction is approximately \(10^{37}\).

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