Consider the cell `Ag(s)|AgBr(s)Br^(c-)(aq)||AgCl(s),Cl^(c-)(aq)|Ag(s)` at `298 K`. The `K_(sp)` of `AgBr` and `AgCl`, respectively are `5xx10^(-13)` and `1xx10^(-10)` . At what ratio of `[Br^(c-)]` and `[Cl^(c-)]` ions, `EMF_(cell)` would be zero ?

Options

A. `1:200`
B. `1:100`
C. `1:500`
D. `200:1`

Explanation

To solve the problem, we need to find the ratio of \([Br^-]\) to \([Cl^-]\) ions at which the EMF of the cell becomes zero. We will use the solubility product constants (\(K_{sp}\)) of AgBr and AgCl, and the Nernst equation for this calculation.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell is represented as:
\[ Ag(s) | AgBr(s) | Br^-(aq) || AgCl(s) | Cl^-(aq) | Ag(s) \]
The reactions occurring at each electrode are:
- Left side: \(AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)\)
- Right side: \(Ag^+(aq) + Cl^-(aq) \rightleftharpoons AgCl(s)\)

2. **Write the Expression for \(K_{sp}\)**:
For AgBr:
\[ K_{sp} = [Ag^+][Br^-] = 5 \times 10^{-13} \]
For AgCl:
\[ K_{sp} = [Ag^+][Cl^-] = 1 \times 10^{-10} \]

3. **Express \([Br^-]\) and \([Cl^-]\)** in terms of \([Ag^+]\):
From the \(K_{sp}\) expressions, we can rearrange to find:
- For AgBr:
\[ [Br^-] = \frac{K_{sp}^{AgBr}}{[Ag^+]} = \frac{5 \times 10^{-13}}{[Ag^+]} \]
- For AgCl:
\[ [Cl^-] = \frac{K_{sp}^{AgCl}}{[Ag^+]} = \frac{1 \times 10^{-10}}{[Ag^+]} \]

4. **Set Up the Nernst Equation**:
The Nernst equation for the cell can be expressed as:
\[ E_{cell} = E^\circ - \frac{0.0591}{n} \log \left(\frac{[Ag^+]_{RHS}}{[Ag^+]_{LHS}}\right) \]
Since we want \(E_{cell} = 0\), we can set the equation to zero:
\[ 0 = E^\circ - \frac{0.0591}{1} \log \left(\frac{[Ag^+]_{RHS}}{[Ag^+]_{LHS}}\right) \]

5. **Substituting \([Ag^+]\)**:
From the earlier expressions for \([Br^-]\) and \([Cl^-]\):
\[ [Ag^+]_{LHS} = \frac{5 \times 10^{-13}}{[Br^-]}, \quad [Ag^+]_{RHS} = \frac{1 \times 10^{-10}}{[Cl^-]} \]
Substitute these into the Nernst equation:
\[ 0 = E^\circ - 0.0591 \log \left(\frac{\frac{1 \times 10^{-10}}{[Cl^-]}}{\frac{5 \times 10^{-13}}{[Br^-]}}\right) \]

6. **Simplifying the Logarithmic Expression**:
\[ 0 = E^\circ - 0.0591 \log \left(\frac{1 \times 10^{-10} \cdot [Br^-]}{5 \times 10^{-13} \cdot [Cl^-]}\right) \]

7. **Setting the Logarithm to Zero**:
For the EMF to be zero, the logarithmic term must equal zero:
\[ \frac{1 \times 10^{-10} \cdot [Br^-]}{5 \times 10^{-13} \cdot [Cl^-]} = 1 \]

8. **Solving for the Ratio**:
Rearranging gives:
\[ [Br^-] = \frac{5 \times 10^{-13}}{1 \times 10^{-10}} \cdot [Cl^-] \]
\[ [Br^-] = 5 \times 10^{-3} \cdot [Cl^-] \]
Thus, the ratio of \([Br^-]\) to \([Cl^-]\) is:
\[ \frac{[Br^-]}{[Cl^-]} = \frac{1}{200} \]

### Final Answer:
The ratio of \([Br^-]\) to \([Cl^-]\) ions at which the EMF of the cell would be zero is:
\[ \text{Ratio} = 1:200 \]