Excess of solid `AgCl` is added to a `0.1 M` solution of `Br^(c-)` ions. `E^(c-)` for half cell is `:`
`AgBr+e^(-) rarr Ag+Br^(c-),E^(c-)=0.095V`
`AgCl+e^(-) rarr Ag+Cl^(c-), E^(c-)=0.222V`
The value of `[Br^(c-)]` ion at equilibrium is `:`
`[` Given `: Antilog (2.152)=142]`

Options

A. `0.0317M`
B. `0.013M`
C. `0.99M`
D. `0.099M`

Explanation

To solve the problem, we need to determine the concentration of bromide ions \([Br^-]\) at equilibrium after adding excess solid AgCl to a 0.1 M solution of bromide ions.

### Step-by-Step Solution:

1. **Identify the Reaction**: When AgCl is added to the solution, it can react with bromide ions to form AgBr and Cl⁻ ions. The reaction can be represented as:
\[ \text{AgCl (s)} + \text{Br}^- \rightleftharpoons \text{AgBr (s)} + \text{Cl}^- \]

2. **Initial Concentrations**:
- Initial concentration of \([Br^-] = 0.1 \, M\)
- Initial concentration of \([Cl^-] = 0 \, M\) (since it is produced from the reaction)

3. **Change in Concentrations**:
- Let \(x\) be the amount of bromide ions that reacts.
- At equilibrium, the concentration of bromide ions will be \(0.1 - x\) and the concentration of chloride ions will be \(x\).

4. **Determine Cell Potential**:
- The cell potential \(E_{cell}\) can be calculated using the reduction potentials provided:
\[ E_{cell} = E_{cathode} - E_{anode} = E_{AgCl} - E_{AgBr} = 0.222 \, V - 0.095 \, V = 0.127 \, V \]

5. **Nernst Equation**:
- The Nernst equation relates the cell potential to the concentrations of the ions:
\[ E_{cell} = \frac{0.059}{n} \log \left( \frac{[Cl^-]}{[Br^-]} \right) \]
where \(n = 1\) (since one electron is involved in the reaction).

6. **Substituting Values**:
- Substitute \(E_{cell} = 0.127 \, V\) and the concentrations:
\[ 0.127 = \frac{0.059}{1} \log \left( \frac{x}{0.1 - x} \right) \]

7. **Solving for the Logarithm**:
- Rearranging gives:
\[ \log \left( \frac{x}{0.1 - x} \right) = \frac{0.127}{0.059} \approx 2.152 \]

8. **Exponentiating**:
- Using the property of logarithms:
\[ \frac{x}{0.1 - x} = 10^{2.152} \approx 142 \]

9. **Setting Up the Equation**:
- This leads to:
\[ x = 142(0.1 - x) \]
\[ x = 14.2 - 142x \]
\[ 143x = 14.2 \]
\[ x \approx 0.09929 \, M \]

10. **Finding \([Br^-]\) at Equilibrium**:
- Now, substituting \(x\) back to find \([Br^-]\):
\[ [Br^-] = 0.1 - x \approx 0.1 - 0.09929 \approx 0.00071 \, M \]

### Final Answer:
The concentration of bromide ions \([Br^-]\) at equilibrium is approximately \(0.00071 \, M\).