Write a balanced redox equation for the reaction of mercury with aqua regia, assuming the products include \(\mathrm{HgCl}_{4^{2-}}\) and \(\mathrm{NO}_{2}(g)\).

We should first write the unbalanced half-reactions for the oxidation and reduction processes that take place during the reaction. We know that mercury reacts with aqua regia and forms \(\mathrm{HgCl}_{4^{2-}}\) and \(\mathrm{NO}_{2}(g)\). So, the general half-reactions are: 1. Oxidation: \(\mathrm{Hg} \to \mathrm{HgCl}_{4^{2-}}\) 2. Reduction: \(\mathrm{HNO}_3 \to \mathrm{NO}_{2}(g)\) Now, we need to balance the half-reactions.

First, we will balance the oxidation half-reaction. In this reaction, the number of chlorines on both sides must be balanced. 1. Oxidation: \(\mathrm{Hg} + 4\mathrm{Cl}^{-} \to \mathrm{HgCl}_{4^{2-}}\) Now, we will balance the reduction half-reaction. Here, we need to balance nitrogen and oxygen atoms: 2. Reduction: \(\mathrm{HNO}_3 \to \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g)\)

In this step, we will balance the charges in both half-reactions by adding electrons. For the oxidation half-reaction, we don't need to add any electrons since it's already balanced. 1. Oxidation: \(\mathrm{Hg} + 4\mathrm{Cl}^{-} \to \mathrm{HgCl}_{4^{2-}}\) In the reduction half-reaction, we have +1 charge on the reactant side and 0 charge on the product side. Thus, we need to add one electron to balance the charges: 2. Reduction: \(\mathrm{HNO}_3 + 1\mathrm{e}^{-} \to \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g)\)

Now that both half-reactions are balanced, we need to combine them. To do this, we need to have the same number of electrons in both half-reactions, which we already have from the previous step. So, we can add the half-reactions together: \(\mathrm{Hg} + 4\mathrm{Cl}^{-} + \mathrm{HNO}_3 + \mathrm{e}^{-} \to \mathrm{HgCl}_{4^{2-}} + \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g) + \mathrm{e}^{-}\) Since the electrons are on both sides of the equation, we can cancel them out: \(\mathrm{Hg} + 4\mathrm{Cl}^{-} + \mathrm{HNO}_3 \to \mathrm{HgCl}_{4^{2-}} + \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g)\)

Remember that aqua regia consists of nitric acid and hydrochloric acid in a 1:3 ratio. We have 4 chloride ions in the reactants, so we need 4 hydrochloric acid molecules to provide these chloride ions: \(\mathrm{Hg} + 4\mathrm{HCl} + \mathrm{HNO}_3 \to \mathrm{HgCl}_{4^{2-}} + \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g)\) Finally, in order to make this a balanced equation, we need to balance the hydrogen ions and the sulfate ions. Therefore, we add water and sulfate ions to both sides of the equation: \(\mathrm{Hg} + 4\mathrm{HCl} + \mathrm{HNO}_3 + 8\mathrm{H}_2\mathrm{O} + 4\mathrm{SO}_{4^{2-}} \to \mathrm{HgCl}_{4^{2-}} + \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g) + 8\mathrm{H}_2\mathrm{O} + 4\mathrm{HSO}_{4^-}\) The balanced redox equation for the reaction of mercury with aqua regia, assuming the products include \(\mathrm{HgCl}_{4^{2-}}\) and \(\mathrm{NO}_{2}(g)\), is: \(\mathrm{Hg} + 4\mathrm{HCl} + \mathrm{HNO}_3 + 8\mathrm{H}_2\mathrm{O} + 4\mathrm{SO}_{4^{2-}} \to \mathrm{HgCl}_{4^{2-}} + \mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_2(g) + 8\mathrm{H}_2\mathrm{O} + 4\mathrm{HSO}_{4^-}\)