1. Write the cell reaction

For the cell $\text{Cu}(s)\,|\,\text{Cu}^{2+}(aq)\,||\,\text{Ag}^+(aq)\,|\,\text{Ag}(s)$

Half-reactions are:

• Anode: $\text{Cu}(s) \rightarrow \text{Cu}^{2+} + 2e^-$
• Cathode: $\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$

Multiplying the silver half-reaction by 2: $2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}(s)$

Overall reaction: $\text{Cu}(s) + 2\text{Ag}^+ (aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$

So, number of electrons transferred is $n=2$

2. Reaction quotient

For the reaction $\text{Cu}(s) + 2\text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2\text{Ag}(s)$

the reaction quotient is $Q=\frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2}$

3. Use Nernst equation

At the same temperature, $E = E^\circ - \frac{0.059}{n}\log Q$

Thus, $E = E^\circ - \frac{0.059}{2}\log \left(\frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2}\right)$

4. For the first cell

Given: $[\text{Cu}^{2+}] = 0.1, \quad [\text{Ag}^+] = 0.01$

So, $Q_1 = \frac{0.1}{(0.01)^2} = \frac{10^{-1}}{10^{-4}} = 10^3$

Hence, $E_1 = E^\circ - \frac{0.059}{2}\log(10^3)$ $E_1 = E^\circ - \frac{0.059}{2}\times 3$ $E_1 = E^\circ - 0.0885$

Given: $E_1 = 0.3095\,\text{V}$

Therefore, $E^\circ = 0.3095 + 0.0885 = 0.3980\,\text{V}$

5. For the second cell

Given: $[\text{Cu}^{2+}] = 0.01, \quad [\text{Ag}^+] = 0.001$

So, $Q_2 = \frac{0.01}{(0.001)^2} = \frac{10^{-2}}{10^{-6}} = 10^4$

Now, $E_2 = E^\circ - \frac{0.059}{2}\log(10^4)$ $E_2 = 0.3980 - \frac{0.059}{2}\times 4$ $E_2 = 0.3980 - 0.1180$ $E_2 = 0.2800\,\text{V}$

6. Convert to required form

We need: $E_2 = \underline{\hspace{1cm}} \times 10^{-2}\,\text{V}$

Since $0.2800\,\text{V} = 28 \times 10^{-2}\,\text{V}$

So the required integer is $\boxed{28}$

7. Comparison with stored answer

Stored correct answer = 28

Our derived answer = 28

Hence, the answer agrees with the stored answer.