GIVEN
$[Mg^{2+}] = 0.01$ M $= 10^{-2}$ M ; $[Ag^+] = 0.001$ M $= 10^{-3}$ M
$E^\circ_{Mg^{2+}/Mg} = -2.37$ V ; $E^\circ_{Ag^+/Ag} = +0.80$ V
$T = 298$ K ; $F = 96500$ C $mol^{-1}$

Step 1 - Identify the anode and the cathode.
By convention, in a cell representation the LEFT half-cell is the ANODE (oxidation) and
the RIGHT half-cell is the CATHODE (reduction).
Anode (LHS, oxidation) : $Mg(s) \longrightarrow Mg^{2+}(aq) + 2e^-$
Cathode (RHS, reduction) : $Ag^+(aq) + e^- \longrightarrow Ag(s) \qquad [\times 2]$

Step 2 - Write the overall cell reaction and find $n$.
$$Mg(s) + 2Ag^+(aq) \longrightarrow Mg^{2+}(aq) + 2Ag(s)$$
Number of electrons transferred, $\mathbf{n = 2}$

Step 3 - Calculate the standard cell potential.
$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$
$$E^\circ_{cell} = (+0.80) - (-2.37)$$
$$\mathbf{E^\circ_{cell} = +3.17\ V}$$
(Positive, so the cell reaction is spontaneous as written.)

Step 4 - Write the Nernst equation for the cell.
$$E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log Q, \qquad
Q = \frac{[Mg^{2+}]}{[Ag^+]^2}$$
(Solids Mg and Ag do not appear in $Q$; their activity is taken as 1.)

Step 5 - Calculate the reaction quotient $Q$.
$$Q = \frac{[Mg^{2+}]}{[Ag^+]^2} = \frac{10^{-2}}{(10^{-3})^2} = \frac{10^{-2}}{10^{-6}} = 10^{4}$$
$$\log Q = \log 10^4 = 4$$

Step 6 - Substitute into the Nernst equation.
$$E_{cell} = 3.17 - \frac{0.059}{2}\times 4$$
$$E_{cell} = 3.17 - (0.0295 \times 4)$$
$$E_{cell} = 3.17 - 0.118$$
$$\mathbf{E_{cell} = 3.052\ V \approx 3.05\ V}$$

Step 7 - Calculate the Gibbs energy change.
$$\Delta G = -nFE_{cell}$$
$$\Delta G = -(2)(96500\ C\ mol^{-1})(3.052\ V)$$
$$\Delta G = -589036\ J\ mol^{-1}$$

FINAL ANSWERS
$$\boxed{E_{cell} = 3.052\ V \approx 3.05\ V}$$
$$\boxed{\Delta G = -589036\ J\ mol^{-1} \approx -589\ kJ\ mol^{-1}}$$

$\Delta G$ is large and negative, confirming that the cell reaction is highly
spontaneous.