The following redox reaction is used to prepare orthotelluric acid: $$ \mathrm{Te}(\mathrm{s})+\mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_{6} \mathrm{TeO}_{6}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ (a) Identify the elements undergoing changes in oxidation state and indicate the initial and final oxidation numbers for these elements. (b) Write and balance the oxidation half-reaction. (c) Write and balance the reduction half-reaction. (d) Combine the half-reactions to produce a balanced redox equation.

Oxidation states, also known as oxidation numbers, serve as a bookkeeping tool in chemistry. They allow us to track electron movement in redox reactions, which is crucial because these reactions involve the transfer of electrons. An oxidation state is a positive or negative number assigned to an atom based on a set of rules. For example, in the Te (s) and H₆TeO₆ equation, Te starts at an oxidation state of 0 and ends at +6. This indicates Te has lost electrons and has been oxidized.

When determining oxidation states, consider that elements in their pure form always have an oxidation state of 0, and the sum of oxidation states in a molecule or compound must equal the overall charge. Understanding how to assign these numbers is essential—it's the first step in identifying which elements are oxidized or reduced in a redox reaction.

Balancing redox equations can seem intimidating, but it can be simplified by dividing the equation into half-reactions – one for oxidation and one for reduction. Each half-reaction must be balanced independently, first for atoms other than oxygen and hydrogen, then for oxygen atoms by adding H₂O, for hydrogen by adding H⁺, and finally for charge by adding electrons. Once both half-reactions are balanced, they can be combined to ensure that the electrons gained in the reduction half-reaction are equal to those lost in the oxidation half-reaction. The result is a balanced redox equation that obeys the laws of conservation of mass and charge. For instance, in our Te and ClO₃^- reaction, water and hydrogen ions are added to balance oxygen and hydrogen atoms, respectively, and electrons are added to balance the charges.

The oxidation half-reaction is where the loss of electrons occurs, representing the oxidation process. For tellurium, in its reaction with ClO₃^-, we see that it goes from an oxidation state of 0 to +6 and therefore loses 6 electrons. To write the half-reaction, we start by indicating this change: Te (s) becomes H₆TeO₆ (aq). Then, we balance the oxygen by adding water molecules, and balance the overall charge by adding the electrons lost by Te on the product side. These steps reveal the underlying changes at the atomic level, showcasing how Te is transformed through the reaction.

The reduction half-reaction focuses on the gain of electrons, reflecting the reduction aspect of a redox process. In our example, we consider chlorine in ClO₃^-, which gains electrons to form Cl₂ gas. To balance the half-reaction, we must ensure the number of atoms of each element and the overall charges on both sides of the reaction are equal. By adding water molecules on the appropriate side, we balance the oxygen atoms. Hydrogen ions are added to compensate for the additional water molecules, and electrons are added on the reactant side to equalize the charge. Through the reduction half-reaction, we can track the transformation of chlorine as it undergoes reduction from ClO₃^- to Cl₂.