Problem 24 - Chemistry The Central Science

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) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow\) $$ \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q) \text { (acidic solution) } $$ (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) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) $$ \mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q) \text { (acidic solution) } $$ (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)

Step-by-Step Solution

Answer

The short answers for each of the reactions are as follows: (a) \(3\mathrm{NO}_{2}^{-}(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14\mathrm{H}^{+}(a q) \longrightarrow 3\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\), with \(Cr_2O_7^{2-}\) as the oxidizing agent and \(NO_2^-\) as the reducing agent. For the remaining reactions (b) to (f), follow the same steps as mentioned above and make sure to add \(\mathrm{H}^{+}\) ions for acidic solutions and \(\mathrm{OH}^{-}\) ions for basic solutions during the balancing process.

Step 1: Assign oxidation numbers

First, assign oxidation numbers to all atoms in the reaction: \(\mathrm{N}\) in \(\mathrm{NO}_{2}^{-}\): +3 \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): +6 \(\mathrm{N}\) in \(\mathrm{NO}_{3}^{-}\): +5 \(\mathrm{Cr}\) in \(\mathrm{Cr}^{3+}\): +3

Step 2: Identify the species being oxidized and reduced

The species being oxidized is \(\mathrm{NO}_{2}^{-}\) (from +3 to +5), and the species being reduced is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (from +6 to +3).

Step 3: Balance the half-reactions

Write separate half-reactions for the oxidation and reduction processes and balance them: Oxidation: \(\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{NO}_{3}^{-}(a q) + 2e^-\) Reduction: \(6e^- + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\)

Step 4: Combine the half-reactions

Multiply the oxidation half-reaction by 3 and add it to the reduction half-reaction: \(3(\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{NO}_{3}^{-}(a q) + 2e^-)\) \(6e^- + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\)

Step 5: Simplify the balanced equation

Add the half-reactions together and cancel the electrons to get the balanced equation: \(3\mathrm{NO}_{2}^{-}(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q)\longrightarrow 3\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\) The balanced equation for reaction (a) is: $$3\mathrm{NO}_{2}^{-}(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q)\longrightarrow 3\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)$$ The oxidizing agent is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), and the reducing agent is \(\mathrm{NO}_{2}^{-}\). Apply the same steps for the remaining reactions (b) to (f). Make sure to add \(\mathrm{H}^{+}\) ions for acidic solutions and \(\mathrm{OH}^{-}\) ions for basic solutions during the balancing process.

Key Concepts

Assigning Oxidation Numbers

Understanding redox reactions begins with assigning oxidation numbers, which help in determining which atoms are oxidized or reduced during the reaction.

Oxidation numbers are like bookkeeping numbers that allow chemists to keep track of electron transfer. They are assigned based on a set of rules: oxygen usually has an oxidation number of -2, hydrogen is typically +1, and the oxidation number of an element in its natural state is always 0. For ions, the oxidation number is the same as the charge of the ion.

For instance, in the reaction involving \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), you assign +3 to nitrogen in \(\mathrm{NO}_{2}^{-}\) and +6 to chromium in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) following the rules. These numbers help identify which atoms gain or lose electrons, and hence indicate the process of oxidation and reduction.

Oxidation and Reduction Identification

Once oxidation numbers are assigned, identifying which species are oxidized and which are reduced is key. Oxidation involves an increase in oxidation number, signifying a loss of electrons, while reduction involves a decrease in oxidation number, indicating a gain of electrons.

In our example, \(\mathrm{NO}_{2}^{-}\) is oxidized since its oxidation number increases from +3 to +5 when it transforms into \(\mathrm{NO}_{3}^{-}\). Conversely, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is reduced because its oxidation number decreases from +6 to +3 as it becomes \(\mathrm{Cr}^{3+}\). This step is crucial because it directs the balance of the equations based on the flow of electrons between the oxidized and reduced species.

Half-Reaction Method

The half-reaction method simplifies complex redox equations by breaking them down into separate oxidation and reduction reactions, which are balanced independently before being recombined.

This method involves adding electrons as reactants in oxidation half-reactions or as products in reduction half-reactions to reflect the transfer of electrons. For instance, the oxidation half-reaction \(\mathrm{NO}_{2}^{-} \rightarrow \mathrm{NO}_{3}^{-}\) is balanced by adding two electrons on the product side to account for the lost electrons.

Balance Charges and Atoms

During balancing, aside from electrons, ensure that both sides have the same number of each type of atom and charges. Add \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions to balance the remaining atoms and charges for reactions in acidic or basic solutions, respectively.

Balancing Equations in Acidic and Basic Solutions

The environment of the reaction (acidic or basic) significantly affects the balancing of redox reactions. For equations in acidic solutions, \(\mathrm{H}^{+}\) ions are added to balance hydrogen atoms and charge. In basic solutions, \(\mathrm{OH}^{-}\) ions are used instead.

For reaction (a) in an acidic solution, we added \(\mathrm{H}^{+}\) to the reduction half-reaction to balance the hydrogen atoms and ensure the net charge of the reactants equals that of the products.

Combining Half-Reactions

After balancing the half-reactions, combine them, ensuring the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction. This equality allows for the cancellation of electrons from the final balanced equation, resulting in a reaction that conserves both mass and charge.

Oxidizing and Reducing Agents

Redox reactions involve two key players: the oxidizing agent, which gets reduced, and the reducing agent, which gets oxidized.

The oxidizing agent, such as \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) in reaction (a), accepts electrons and subsequently undergoes reduction. On the other hand, the reducing agent, like \(\mathrm{NO}_{2}^{-}\), donates electrons and undergoes oxidation.

Importance of Identifying Agents

Identifying the oxidizing and reducing agents not only helps in understanding the reaction's direction but also in applications where control of oxidation or reduction is required, such as in corrosion prevention, manufacturing of batteries, or organic synthesis. The agent's identity aids in predicting and controlling the chemical behavior within a system.