Question

Write plausible half-equations and a balanced oxidation-reduction equation for the disproportionation of \(\mathrm{XeF}_{4}\) to \(\mathrm{Xe}\) and \(\mathrm{XeO}_{3}\) in aqueous acidic solution. Xe and \(\mathrm{XeO}_{3}\) are produced in a 2: 1 mol ratio, and \(\mathrm{O}_{2}(\mathrm{g})\) is also produced.

Short answer

Based on the computed steps, the redox equation for the disproportionation of \(\mathrm{XeF}_{4}\) in aqueous acidic solution is \(3\mathrm{XeF}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Xe} + \mathrm{XeO}_3+6\mathrm{HF} + \mathrm{O_2}\).

Step-by-step solution

Determination of Oxidation Numbers

The first step is to identify the oxidation state of Xenon in \(\mathrm{Xe}\), \(\mathrm{XeF}_{4}\), and \(\mathrm{XeO}_{3}\). For \(\mathrm{Xe}\), it's 0 as it is in the elemental state. In \(\mathrm{XeF}_{4}\), it is +4 as Fluorine has an oxidation state of -1. In \(\mathrm{XeO}_{3}\), Xenon has an oxidation state of +6 due to Oxygen's oxidation state being -2.

Deriving Half-Reactions

Next, we can construct the half-reactions. From \(\mathrm{XeF}_{4}\) (Xenon in +4 state) to \(\mathrm{Xe}\) (Xenon in 0 state), it's a reduction half-reaction: \(\mathrm{XeF}_{4} + 4H_{2}O \rightarrow \mathrm{Xe} + 4HF + 4OH-\). From \(\mathrm{XeF}_{4}\) (Xenon in +4 state) to \(\mathrm{XeO}_{3}\) (Xenon in +6 state) it's an oxidation half-reaction: \(\mathrm{XeF}_{4} + 4H^{+} \rightarrow \mathrm{XeO}_{3} + 4HF\).

Balance the Half-Reactions

In order to balance the half-reactions, electrons are added where required. For the reduction half-reaction: \(\mathrm{XeF}_{4} + 4\mathrm{H}_{2}\mathrm{O} + 4e^{-} \rightarrow \mathrm{Xe} + 4\mathrm{HF} + 4\mathrm{OH}^{-}\) and for the oxidation half-reaction: \(\mathrm{XeF}_{4} + 4\mathrm{H}^{+} +2e^{-} \rightarrow \mathrm{XeO}_{3} + 4\mathrm{HF}\).

Combine the Half-Reactions

Now we can combine the half-reactions together. To balance the charge, we need multiples of each half-reaction. The reduction reaction must take place twice for every one instance of the oxidation reaction to reach equilibrium, as described by the question. This also aligns with the given 2:1 mol ratio of \(\mathrm{Xe}:\mathrm{XeO}_{3}\) and results in the following equation: \(3\mathrm{XeF}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Xe} + \mathrm{XeO}_3+6\mathrm{HF} + \mathrm{O_2} + 2\mathrm{OH}⁻+ 4\mathrm{H}^+\).

Simplify and Final Check

In the last step we simplify the equation: \(3\mathrm{XeF}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Xe} + \mathrm{XeO}_3+6\mathrm{HF} + \mathrm{O_2}\). Now the numbers of atoms and charges, and the mol ratio are all balanced for each element in the equation.

Key concepts

Oxidation-Reduction Equations

Oxidation-reduction equations, commonly known as redox equations, are chemical equations illustrating the transfer of electrons between substances. These transfers result in changes in the oxidation states of the atoms involved. In a redox process, one species is oxidized (loses electrons) while another is reduced (gains electrons).

To write a redox equation, the individual oxidation and reduction processes are first expressed as half-reactions. Each half-reaction shows only one process, either the oxidation or the reduction. The half-reactions are then balanced for mass and charge, and finally combined to form a balanced redox equation that conserves both mass and charge overall.

Oxidation States

Oxidation states, also known as oxidation numbers, provide a way to keep track of the electrons in compounds during a redox process. These numerical values are indicative of the number of electrons an atom can gain, lose, or share during chemical reactions. For instance, in the disproportionation of \(\mathrm{XeF}_{4}\), the oxidation states of Xenon are \(0\), \(+4\), and \(+6\) in \(\mathrm{Xe}\), \(\mathrm{XeF}_{4}\), and \(\mathrm{XeO}_{3}\) respectively. It's important to remember that the sum of the oxidation states for all atoms in a neutral compound must be zero, and it must be equal to the overall charge for ions.

Half-Reactions

Half-reactions are the bread and butter of understanding redox processes. They split the overall reaction into two separate equations, each representing one component of the redox pair. In a disproportionation reaction like the one involving \(\mathrm{XeF}_{4}\), we have both an oxidation half-reaction, where Xenon increases its oxidation state, and a reduction half-reaction, where Xenon decreases its oxidation state.

To balance half-reactions, specially in acidic solutions, it's common to add \(\mathrm{H}^+\) ions to balance hydrogen and water to balance oxygen, finishing with electrons to balance the charges. In our example, the electrons balance perfectly when you combine the half-reactions in accordance with the stoichiometric coefficients that reflect the substance's mole ratio.

Balancing Chemical Equations

Balancing chemical equations is critical to accurately represent what occurs during a chemical reaction. The law of conservation of mass dictates that matter is neither created nor destroyed. Hence, the number of each type of atom on the reactant side must be the same as on the product side. In redox reactions, balancing also includes ensuring that the number of electrons lost equals the number of electrons gained to preserve charge neutrality. This might require multiplying the half-reactions by appropriate factors before combining them, as seen in the disproportionation of \(\mathrm{XeF}_{4}\) into \(\mathrm{Xe}\) and \(\mathrm{XeO}_{3}\).

Acidic Solution Chemistry

Redox reactions occurring in acidic solutions often require the presence of \(\mathrm{H}^+\) ions. These ions participate in balancing the half-reactions for charge and mass. For example, in our disproportionation reaction, the \(\mathrm{H}^+\) ions are necessary to balance out the electrons transferred in the oxidation of \(\mathrm{XeF}_{4}\) to \(\mathrm{XeO}_{3}\). Acidic solution chemistry is also crucial when balancing oxygen atoms. Water (\(\mathrm{H}_2\mathrm{O}\)) is typically used to balance the oxygen, and then \(\mathrm{H}^+\) is added or removed to finalize the balance, ensuring that reactions in acidic solutions are correctly represented.