Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in basic solution. Identify the oxidizing agent and reducing agent in cach reaction. (a) Production of chlorite ions from dichlorine heptoxide: \(\mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{2}{ }^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})\) (b) Action of permanganate ions on sulfide ions: \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{S}(\mathrm{s})+\mathrm{MnO}_{2}(\mathrm{~s})\) (c) Reaction of hydrazine with chlorate ions: \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{ClO}_{3}{ }^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cl}^{-}\)(aq) (d) Reaction of plumbate ions and hypochlorite ions: \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq}) \longrightarrow \mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{Cl}^{-}(\mathrm{aq})\)

Step by step solution

01

Write the half-reactions for (a)

Divide the equation into two half-reactions, one for oxidation and one for reduction.\[\text{Reduction: } \mathrm{Cl}_{2}\mathrm{O}_{7} + \mathrm{e}^{-} \rightarrow \mathrm{ClO}_{2}^{-}\]\[\text{Oxidation: } \mathrm{H}_{2}\mathrm{O}_{2} \rightarrow \mathrm{O}_{2} + \mathrm{e}^{-}\]

02

Balance the half-reactions for atoms and charge in (a)

For the reduction half-reaction, balance chlorine and oxygen by adding water and hydroxide ions. For the oxidation half-reaction, balance oxygen by adding water and hydroxide ions, then balance hydrogen and finally add electrons to balance charge. The balanced half-reactions in basic solution will look like this:\[\text{Reduction: } \mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{e}^{-} + 4\mathrm{OH}^{-} \rightarrow 2\mathrm{ClO}_{2}^{-} + 3\mathrm{H}_{2}\mathrm{O}\]\[\text{Oxidation: } \mathrm{H}_{2}\mathrm{O}_{2} + 2\mathrm{OH}^{-} \rightarrow \mathrm{O}_{2} + 2\mathrm{H}_{2}\mathrm{O} + 2\mathrm{e}^{-}\]

03

Combine and balance the overall equation for (a)

Combine the oxidation and reduction half-reactions ensuring that the number of electrons canceled.\[\mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{H}_{2}\mathrm{O}_{2} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{ClO}_{2}^{-} + 3\mathrm{H}_{2}\mathrm{O} + \mathrm{O}_{2}\]Check all atoms and charges are balanced, then identify the oxidizing agent as \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) and the reducing agent as \(\mathrm{H}_{2}\mathrm{O}_{2}\).

04

Repeat Steps 1-3 for reactions (b), (c), and (d)

Follow similar steps as for part (a) for the remaining reactions to balance each and identify oxidizing and reducing agents for each reaction.

05

Final Balancing and Identifying for (b)

The balanced redox reaction for (b) is:\[2\mathrm{MnO}_{4}^{-} + 5\mathrm{S}^{2-} + 6\mathrm{OH}^{-} \rightarrow 2\mathrm{MnO}_{2} + 5\mathrm{S} + 6\mathrm{H}_{2}\mathrm{O}\]The oxidizing agent is \(\mathrm{MnO}_{4}^{-}\) and the reducing agent is \(\mathrm{S}^{2-}\).

06

Final Balancing and Identifying for (c)

The balanced redox reaction for (c) is:\[\mathrm{N}_{2}\mathrm{H}_{4} + 2\mathrm{ClO}_{3}^{-} \rightarrow 2\mathrm{NO} + 2\mathrm{Cl}^{-} + 4\mathrm{H}_{2}\mathrm{O}\]The oxidizing agent is \(\mathrm{ClO}_{3}^{-}\) and the reducing agent is \(\mathrm{N}_{2}\mathrm{H}_{4}\).

07

Final Balancing and Identifying for (d)

The balanced redox reaction for (d) is:\[2\mathrm{ClO}^{-} + \mathrm{Pb}(\text{OH})_{4}^{2-} + 2\mathrm{OH}^{-} \rightarrow \mathrm{PbO}_{2} + 2\mathrm{Cl}^{-} + 4\mathrm{H}_{2}\mathrm{O}\]The oxidizing agent is \(\mathrm{ClO}^{-}\) and the reducing agent is \(\mathrm{Pb}(\text{OH})_{4}^{2-}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation and Reduction Half-Reactions

Understanding oxidation and reduction half-reactions is fundamental to solving redox equations. Oxidation refers to the loss of electrons, which raises the oxidation state of the element undergoing oxidation. Conversely, reduction involves the gain of electrons, lowering the oxidation state of the element being reduced.

In a redox reaction, we split the overall process into its oxidation and reduction parts to tackle the balancing act systematically. For instance, in the textbook exercise with dichlorine heptoxide and hydrogen peroxide reacting to form chlorite ions and oxygen gas, each component is separated into its respective half-reaction.

The reduction half-reaction shows dichlorine heptoxide gaining electrons to become chlorite ions. Meanwhile, the oxidation half-reaction reveals hydrogen peroxide losing electrons to form oxygen gas. By identifying these processes separately, we can balance each half-reaction for atoms and charge, which is a strategy that simplifies the overall balancing of redox reactions.

Oxidizing and Reducing Agents

Identifying the oxidizing and reducing agents in a chemical reaction is crucial since these agents drive the redox process. The oxidizing agent is the species that gets reduced by accepting electrons, while the reducing agent is the species that gets oxidized by donating electrons.

In the provided exercise examples, we can pinpoint the oxidizing agent as dichlorine heptoxide (for reaction (a)), permanganate ions (for reaction (b)), and chlorate ions (for reaction (c)). These species each accept electrons and, in the process, oxidize another species in the reaction. On the other hand, the reducing agents, like hydrogen peroxide in reaction (a), sulfide ions in reaction (b), and hydrazine in reaction (c), supply electrons and become oxidized themselves. By understanding the role these agents play, students can better determine the direction of electron flow in redox reactions.

Chemical Reactions in Basic Solution

When balancing redox reactions in a basic solution, an additional step is required compared to balancing in an acidic environment. After separating and balancing the half-reactions for atoms other than oxygen and hydrogen, we then balance the oxygens by adding water molecules and the hydrogens by adding hydroxide ions (OH^-) to the reactions.

Next, we balance the charges by adding electrons. In the end, we ensure that the added OH^- ions and water molecules are accounted for on both sides of the reaction, keeping in mind the basic environment. This approach to balancing reactions in a basic solution hinges on the understanding that water and hydroxide ions are integral to maintaining the basic nature of the solution.

Stoichiometry of Redox Reactions

Stoichiometry in redox reactions pertains to the quantitative relationship between reactants and products. It involves ensuring that the number of atoms and the charge is balanced on both sides of the reaction. This process requires careful adjustment of the coefficients in front of each species in the reaction.

After balancing the half-reactions separately, we then combine them to cancel out the electrons and scale the stoichiometric coefficients to attain an overall balanced equation. For instance, when balancing reaction (b) from the exercise, it’s critical to multiply the half-reactions by appropriate factors such that the number of electrons lost and gained is equal, thereby achieving a balanced redox equation. Mastery of stoichiometry is crucial as it not only informs the conservation of mass and charge but also allows precise calculations when working with chemical reactions.