Let’s analyze the situation step by step.

We have 1 M aqueous solutions of:

$$\text{AgNO}_3$$ with $$\text{Ag}^+/ \text{Ag} \quad E^0 = +0.80\,V$$

$$\text{Hg}_2(\text{NO}_3)_2$$ with $$\text{Hg}_2^{2+}/ \text{Hg} \quad E^0 = +0.79\,V$$

$$\text{Cu(NO}_3)_2$$ with $$\text{Cu}^{2+}/ \text{Cu} \quad E^0 = +0.24\,V$$

$$\text{Mg(NO}_3)_2$$ with $$\text{Mg}^{2+}/ \text{Mg} \quad E^0 = -2.37\,V$$

In an electrolytic cell with inert electrodes, reduction occurs at the cathode. The ion with the highest (most positive) reduction potential will be reduced (or deposited) first. Therefore, as we increase the applied voltage (making the cathode sufficiently negative), the metals will begin to deposit in the order of their reduction potentials.

The order of reduction (deposition) will be:

First, silver ($$+0.80\,V$$)

Next, mercury ($$+0.79\,V$$)

Then, copper ($$+0.24\,V$$)

So, Statement (I) which says “With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu” is correct.

For magnesium, the reduction potential is very negative ($$-2.37\,V$$). In aqueous solution, before reaching such a negative potential, water will be reduced. The cathodic reaction for water typically is:

$$2\,\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\,\text{OH}^-$$

which leads to the evolution of hydrogen gas, not oxygen. (Oxygen is produced at the anode during the oxidation of water.)

Hence, Statement (II) erroneously states that “oxygen gas will be evolved at the cathode” when in fact, if magnesium were to be reduced (which it won’t be), water reduction would yield hydrogen gas. Thus, Statement (II) is incorrect.

Summary:

Statement (I) is correct.

Statement (II) is incorrect.

Therefore, the most appropriate answer is:

Option C: Statement I is correct but Statement II is incorrect.