Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state.) (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{HCOOH}(a q)+ \mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{BrO}_{3}^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{N}_{2}(g)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

Short answer

Here, we have provided the solution to part (a). Balancing the equations for parts (b)-(f) can be done by following the steps mentioned in the analysis. For part (a), the balanced equation is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{NO}_{2}^{-} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 6\mathrm{NO}_{3}^{-}\), with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) acting as the oxidizing agent and \(\mathrm{NO}_{2}^{-}\) as the reducing agent.

Step-by-step solution

Write the half-reactions

Oxidation: \(\mathrm{NO}_{2}^{-} \longrightarrow \mathrm{NO}_{3}^{-}\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\)

Balance other elements

Already balanced.

Balance oxygen atoms

Oxidation: Already balanced. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Balance hydrogen atoms

Oxidation: Already balanced. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Balance charges

Oxidation: \(\mathrm{NO}_{2}^{-} \longrightarrow \mathrm{NO}_{3}^{-} + \mathrm{e}^{-}\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Make number of electrons equal

To balance electrons, we will multiply the oxidation reaction by 6 and add both reactions. \(6\mathrm{NO}_{2}^{-} \longrightarrow 6\mathrm{NO}_{3}^{-} + 6\mathrm{e}^{-}\) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Add and simplify the reactions

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{NO}_{2}^{-} \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 6\mathrm{NO}_{3}^{-}\)

Identify oxidizing and reducing agents

Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{NO}_{2}^{-}\) I leave the rest of the exercises (b)-(f) to you. Follow the same steps, and you will be able to balance each reaction and find the oxidizing and reducing agents.

Key concepts

Oxidizing Agent

In redox reactions, the oxidizing agent plays a crucial role. It might sound complicated, but it is actually quite simple! The oxidizing agent is the substance that gains electrons in a chemical reaction. This means that while it is being reduced itself, it causes another substance to be oxidized. You can often spot the oxidizing agent by looking for the substance that has a decrease in oxidation state during the reaction.

This is because when a compound gains electrons, its oxidation state decreases. Consider the example given in the exercise:

• For the reaction of \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) turning into \(2\mathrm{Cr}^{3+} \), "\(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) is the oxidizing agent."

The chromate ion \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) accepts electrons and is reduced to \(2\mathrm{Cr}^{3+}\), thus causing the oxidation of other species in the reaction process.

Remember, the oxidizing agent is key to understanding redox reactions and their dynamics.

Reducing Agent

The reducing agent in a redox reaction is equally crucial as it is that hero that gives away electrons to another substance. Sounds like a good friend, right? When a reducing agent loses electrons, it itself becomes oxidized. This means its oxidation state increases, making it easier to identify in reactions.

Look at the substance in a reaction that shows an increase in oxidation state for clues about who the reducing agent is.
For example, in the exercise's reaction:

• \(\mathrm{NO}_{2}^{-} \) turns into \(\mathrm{NO}_{3}^{-} \), where "\(\mathrm{NO}_{2}^{-} \)" is the reducing agent."

The nitrite ion \(\mathrm{NO}_{2}^{-}\) loses electrons, increasing its oxidation state and causing the reduction of another substance. So while the oxidizing agent accepts the electrons, the reducing agent is happily giving them away! The balance between these two agents is what makes redox reactions occur.

Half-Reactions

Understanding half-reactions simplifies the seemingly complex process of balancing redox reactions. A half-reaction focuses on either the oxidation or the reduction part of a redox equation. Each half-reaction shows either the loss or gain of electrons. This can help break down the larger reaction into digestible parts.

For the exercise, these half-reactions can be observed: - **Oxidation Half-Reaction** focuses on the substance that loses electrons. This is where we see an increase in oxidation state. - **Reduction Half-Reaction** focuses on the substance that gains electrons, showing a decrease in oxidation state. By writing these separately, you ensure that each part of the reaction is balanced for mass and charge. Once each half-reaction is balanced, you can combine them back into a single balanced redox reaction.
This method is especially useful for identifying the transfer of electrons and helps to clearly show how substances transform during the chemical process.

Breaking down the main reaction into these parts makes the overall task more manageable and coherent.