`Ag_((s))|Ag_((aq))^(+)(0.01M)||Ag_((aq))^(+)(0.1M)|Ag_((s))E_(Ag_((s))//Ag_((aq)))^(@)=0.80`volt

Options

A. Cell cannot function as anode and cathode are of same material
B. `E_(cell)=0.0592V`
C. `E_(cell)=0.80V`
D. `E_(cell)=0.0296V`

Explanation

To solve the problem, we need to calculate the cell potential (E_cell) using the Nernst equation. The Nernst equation relates the cell potential to the standard electrode potential and the concentrations of the reactants and products involved in the electrochemical reaction.

### Step-by-Step Solution:

1. **Identify the components of the cell**:
- The cell is represented as:
\[ \text{Ag(s)} | \text{Ag}^+ (0.01M) || \text{Ag}^+ (0.1M) | \text{Ag(s)} \]
- Here, we have two silver electrodes with different concentrations of Ag⁺ ions.

2. **Standard Electrode Potential (E°)**:
- Given \( E^\circ = 0.80 \, \text{V} \).

3. **Identify concentrations**:
- Concentration of Ag⁺ at the anode (C1) = 0.01 M
- Concentration of Ag⁺ at the cathode (C2) = 0.1 M

4. **Determine the number of electrons transferred (n)**:
- For the reduction of Ag⁺ to Ag, n = 1 (since one electron is involved in the half-reaction).

5. **Apply the Nernst equation**:
- The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{C1}{C2} \right) \]
- Substituting the values:
\[ E = 0.80 - \frac{0.0591}{1} \log \left( \frac{0.01}{0.1} \right) \]

6. **Calculate the logarithmic term**:
- Calculate \( \frac{C1}{C2} = \frac{0.01}{0.1} = 0.1 \)
- Thus, \( \log(0.1) = -1 \).

7. **Substitute back into the Nernst equation**:
- Now substituting into the equation:
\[ E = 0.80 - 0.0591 \times (-1) \]
\[ E = 0.80 + 0.0591 \]
\[ E = 0.8591 \, \text{V} \]

8. **Final answer**:
- The cell potential \( E_{cell} \) is approximately \( 0.86 \, \text{V} \).