GIVEN
$\Delta G^\circ = -43500$ J $mol^{-1}$ ; $T = 25\ ^\circ C = 298$ K ; $F = 96500$ C $mol^{-1}$
$p_{H_2} = 0.4$ atm ; $[H^+] = 0.1$ M ; $[Cl^-] = 0.2$ M

Step 1 - Find the number of electrons transferred, $n$.
Splitting the reaction into half reactions:
Oxidation (anode) : $H_2(g) \longrightarrow 2H^+ + 2e^-$
Reduction (cathode) : $2AgCl(s) + 2e^- \longrightarrow 2Ag(s) + 2Cl^-$
$$\mathbf{n = 2}$$

Step 2 - Calculate $E^\circ_{cell}$ from $\Delta G^\circ$.
$$\Delta G^\circ = -nFE^\circ_{cell}$$
$$E^\circ_{cell} = \frac{-\Delta G^\circ}{nF} = \frac{-(-43500)}{2 \times 96500}$$
$$E^\circ_{cell} = \frac{43500}{193000}$$
$$\mathbf{E^\circ_{cell} = 0.2254\ V}$$

Step 3 - Write the reaction quotient $Q$.
Solids ($AgCl$ and $Ag$) are omitted; the gas appears as its partial pressure.
$$Q = \frac{[H^+]^2[Cl^-]^2}{p_{H_2}}$$

Step 4 - Substitute the given values into $Q$.
$$Q = \frac{(0.1)^2 (0.2)^2}{0.4} = \frac{(1\times10^{-2})(4\times10^{-2})}{0.4}$$
$$Q = \frac{4 \times 10^{-4}}{4 \times 10^{-1}} = 1 \times 10^{-3}$$
$$\log Q = \log 10^{-3} = -3$$

Step 5 - Apply the Nernst equation.
$$E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log Q$$
$$E_{cell} = 0.2254 - \frac{0.059}{2} \times (-3)$$
$$E_{cell} = 0.2254 - (0.0295 \times -3)$$
$$E_{cell} = 0.2254 + 0.0885$$

Step 6 - Final answer.
$$\boxed{E_{cell} = 0.3139\ V \approx 0.314\ V}$$

Note: $E_{cell} > E^\circ_{cell}$ here because $Q < 1$ (the product concentrations are
low), which by Le Chatelier's principle drives the forward reaction and raises the cell
potential.