For the galvanic cell: `Ag|AgCI(s)|KCI(0.2M)||KBr(0.001M)|AgBr(s)|Ag`, calculate the EMF generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at `25^(@)C`.
`[K_(sp(AgCI)) = 2.8 xx 10^(-10), K_(sp(AgBr)) = 3.3 xx 10^(-13)]`

Explanation

To calculate the EMF generated for the galvanic cell represented as `Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag`, we will follow these steps:

### Step 1: Write the solubility product expressions for AgCl and AgBr.
The solubility product (Ksp) expressions are given as:
- For AgCl:
\[ K_{sp(AgCl)} = [Ag^+][Cl^-] \]
- For AgBr:
\[ K_{sp(AgBr)} = [Ag^+][Br^-] \]

### Step 2: Substitute the known values into the Ksp expressions.
Given:
- \( K_{sp(AgCl)} = 2.8 \times 10^{-10} \)
- \( [Cl^-] = 0.2 \, M \)

From the Ksp expression for AgCl:
\[ 2.8 \times 10^{-10} = [Ag^+][0.2] \]
Solving for \([Ag^+]\):
\[ [Ag^+] = \frac{2.8 \times 10^{-10}}{0.2} = 1.4 \times 10^{-9} \, M \]

Now for AgBr:
- \( K_{sp(AgBr)} = 3.3 \times 10^{-13} \)
- \( [Br^-] = 0.001 \, M \)

From the Ksp expression for AgBr:
\[ 3.3 \times 10^{-13} = [Ag^+][0.001] \]
Solving for \([Ag^+]\):
\[ [Ag^+] = \frac{3.3 \times 10^{-13}}{0.001} = 3.3 \times 10^{-10} \, M \]

### Step 3: Use the Nernst equation to calculate the EMF.
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]
At 25°C (298 K), we can simplify it to:
\[ E_{cell} = E^0_{cell} - 0.0591/n \log Q \]
Where:
- \( n = 1 \) (one electron transfer)
- \( Q = \frac{[Ag^+]_{AgBr}}{[Ag^+]_{AgCl}} = \frac{3.3 \times 10^{-10}}{1.4 \times 10^{-9}} \)

Calculating \( Q \):
\[ Q = \frac{3.3 \times 10^{-10}}{1.4 \times 10^{-9}} = 0.2357 \]

### Step 4: Calculate the EMF.
Substituting into the Nernst equation:
\[ E_{cell} = 0 - \frac{0.0591}{1} \log(0.2357) \]
Calculating \( \log(0.2357) \):
\[ \log(0.2357) \approx -0.627 \]
Now substituting back:
\[ E_{cell} = -0.0591 \times (-0.627) = 0.0371 \, V \]

### Step 5: Assign polarity to the electrodes.
Since the calculated EMF is positive, the cell reaction is spontaneous. The electrode where reduction occurs (AgBr) is the cathode, and the electrode where oxidation occurs (AgCl) is the anode.

### Final Answer:
- EMF (E_cell) = 0.0371 V
- Anode: Ag|AgCl (oxidation)
- Cathode: AgBr|Ag (reduction)