1. Find concentrations after electrolysis

We have two electrolytic cells.

(a) $\mathrm{Cu}^{2+}(1.5\,\mathrm{M},1\,\mathrm{L})/\mathrm{Cu}$

Initial moles of $\mathrm{Cu}^{2+}$: $1.5\times 1=1.5\text{ mol}$

Reaction at cathode: $\mathrm{Cu}^{2+}+2e^-\rightarrow \mathrm{Cu}$

Given charge passed $=1$ Faraday $=1$ mol $e^-$. So moles of $\mathrm{Cu}^{2+}$ reduced: $\frac{1}{2}=0.5\text{ mol}$

Remaining moles of $\mathrm{Cu}^{2+}$: $1.5-0.5=1.0\text{ mol}$

Since volume is $1\,\mathrm{L}$, final concentration: $[\mathrm{Cu}^{2+}]=1.0\,\mathrm{M}$

(b) $\mathrm{Ag}^{+}(0.2\,\mathrm{M},1\,\mathrm{L})/\mathrm{Ag}$

Initial moles of $\mathrm{Ag}^{+}$: $0.2\times 1=0.2\text{ mol}$

Reaction: $\mathrm{Ag}^{+}+e^-\rightarrow \mathrm{Ag}$

Given charge passed $=0.1$ Faraday $=0.1$ mol $e^-$. Therefore moles of $\mathrm{Ag}^+$ reduced: $0.1\text{ mol}$

Remaining moles of $\mathrm{Ag}^{+}$: $0.2-0.1=0.1\text{ mol}$

Thus final concentration: $[\mathrm{Ag}^{+}]=0.1\,\mathrm{M}$

2. Construct the electrochemical cell

Standard reduction potentials: $E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}}=0.80\,\mathrm{V}$ $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=0.34\,\mathrm{V}$

Since silver has higher reduction potential, it acts as cathode.

Cell reaction: $\mathrm{Cu}+2\mathrm{Ag}^+\rightarrow \mathrm{Cu}^{2+}+2\mathrm{Ag}$

So, $E^\circ_{\text{cell}}=0.80-0.34=0.46\,\mathrm{V}$

3. Apply Nernst equation

For the reaction $\mathrm{Cu}+2\mathrm{Ag}^+\rightarrow \mathrm{Cu}^{2+}+2\mathrm{Ag}$ we have $n=2$ and $Q=\frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2}$

Using final concentrations: $Q=\frac{1.0}{(0.1)^2}=100$

Nernst equation at $298\,\mathrm{K}$: $E=E^\circ-\frac{0.0591}{n}\log Q$

Using the given approximation $\frac{2\cdot 303RT}{F}=0.06\,\mathrm{V}$, so $\frac{2.303RT}{F}=0.03\,\mathrm{V}$

Hence for $n=2$, $E=0.46-\frac{0.06}{2}\log(100)$ $E=0.46-0.03\times 2$ $E=0.46-0.06=0.40\,\mathrm{V}$

Therefore, $E=400\,\mathrm{mV}$

4. Comparison with stored answer

Derived answer: $400\,\mathrm{mV}$. Stored correct answer: $400$.

They match.