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

Explanation

To solve the problem, we need to determine the ratio of the concentrations of Br⁻ and Cl⁻ ions at which the EMF of the cell becomes zero. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Cell Setup
The cell consists of two half-cells:
1. Left half-cell: Ag(s) | AgBr(s) | Br⁻(aq)
2. Right half-cell: AgCl(s) | Cl⁻(aq) | Ag(s)

### Step 2: Write the Nernst Equation
The Nernst equation for the cell potential (E_cell) is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \left( \frac{[Ag^+]_{left}}{[Ag^+]_{right}} \right) \]
Where:
- \(E^0_{cell}\) is the standard cell potential.
- \(n\) is the number of moles of electrons transferred (which is 1 for Ag to Ag⁺ conversion).

### Step 3: Set EMF to Zero
To find the ratio at which the EMF is zero:
\[ 0 = E^0_{cell} - \frac{0.0591}{n} \log \left( \frac{[Ag^+]_{left}}{[Ag^+]_{right}} \right) \]
This implies:
\[ \frac{[Ag^+]_{left}}{[Ag^+]_{right}} = 1 \]
Thus, the concentrations of Ag⁺ on both sides are equal.

### Step 4: Relate Ksp and Ion Concentrations
From the solubility product (Ksp) expressions:
- For AgBr: \( K_{sp} = [Ag^+][Br^-] \)
- For AgCl: \( K_{sp} = [Ag^+][Cl^-] \)

Since \([Ag^+]_{left} = [Ag^+]_{right}\), we can denote this common concentration as \([Ag^+]\).

### Step 5: Set Up the Equations
Using the Ksp expressions:
\[ K_{sp, AgBr} = [Ag^+][Br^-] \quad \text{and} \quad K_{sp, AgCl} = [Ag^+][Cl^-] \]
We can rearrange these to find:
\[ [Br^-] = \frac{K_{sp, AgBr}}{[Ag^+]} \quad \text{and} \quad [Cl^-] = \frac{K_{sp, AgCl}}{[Ag^+]} \]

### Step 6: Find the Ratio of Ion Concentrations
Now, we can find the ratio of \([Cl^-]\) to \([Br^-]\):
\[ \frac{[Cl^-]}{[Br^-]} = \frac{K_{sp, AgCl}}{K_{sp, AgBr}} \]

### Step 7: Substitute the Ksp Values
Given:
- \( K_{sp, AgBr} = 5 \times 10^{-13} \)
- \( K_{sp, AgCl} = 1 \times 10^{-10} \)

Substituting these values:
\[ \frac{[Cl^-]}{[Br^-]} = \frac{1 \times 10^{-10}}{5 \times 10^{-13}} = \frac{1}{0.005} = 200 \]

### Step 8: Find the Required Ratio
To find the ratio of \([Br^-]\) to \([Cl^-]\):
\[ \frac{[Br^-]}{[Cl^-]} = \frac{1}{200} \]

### Final Answer
Thus, the ratio of \([Br^-]\) to \([Cl^-]\) at which the EMF of the cell is zero is:
\[ 1 : 200 \]

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