Consider the cell `AG|AgBr(s)|Br^(-)||AgCI(s)|CI^(-)|Ag` at `25^(@)C`. The solubility product constants of `AgBr & AgCI` are respectively `5 xx 10^(-13) & 1 xx 10^(-10)`. For what ratio of the concentration of `Br^(-) & CI^(-)` ions would the emf of the cell be zero?

Explanation

To solve the problem, we need to find the ratio of the concentrations of \( \text{Br}^- \) and \( \text{Cl}^- \) ions when the EMF of the cell is zero. We will use the Nernst equation and the solubility product constants (Ksp) for \( \text{AgBr} \) and \( \text{AgCl} \).

### Step-by-Step Solution:

1. **Identify the Cell Representation**:
The cell is represented as:
\[ \text{Ag} | \text{AgBr}(s) | \text{Br}^- || \text{AgCl}(s) | \text{Cl}^- | \text{Ag} \]

2. **Write the Nernst Equation**:
The Nernst equation for the cell can be expressed as:
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Ag}^+]_{\text{right}}}{[\text{Ag}^+]_{\text{left}}} \right) \]
where \( n \) is the number of electrons transferred in the cell reaction.

3. **Set EMF to Zero**:
We want to find the ratio when the EMF of the cell is zero:
\[ 0 = E^0_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Ag}^+]_{\text{right}}}{[\text{Ag}^+]_{\text{left}}} \right) \]
Since we are looking for the condition when \( E_{\text{cell}} = 0 \), we can rearrange this to:
\[ \frac{[\text{Ag}^+]_{\text{right}}}{[\text{Ag}^+]_{\text{left}}} = 10^{\frac{n E^0_{\text{cell}}}{0.0591}} \]

4. **Calculate \( [\text{Ag}^+] \) from Ksp**:
For \( \text{AgBr} \):
\[ K_{sp}(\text{AgBr}) = [\text{Ag}^+][\text{Br}^-] = 5 \times 10^{-13} \]
Therefore,
\[ [\text{Ag}^+]_{\text{left}} = \frac{5 \times 10^{-13}}{[\text{Br}^-]} \]

For \( \text{AgCl} \):
\[ K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-] = 1 \times 10^{-10} \]
Therefore,
\[ [\text{Ag}^+]_{\text{right}} = \frac{1 \times 10^{-10}}{[\text{Cl}^-]} \]

5. **Substitute into the Nernst Equation**:
Substitute the expressions for \( [\text{Ag}^+] \) into the Nernst equation:
\[ 0 = -\frac{0.0591}{n} \log \left( \frac{\frac{1 \times 10^{-10}}{[\text{Cl}^-]}}{\frac{5 \times 10^{-13}}{[\text{Br}^-]}} \right) \]

6. **Simplify the Logarithm**:
This simplifies to:
\[ 0 = -\frac{0.0591}{n} \log \left( \frac{1 \times 10^{-10} \cdot [\text{Br}^-]}{5 \times 10^{-13} \cdot [\text{Cl}^-]} \right) \]
Rearranging gives us:
\[ 1 \times 10^{-10} \cdot [\text{Br}^-] = 5 \times 10^{-13} \cdot [\text{Cl}^-] \]

7. **Find the Ratio**:
Rearranging gives us the ratio:
\[ \frac{[\text{Br}^-]}{[\text{Cl}^-]} = \frac{5 \times 10^{-13}}{1 \times 10^{-10}} = \frac{5}{1000} = \frac{1}{200} \]

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