The EMF of cell
`Ag | AgCl, 0.05 M KCl || 0.05 M AgNO_(3) |Ag` is 0.788 Volt.
Find the solubility product of AgCl`.

Explanation

To find the solubility product (Ksp) of AgCl from the given EMF of the cell, we can follow these steps:

### Step 1: Write the cell reaction and identify the components
The cell is represented as:
\[ \text{Ag | AgCl, 0.05 M KCl || 0.05 M AgNO}_3 \text{| Ag} \]

In this cell:
- The anode reaction involves the oxidation of silver (Ag) to silver ions (Ag⁺).
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