The `EMF` of the cell `:`
`Ag|Ag_(2)CrO_(4)(s),K_(2)CrO_(4)(0.1 M)||AgNO_(3)(0.1M)||Ag` is `206.5mV.` Calculate the solubility of `Ag_(2)CrO_(4)` in `1M Na_(2)CrO_(4)` solution.

Explanation

To solve the problem, we need to calculate the solubility of \( \text{Ag}_2\text{CrO}_4 \) in a \( 1 \, \text{M} \, \text{Na}_2\text{CrO}_4 \) solution using the provided EMF of the cell. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Cell Reaction
The cell is a concentration cell involving silver ions. At the anode, silver is oxidized to silver ions, and at the cathode, silver ions are reduced to silver metal. The half-reactions can be written as:
- Anode: \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)
- Cathode: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

### Step 2: Write the Nernst Equation
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