The e.f.m. of the cell `Ag|Ag^(+)(0.1M)||Ag^(+)(1M)|Ag` at 298K is

Options

A. 0.0059V
B. 0.059V
C. 5.9V
D. 0.59V

Explanation

To find the electromotive force (e.m.f.) of the cell `Ag|Ag^(+)(0.1M)||Ag^(+)(1M)|Ag` at 298K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- At the anode (oxidation): \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)
- At the cathode (reduction): \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

2. **Determine the Number of Electrons Transferred (n)**:
- In this case, 1 electron is transferred in both half-reactions. Therefore, \( n = 1 \).

3. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \]
where \( E^\circ_{\text{cell}} \) is the standard cell potential, \( n \) is the number of moles of electrons transferred, and \( Q \) is the reaction quotient.

4. **Calculate the Standard Cell Potential (E°cell)**:
- Since both electrodes are the same metal (Ag), the standard cell potential \( E^\circ_{\text{cell}} = 0 \) V.

5. **Calculate the Reaction Quotient (Q)**:
- The reaction quotient \( Q \) is given by the ratio of the concentrations of the products to the reactants:
\[ Q = \frac{[\text{Ag}^+]_{\text{cathode}}}{[\text{Ag}^+]_{\text{anode}}} = \frac{1 \, \text{M}}{0.1 \, \text{M}} = 10 \]

6. **Substitute Values into the Nernst Equation**:
\[ E_{\text{cell}} = 0 - \frac{0.059}{1} \log(10) \]
- Since \( \log(10) = 1 \):
\[ E_{\text{cell}} = -0.059 \, \text{V} \]

7. **Final Calculation**:
- The e.m.f. of the cell is:
\[ E_{\text{cell}} = -0.059 \, \text{V} \]
- Since we are interested in the absolute value, we can state:
\[ E_{\text{cell}} = 0.059 \, \text{V} \]

### Conclusion:
The e.m.f. of the cell `Ag|Ag^(+)(0.1M)||Ag^(+)(1M)|Ag` at 298K is **0.059 V**.