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Chapter 20: Problem 22

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) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)-\cdots\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(\mathrm{s})+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{MnO}_{4}^{-(a q)}+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(a q)\) (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

Expert verified
(a) Short answer: Balanced equation: \(2Cr_{2}O_{7}^{2-} + 6NO_{2}^{-} + 14H^{+} \rightarrow 2Cr^{3+} + 6NO_{3}^{-} + 7H_{2}O\). Oxidizing agent: \(Cr_{2}O_{7}^{2-}\), Reducing agent: \(NO_{2}^{-}\) (b) Short answer: Balanced equation: \(3S + 2HNO_{3} \rightarrow 3H_{2}SO_{3} + N_{2}O\). Oxidizing agent: \(HNO_{3}\), Reducing agent: \(S\) (c) Short answer: Balanced equation: \(2Cr_{2}O_{7}^{2-} + 6CH_{3}OH + 14H^{+} \rightarrow 2Cr^{3+} + 6HCO_{2}H + 7H_{2}O\). Oxidizing agent: \(Cr_{2}O_{7}^{2-}\), Reducing agent: \(CH_{3}OH\) (d) Short answer: Balanced equation: \(MnO_{4}^{-} + 5Cl^{-} + 8H^{+} \rightarrow Mn^{2+} + 2.5Cl_{2} + 4H_{2}O\). Oxidizing agent: \(MnO_{4}^{-}\), Reducing agent: \(Cl^{-}\) (e) Short answer: Balanced equation: \(2Al + 3NO_{2}^{-} \rightarrow 2AlO_{2}^{-} + 3NH_{4}^{+}\). Oxidizing agent: \(NO_{2}^{-}\), Reducing agent: \(Al\) (f) Short answer: Balanced equation: \(2H_{2}O_{2} + 4ClO_{2} \rightarrow 6ClO_{2}^{-} + 4H_{2}O\). Oxidizing agent: \(ClO_{2}\), Reducing agent: \(H_{2}O_{2}\)

Step by step solution

01

Identify the change in oxidation states

For NO鈧傗伝, the nitrogen has an oxidation state of +3, and for NO鈧冣伝, it has an oxidation state of +5. For Cr鈧侽鈧嚶测伝, chromium has an oxidation state of +6, and for Cr鲁鈦, it has an oxidation state of +3.
02

Balance half-reactions

Write two half-reactions: 1) Balance the nitrogen atoms: NO鈧傗伝 鈫 NO鈧冣伝 2) Balance the chromium atoms: Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦 Now, balance the electrons: 1) NO鈧傗伝 鈫 NO鈧冣伝 + 2e鈦 2) Cr鈧侽鈧嚶测伝 + 6e鈦 鈫 2Cr鲁鈦
03

Combine and balance the final equation

To combine these half-reactions, we need to balance the electrons. We'll multiply the first half-reaction by 3 and then add the half-reactions: 3(NO鈧傗伝 鈫 NO鈧冣伝 + 2e鈦) Cr鈧侽鈧嚶测伝 + 6e鈦 鈫 2Cr鲁鈦 ------------------------ 2Cr鈧侽鈧嚶测伝 + 6NO鈧傗伝 鈫 2Cr鲁鈦 + 6NO鈧冣伝 Finally, balance the solutions charge by adding H鈦 ions: 2Cr鈧侽鈧嚶测伝 + 6NO鈧傗伝 + 14H鈦 鈫 2Cr鲁鈦 + 6NO鈧冣伝 + 7H鈧侽
04

Identify the oxidizing and reducing agents

The oxidizing agent is the species that is being reduced and causing the oxidation. In this case, it's Cr鈧侽鈧嚶测伝. The reducing agent is the species that is being oxidized and causing the reduction. In this case, it's NO鈧傗伝. For the other reactions, we'll follow the same steps: (b) S(s) + HNO鈧(aq)鈫 H鈧係O鈧(aq) + N鈧侽(g) (acidic solution) Half-reactions: S + 2e鈦 鈫 S虏鈦 2HNO鈧 + 6e鈦 鈫 N鈧侽 + 3H鈧侽 Balanced equation: 3S + 2HNO鈧 鈫 3H鈧係O鈧 + N鈧侽 Oxidizing agent: HNO鈧 Reducing agent: S (c) Cr鈧侽鈧嚶测伝(aq) + CH鈧僌H(aq) 鈫 HCO鈧侶(aq) + Cr鲁鈦(aq) (acidic solution) Half-reactions: Cr鈧侽鈧嚶测伝 + 6e鈦 + 14H鈦 鈫 2Cr鲁鈦 + 7H鈧侽 CH鈧僌H 鈫 HCO鈧侶 + 2e鈦 Balanced equation: 2Cr鈧侽鈧嚶测伝 + 6CH鈧僌H + 14H鈦 鈫 2Cr鲁鈦 + 6HCO鈧侶 + 7H鈧侽 Oxidizing agent: Cr鈧侽鈧嚶测伝 Reducing agent: CH鈧僌H (d) MnO鈧勨伝(aq) + Cl鈦(aq) 鈫 Mn虏鈦(aq) + Cl鈧(aq) (acidic solution) Half-reactions: MnO鈧勨伝 + 5e鈦 + 8H鈦 鈫 Mn虏鈦 + 4H鈧侽 2Cl鈦 鈫 Cl鈧 + 2e鈦 Balanced equation: MnO鈧勨伝 + 5Cl鈦 + 8H鈦 鈫 Mn虏鈦 + 2.5Cl鈧 + 4H鈧侽 Oxidizing agent: MnO鈧勨伝 Reducing agent: Cl鈦 (e) NO鈧傗伝(aq) + Al(s) 鈫 NH鈧勨伜(aq) + AlO鈧傗伝(aq) (basic solution) Balance under basic conditions with OH鈦 and H鈧侽: Half-reactions: NO鈧傗伝 + 3H鈧侽 鈫 NH鈧勨伜 + 4OH鈦 Al 鈫 AlO鈧傗伝 + 3H鈧侽 + 4e鈦 Balanced equation: 2Al + 3NO鈧傗伝 鈫 2AlO鈧傗伝 + 3NH鈧勨伜 Oxidizing agent: NO鈧傗伝 Reducing agent: Al (f) H鈧侽鈧(aq) + ClO鈧(aq) 鈫 ClO鈧傗伝(aq) + O鈧(g) (basic solution) Balance under basic conditions with OH鈦 and H鈧侽: Half-reactions: H鈧侽鈧 + 2OH鈦 鈫 2H鈧侽 + O鈧 2ClO鈧 + 2OH鈦 鈫 2ClO鈧傗伝 + H鈧侽 + O鈧 Balanced equation: 2H鈧侽鈧 + 4ClO鈧 鈫 6ClO鈧傗伝 + 4H鈧侽 Oxidizing agent: ClO鈧 Reducing agent: H鈧侽鈧

