At what `[[Br]^(-)]/sqrt([CO_(3)^(2-)]]` does the following cell have its reaction at equilibrium?
`Ag(s) | Ag_(2)CO_(3)(s)| Na_(2)CO_(3)(aq)|| KBr(aq) | AgBr(s)| Ag(s)`
`K_z9sp) = 8 xx 10^(-12)` for Ag_(2)CO^(3)` and `K_(sp) = 4 xx 10^(-13)` for AgBr

Options

A. `sqrt1xx10^(-7)`
B. `sqrt2xx10^(-7)`
C. `sqrt3xx10^(-7)`
D. `sqrt4xx10^(-7)`

Explanation

To solve the problem, we need to find the ratio of \([Br^-]\) to \(\sqrt{[CO_3^{2-}]}\) at equilibrium for the given electrochemical cell. The cell reaction involves the solubility products (Ksp) of silver carbonate (\(Ag_2CO_3\)) and silver bromide (\(AgBr\)).

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The dissolution of \(Ag_2CO_3\) can be represented as:
\[ Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq) \]
- The dissolution of \(AgBr\) can be represented as:
\[ AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq) \]

2. **Write the Ksp Expressions**:
- For \(Ag_2CO_3\):
\[ K_{sp} = [Ag^+]^2[CO_3^{2-}] = 8 \times 10^{-12} \]
- For \(AgBr\):
\[ K_{sp} = [Ag^+][Br^-] = 4 \times 10^{-13} \]

3. **Set Up the Equilibrium Expression**:
- At equilibrium, we can express the concentrations of ions in terms of \(x\):
- Let \([Ag^+] = x\)
- From the \(Ag_2CO_3\) dissolution, \([CO_3^{2-}] = \frac{K_{sp}}{x^2}\)
- From the \(AgBr\) dissolution, \([Br^-] = \frac{K_{sp}}{x}\)

4. **Relate the Concentrations**:
- We need to find:
\[ \frac{[Br^-]}{\sqrt{[CO_3^{2-}]}} \]
- Substitute the expressions for \([Br^-]\) and \([CO_3^{2-}]\):
\[ \frac{[Br^-]}{\sqrt{[CO_3^{2-}]}} = \frac{\frac{K_{sp, AgBr}}{x}}{\sqrt{\frac{K_{sp, Ag_2CO_3}}{x^2}}} \]
- Simplifying this gives:
\[ = \frac{K_{sp, AgBr}}{x} \cdot \frac{x}{\sqrt{K_{sp, Ag_2CO_3}}} = \frac{K_{sp, AgBr}}{\sqrt{K_{sp, Ag_2CO_3}}} \]

5. **Substitute the Values**:
- Substitute the known \(K_{sp}\) values:
\[ = \frac{4 \times 10^{-13}}{\sqrt{8 \times 10^{-12}}} \]

6. **Calculate the Result**:
- First, calculate \(\sqrt{8 \times 10^{-12}}\):
\[ \sqrt{8} = 2\sqrt{2} \quad \text{and} \quad \sqrt{10^{-12}} = 10^{-6} \]
\[ \sqrt{8 \times 10^{-12}} = 2\sqrt{2} \times 10^{-6} \]
- Now substitute back:
\[ = \frac{4 \times 10^{-13}}{2\sqrt{2} \times 10^{-6}} = \frac{4}{2\sqrt{2}} \times 10^{-7} = \frac{2}{\sqrt{2}} \times 10^{-7} = \sqrt{2} \times 10^{-7} \]

### Final Answer:
\[ \frac{[Br^-]}{\sqrt{[CO_3^{2-}]}} = \sqrt{2} \times 10^{-7} \]