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

Step by step solution

01

Determine the oxidation states

We assign oxidation states for each atom in the reaction: \(NO_2^-\): N has an oxidation state of +3. \(Cr_2O_7^{2-}\): Cr has an oxidation state of +6. \(Cr^{3+}\): Cr has an oxidation state of +3. \(NO_3^-\): N has an oxidation state of +5.

02

Write half-reactions

Divide the equation into two half-reactions: \(NO_2^-(aq) \rightarrow NO_3^-(aq)\) \(Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq)\)

03

Balance half-reactions

Balance the half-reactions for the number of atoms and charge, and combine them back into the balanced redox equation: \(3NO_2^-(aq) + 3H_2O(l) \rightarrow 3NO_3^-(aq) + 6H^+(aq) + 6e^-\) \(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)\) Adding the two-balanced half-reactions gives: \(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 3NO_2^-(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) + 3NO_3^-(aq) + 6H^+(aq)\)

04

Identify oxidizing and reducing agents

The reducing agent is the species being oxidized, and the oxidizing agent is the species being reduced in the reaction: Reducing agent: \(NO_2^-(aq)\) Oxidizing agent: \(Cr_2O_7^{2-}(aq)\) #Analysis# and #Solution# should be done for exercises (b), (c), (d), (e), and (f) in the same format as described above.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Oxidation State Determination

Understanding redox reactions begins with the ability to determine oxidation states of elements within compounds. The oxidation state is essentially a hypothetical charge that an atom would have if all bonds to atoms of different elements were fully ionic. For example, in water (H_2O), the hydrogen atoms each carry an oxidation state of +1, while the oxygen has an oxidation state of -2. When determining oxidation states, there are a few rules we must follow:

• The oxidation state of an atom in its elemental form is zero.
• For a monoatomic ion, the oxidation state is equal to the charge of the ion.
• Oxygen typically has an oxidation state of -2, except in peroxides like H_2O_2 where it's -1, or in compounds with fluorine.
• Hydrogen usually has an oxidation state of +1 when bonded to non-metals, and -1 when bonded to metals.
• The sum of oxidation states for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

These rules help us determine changes in oxidation state in a reaction to identify which elements are oxidized and which are reduced.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. It divides the overall reaction into two separate half-reactions: one for oxidation and one for reduction. Each half-reaction is balanced separately, where we consider both the conservation of charge and the conservation of mass.

The key steps in this method include:

1. Assign oxidation numbers to determine what is oxidized and what is reduced.
2. Separate the equation into two half-reactions.
3. Balance the atoms in each half-reaction. Begin with elements other than O and H, balance O using H_2O, and H using H^+ (in acidic solutions) or OH^- (in basic solutions).
4. Balance the charge by adding electrons (e^-) to the more positive side to make the charge equal on both sides of the half-reaction.
5. Equalize the number of electrons transferred in both half-reactions by multiplying the half-reactions by appropriate coefficients.
6. Add the half-reactions back together and simplify to get the overall balanced equation.

This method ensures that the final equation reflects the conservation of both mass and charge.

Identifying Oxidizing and Reducing Agents

In redox reactions, we come across substances that cause oxidation and reduction, known as oxidizing and reducing agents, respectively. The reducing agent is the species that donates electrons to another species; it gets oxidized in the process. Conversely, the oxidizing agent is the species that accepts electrons; it gets reduced as the reaction proceeds.

To identify these agents, look for changes in oxidation states of the reactants. Here are some key points to remember:

• The substance that gets oxidized (increases in oxidation state) is the reducing agent.
• The substance that gets reduced (decreases in oxidation state) is the oxidizing agent.
• Focus on the reactants when searching for oxidizing and reducing agents, as their roles are defined before the reaction occurs.

For example, in the reaction step mentioned, NO_2^- is the reducing agent because its oxidation state increases from +3 to +5 upon becoming NO_3^-, and Cr_2O_7^{2-} is the oxidizing agent as it reduces its oxidation state from +6 to +3 when it turns into Cr^{3+}. Identifying these agents is crucial for understanding the directional flow of electrons in the reaction.