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Key Concepts

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

Oxidation States
Understanding oxidation states is crucial for balancing redox reactions. An oxidation state, or oxidation number, is a signed integer representing the number of electrons an atom either gains or loses to form a chemical bond. When an atom loses electrons, we say it has been oxidized, which increases its oxidation state. Conversely, when an atom gains electrons, it is reduced, and its oxidation state decreases.

In the given exercise, identifying the change in oxidation states is the first step toward solving redox reactions. For instance, in reaction (a), nitrogen's oxidation state changes from +3 in NO鈧傗伝 to +5 in NO鈧冣伝, indicating an oxidation process. Contrastingly, chromium's oxidation state decreases from +6 in Cr鈧侽鈧嚶测伝 to +3 in Cr鲁鈦, signifying a reduction process.

Remember that some atoms can have an 'atypical' oxidation state, such as oxygen in hydrogen peroxide (H鈧侽鈧). Normally, oxygen has an oxidation state of -2, but in H鈧侽鈧, it is -1. Recognizing atypical states is necessary for accurately balancing equations and determining the oxidizing and reducing agents in redox reactions.
Half-Reaction Method
The half-reaction method is a systematic approach for balancing redox reactions. It involves separating the overall chemical equation into two half-reactions: one for oxidation and another for reduction. Each half-reaction represents the fate of one of the species involved in the redox process.

As seen in the second step of solving the exercise, each half-reaction must be balanced separately. This involves not just balancing the elements involved, but also ensuring that the charges on both sides are equal, primarily by adding electrons. Balancing half-reactions for reaction (a) goes like this:
  1. NO鈧傗伝 鈫 NO鈧冣伝 (oxidation)
  2. Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦 (reduction)
After balancing for atoms and charge, we then add the half-reactions back together and balance the overall equation for mass and charge, ensuring the number of electrons lost in oxidation equals those gained in reduction. The result is a stoichiometrically and charge consistent balanced redox equation.
Oxidizing and Reducing Agents
Finalizing the redox balancing process includes identifying oxidizing and reducing agents. The oxidizing agent is a substance that takes electrons from another substance during the redox reaction, thus being reduced itself. In contrast, the reducing agent donates electrons and becomes oxidized.

In part (a) of the exercise, Cr鈧侽鈧嚶测伝 acts as the oxidizing agent as its oxidation state is reduced from +6 to +3. NO鈧傗伝 is the reducing agent, its oxidation state increases from +3 to +5. Identifying these agents is essential in understanding the redox process, as it provides insights into the reaction's direction and the roles each species plays.

The oxidizing and reducing agents drive the chemical changes in redox reactions. Their identification not only enables us to balance reactions but also to predict the products and reactants in various chemical environments, be it acidic or basic.

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Most popular questions from this chapter

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l)--\rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (b) \(\mathrm{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots \rightarrow \mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)--\rightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g) $$ (d) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s)--\rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to \(\mathrm{Sn}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\left.\mathrm{I}^{-}\right)\), (b) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{I}^{-}\left(\right.\) to form \(\left.\mathrm{I}_{2}\right),(\mathrm{c})\) reduction of \(\mathrm{Ce}^{4+}\) to \(\mathrm{Ce}^{3+}\) by \(\mathrm{H}_{2} \mathrm{O}_{2}\), (d) reduction of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) by \(\mathrm{Sn}^{2+}\) (to form \(\mathrm{Sn}^{4+}\) ).

Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are \(\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\) (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-}--\rightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two cell compartments have \(\left[\mathrm{Cl}^{-} \mathrm{J}=0.0150 \mathrm{M}\right.\) and \(\left[\mathrm{Cl}^{-}\right]=2.55 M\), respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl \(^{-}\) ] will increase, decrease, or stay the same as the cell operates.
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