For the cell
`Cu (s) | Cu ^(2+) (aq) (0.1 M) || Ag ^(+) (aq) (0.01 M) | Ag (s)`
the cell potential `E _(1) = 0.3095 V`
For the cell
`Cu (s) | Cu ^(+2) (aq) (0.01 M) || Ag ^(+) (aq)( 0. 001 M) | Ag (s)`
the cell potential = _______ `xx 10 ^(-2) V.`
(Round off the Nearest Integer).
[Use : `(2.303 RT)/(F) =0.059`]

Answer

To find the cell potential for the second cell given, we will use the Nernst equation. Let's break down the steps:

### Step 1: Understand the Nernst Equation
The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential and the concentrations of the reactants and products:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
where:
- \(E\) is the cell potential,
- \(E^\circ\) is the standard cell potential,
- \(n\) is the number of moles of electrons transferred in the reaction.

### Step 2: Identify the Components of the Cell
For the first cell:
\[ \text{Cu (s)} | \text{Cu}^{2+} (aq) (0.1 M) || \text{Ag}^{+} (aq) (0.01 M) | \text{Ag (s)} \]
- Anode reaction: \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \)
- Cathode reaction: \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \)

From the first cell, we know \(E_1 = 0.3095 \, V\).

### Step 3: Calculate the Standard Cell Potential \(E^\circ\)
Using the Nernst equation for the first cell:
\[ E_1 = E^\circ - \frac{0.0591}{2} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2} \right) \]
Substituting the known values:
\[ 0.3095 = E^\circ - \frac{0.0591}{2} \log \left( \frac{0.1}{(0.01)^2} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{0.1}{(0.01)^2} \right) = \log \left( \frac{0.1}{0.0001} \right) = \log(1000) = 3 \]
Now substituting this back:
\[ 0.3095 = E^\circ - \frac{0.0591}{2} \cdot 3 \]
\[ 0.3095 = E^\circ - 0.08865 \]
\[ E^\circ = 0.3095 + 0.08865 = 0.39815 \, V \]

### Step 4: Calculate the Cell Potential for the Second Cell
For the second cell:
\[ \text{Cu (s)} | \text{Cu}^{2+} (aq) (0.01 M) || \text{Ag}^{+} (aq) (0.001 M) | \text{Ag (s)} \]
Using the same Nernst equation:
\[ E_2 = E^\circ - \frac{0.0591}{2} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2} \right) \]
Substituting the values:
\[ E_2 = 0.39815 - \frac{0.0591}{2} \log \left( \frac{0.01}{(0.001)^2} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{0.01}{(0.001)^2} \right) = \log \left( \frac{0.01}{0.000001} \right) = \log(10000) = 4 \]
Now substituting this back:
\[ E_2 = 0.39815 - \frac{0.0591}{2} \cdot 4 \]
\[ E_2 = 0.39815 - 0.1182 \]
\[ E_2 = 0.27995 \, V \]

### Step 5: Convert to Required Format
To express \(E_2\) in the form \(xx \times 10^{-2} V\):
\[ E_2 = 27.995 \times 10^{-2} V \approx 28 \times 10^{-2} V \]

### Final Answer
The cell potential for the second cell is:
\[ \boxed{28} \]