Question

Use the half-reaction method to balance each redox reaction occurring in acidic aqueous solution. (a) \(\mathrm{ClO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Al}^{3+}(a q)\) (c) \(\mathrm{Br}_{2}(a q)+\mathrm{Sn}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{Br}^{-}(a q)\)

Short answer

For reaction (a), the balanced equation is \(\mathrm{ClO}_{4}^{-} + 2\mathrm{Cl}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{Cl}_{2}\). For reactions (b) and (c), follow similar balancing steps, ensuring that the number of electrons lost in oxidation equals the electrons gained in reduction.

Step-by-step solution

Identify the Oxidation and Reduction Half-reactions for Reaction (a)

For \(\mathrm{ClO}_{4}^{-} + \mathrm{Cl}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{Cl}_{2}\), identify the two half-reactions: Oxidation: \(\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}\) Reduction: \(\mathrm{ClO}_{4}^{-} \longrightarrow \mathrm{ClO}_{3}^{-}\)

Balance the Atoms Other Than Oxygen and Hydrogen for Reaction (a)

For Oxidation: Chlorine is already balanced. For Reduction: Chlorine is already balanced.

Balance the Oxygen Atoms with Water for Reaction (a)

Reduction half: No need to balance oxygen as both sides have the same number of oxygen atoms (4 each).

Balance the Hydrogen Atoms with H+ for Reaction (a)

Reduction half: No hydrogen atoms are involved, so there's no need for balancing.

Balance the Charge with Electrons for Reaction (a)

Oxidation half-reaction: \(\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2} + 2e^{-}\) Reduction half-reaction: \(\mathrm{ClO}_{4}^{-} + 2e^{-} \longrightarrow \mathrm{ClO}_{3}^{-}\)

Multiply the Half-reactions by the Appropriate Factors for Reaction (a)

To balance the electrons, multiply the reduction half-reaction by 1 (as there are 2 electrons in both half-reactions).

Add the Half-reactions to Get the Overall Balanced Equation for Reaction (a)

Combine the half-reactions: \(2\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2} + 2e^{-}\) \(\mathrm{ClO}_{4}^{-} + 2e^{-} \longrightarrow \mathrm{ClO}_{3}^{-}\) to get the balanced reaction: \(\mathrm{ClO}_{4}^{-} + 2\mathrm{Cl}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{Cl}_{2}\)

Identify the Oxidation and Reduction Half-reactions for Reaction (b)

For \(\mathrm{MnO}_{4}^{-} + \mathrm{Al} \longrightarrow \mathrm{Mn}^{2+} + \mathrm{Al}^{3+}\), identify the two half-reactions: Oxidation: \(\mathrm{Al} \longrightarrow \mathrm{Al}^{3+}\) Reduction: \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}\)

Balance for Reaction (b)

Balance atoms and charge for both half-reactions following similar steps 2 through 7. This will involve adding \(H_2O\), \(H^+\), and electrons to balance oxygen, hydrogen, and charge respectively in the reduction half-reaction.

Add the Balanced Half-reactions for Reaction (b)

Combine the balanced half-reactions to get the overall equation for reaction (b).

Identify the Oxidation and Reduction Half-reactions for Reaction (c)

For \(\mathrm{Br}_{2} + \mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+} + \mathrm{Br}^{-}\), identify the two half-reactions: Oxidation: \(\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+}\) Reduction: \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}\)

Balance for Reaction (c)

Balance atoms and charge for both half-reactions following similar steps 2 through 7. This will probably involve adding \(H_2O\), \(H^+\), and electrons to balance oxygen (if needed), hydrogen, and charge respectively in the reduction half-reaction.

Add the Balanced Half-reactions for Reaction (c)

Combine the balanced half-reactions to get the overall equation for reaction (c).

Key concepts

Balancing Redox Reactions

Balancing redox reactions, which include both reduction and oxidation processes, can initially seem complex. But, breaking down the equation into separate half-reactions simplifies the task dramatically. The half-reaction method systematically balances the redox reaction in an acidic or basic medium.

In the example of \(\mathrm{ClO}_{4}^{-} + \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + \mathrm{Cl}_{2}\), we initially separate the reaction into its oxidation and reduction components. Special attention is given to balancing the atoms of each element and the overall charge. It's important to balance elements one at a time and save hydrogen and oxygen for last, as they can be balanced using \(H^+\) ions and water (\(H_2O\)) in acidic solutions.

Finally, after both half-reactions are balanced, they're combined ensuring that the number of electrons lost in the oxidation half equals the number of electrons gained in the reduction half. This comprehensive approach ensures a correct and balanced overall redox equation.

Oxidation Half-Reaction

An oxidation half-reaction details the loss of electrons from a species, signifying it is being oxidized. To properly balance an oxidation half-reaction, one must ensure that the number of each type of atom is equal on both sides, except for oxygen and hydrogen, which are balanced after all other elements.

For instance, in the oxidation half-reaction \(\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}\), chlorine atoms are already balanced, but the charge is not. This discrepancy is resolved by adding electrons to the right side to equalize the charge. In this case, two electrons (\(2e^{-}\)) are required to compensate for the zero charge on the left turning into a one plus charge for each chlorine atom that's been transformed into diatomic chlorine gas (\(\mathrm{Cl}_{2}\)).

Example of Oxidation Half-Reaction Balancing

Starting with \(\mathrm{Cl}^{-}\), the balanced equation would be \(\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2e^{-}\) indicating the loss of two electrons, which is characteristic of an oxidation process.

Reduction Half-Reaction

In contrast to oxidation, a reduction half-reaction illustrates the gain of electrons by a species. During the balancing process, one ensures that along with the atoms, the charges on both sides of the reaction are equal. The accumulation of electrons on the reactant side indicates a reduction.

Consider the reduction half-reaction from the given example, \(\mathrm{ClO}_{4}^{-} \longrightarrow \mathrm{ClO}_{3}^{-}\). Here, the chloride atoms are balanced, and, unlike the oxidation half-reaction, there is no need to balance oxygen or hydrogen but rather, the charge. Two electrons (\(2e^{-}\)) are added to the left side to counterbalance the loss of an oxygen atom during the transformation from perchlorate to chlorate.

Example of Reduction Half-Reaction Balancing

For the reaction \(\mathrm{ClO}_{4}^{-} + 2e^{-} \longrightarrow \mathrm{ClO}_{3}^{-}\), we can see the gain of two electrons, highlighting the defining nature of a reduction reaction.