Consider the following reaction,
`Zn(s)+Cu^(2+) (0.1 M) rarr Zn^(2+) (1 M)+Cu(s)` above reaction, taking place in a cell, `E_("cell")^(@)` is `1.10 V. E_("cell")` for the cell will be `(2.303 (RT)/(F)=0.0591)`

Options

A. `1.07` volt
B. `0.82` volt
C. `2.14` volt
D. `1.80` volt

Explanation

To solve the problem, we will use the Nernst equation to calculate the cell potential \( E_{\text{cell}} \) for the given reaction. Here are the steps involved:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

### Step 2: Identify \( E^{\circ}_{\text{cell}} \) and \( n \)
From the question, we know:
- \( E^{\circ}_{\text{cell}} = 1.10 \, \text{V} \)
- The reaction involves the transfer of 2 electrons (as zinc is oxidized to \( \text{Zn}^{2+} \) and copper ions are reduced to copper metal).

### Step 3: Determine the Concentrations
From the reaction:
\[ \text{Zn(s)} + \text{Cu}^{2+} (0.1 \, \text{M}) \rightarrow \text{Zn}^{2+} (1 \, \text{M}) + \text{Cu(s)} \]
- The concentration of \( \text{Zn}^{2+} \) is \( 1 \, \text{M} \).
- The concentration of \( \text{Cu}^{2+} \) is \( 0.1 \, \text{M} \).

### Step 4: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
\[ E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]

### Step 5: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

### Step 6: Substitute the Logarithm Value
Now substitute this value back into the equation:
\[ E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \cdot 1 \]
\[ E_{\text{cell}} = 1.10 - 0.02955 \]

### Step 7: Final Calculation
Now perform the final calculation:
\[ E_{\text{cell}} = 1.10 - 0.02955 = 1.07045 \approx 1.07 \, \text{V} \]

### Conclusion
Thus, the cell potential \( E_{\text{cell}} \) is approximately \( 1.07 \, \text{V} \).