Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Short Answer

Expert verified
Solutions: a. 6I⁻ + ClO⁻ → 2I₃⁻ + Cl⁻ b. As₂O₃ + 6NO₃⁻ → 2H₃AsO₄ + 6NO c. 10Br⁻ + 2MnO₄⁻ → 5Br₂ + 2Mn²⁺ d. 3CH₃OH + Cr₂O₇²⁻ → 3CH₂O + 2Cr³⁺

Step by step solution

01

Identify half-reactions

: Oxidation: I⁻ → I₃⁻ Reduction: ClO⁻ → Cl⁻
02

Balance atoms

: Oxidation: 3I⁻ → I₃⁻ Reduction: ClO⁻ → Cl⁻ (no changes needed)
03

Balance charges

: Oxidation: Already balanced. Reduction: Add 2 electrons to the right side: ClO⁻ → Cl⁻ + 2e⁻
04

Combine half-reactions

: To combine, multiply the oxidation half-reaction by 2, so that the number of electrons exchanged is the same in both half-reactions. 2(3I⁻ → I₃⁻) + (ClO⁻ → Cl⁻ + 2e⁻) 6I⁻ + ClO⁻ → 2I₃⁻ + Cl⁻ + 2e⁻
05

Simplify and Check

: 6I⁻ + ClO⁻ → 2I₃⁻ + Cl⁻ The reaction is balanced. b. As₂O₃(s) + NO₄⁻(aq) → H₃AsO₄(aq) + NO(g)
06

Identify half-reactions

: Oxidation: As₂O₃ → H₃AsO₄ Reduction: NO₄⁻ → NO
07

Balance atoms

: Oxidation: As₂O₃ → 2H₃AsO₄ Reduction: NO₃⁻ → NO (no changes needed)
08

Balance charges

: Oxidation: Add 6 electrons to the right side: As₂O₃ → 2H₃AsO₄ + 6e⁻ Reduction: Add 2 electrons to the right side: NO₃⁻ → NO + e⁻
09

Combine half-reactions

: To combine, multiply the reduction half-reaction by 6, so that the number of electrons exchanged is the same in both half-reactions. (As₂O₃ → 2H₃AsO₄ + 6e⁻) + 6(NO₃⁻ → NO + e⁻) As₂O₃ + 6NO₃⁻ → 2H₃AsO₄ + 6NO + 6e⁻
10

Simplify and Check

: As₂O₃ + 6NO₃⁻ → 2H₃AsO₄ + 6NO The reaction is balanced. c. Br⁻(aq) + MnO₄⁻(aq) → Br₂(l) + Mn²⁺(aq)
11

Identify half-reactions

: Oxidation: Br⁻ → Br₂ Reduction: MnO₄⁻ → Mn²⁺
12

Balance atoms

: Oxidation: 2Br⁻ → Br₂ Reduction: MnO₄⁻ → Mn²⁺ (no changes needed)
13

Balance charges

: Oxidation: Add 2 electrons to the right side: 2Br⁻ → Br₂ + 2e⁻ Reduction: Add 5 electrons to the left side: 5e⁻ + MnO₄⁻ → Mn²⁺
14

Combine half-reactions

: To combine, multiply the oxidation half-reaction by 5 and the reduction one by 2, so that the number of electrons exchanged is the same in both half-reactions. 5(2Br⁻ → Br₂ + 2e⁻) + 2(5e⁻ + MnO₄⁻ → Mn²⁺) 10Br⁻ + 2MnO₄⁻ → 5Br₂ + 2Mn²⁺ + 10e⁻
15

Simplify and Check

: 10Br⁻ + 2MnO₄⁻ → 5Br₂ + 2Mn²⁺ The reaction is balanced. d. CH₃OH(aq) + Cr₂O₇²⁻(aq) → CH₂O(aq) + Cr³⁺(aq)
16

Identify half-reactions

: Oxidation: CH₃OH → CH₂O Reduction: Cr₂O₇²⁻ → Cr³⁺
17

Balance atoms

: Oxidation: CH₃OH → CH₂O (no changes needed) Reduction: Cr₂O₇²⁻ → 2Cr³⁺
18

Balance charges

: Oxidation: Add 2 electrons to the right side: CH₃OH → CH₂O + 2e⁻ Reduction: Add 6 electrons to the left side: 6e⁻ + Cr₂O₇²⁻ → 2Cr³⁺
19

Combine half-reactions

: To combine, multiply the oxidation half-reaction by 3, so that the number of electrons exchanged is the same in both half-reactions. 3(CH₃OH → CH₂O + 2e⁻) + (6e⁻ + Cr₂O₇²⁻ → 2Cr³⁺) 3CH₃OH + Cr₂O₇²⁻ → 3CH₂O + 2Cr³⁺ + 6e⁻
20

Simplify and Check

: 3CH₃OH + Cr₂O₇²⁻ → 3CH₂O + 2Cr³⁺ The reaction is balanced.

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Key Concepts

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

Half-Reaction Method
The half-reaction method is a systematic approach for balancing redox (oxidation-reduction) reactions, which occur in all forms of chemistry, especially in acidic and basic solutions. The method involves separating the oxidation and reduction parts of a reaction to easily balance the atoms and charges involved.

During the oxidation process, an element loses electrons and increases its oxidation state, whereas reduction signifies the gain of electrons and a decrease in oxidation state. Once the half-reactions are identified, they are balanced individually for mass and charge. In acidic solutions, hydrogen ions (H+) and water (H2O) are often added to balance oxygen and hydrogen atoms. Finally, the half-reactions are recombined, and electrons are added or subtracted to balance the overall charge, ensuring that the total number of electrons lost in oxidation equals the number of electrons gained in reduction.

For instance, in a reaction where bromide ions (Br-) are oxidized to molecular bromine (Br2) and permanganate ions (MnO4-) are reduced to manganese ions (Mn2+), it is crucial to balance not only the elements involved but also the charges, to achieve a completely balanced redox reaction.
Oxidation Number
An oxidation number, also known as an oxidation state, is an indicator of the degree of oxidation of an atom in a chemical compound. This concept plays an integral role in the half-reaction method of balancing redox reactions. Understanding oxidation numbers allows students to identify which species are oxidized and reduced in the reaction.

For a given atom, the oxidation number is based on a set of rules, such as the oxidation state of an atom in a pure element is zero, oxygen is usually -2 except in peroxides where it's -1, and hydrogen is +1 when bonded to nonmetals and -1 when bonded to metals. In a balanced reaction, the sum of the oxidation numbers must remain constant; a change in oxidation number indicates a redox process. For instance, the oxidation number of iodide (I-) increases from -1 to 0 when it forms triiodide (I3-), signifying oxidation, while the oxidation number of chlorate ion (ClO-) decreases from +1 to -1 when it forms chloride ion (Cl-), indicating reduction.
Acidic Solution Redox
Redox reactions in acidic solutions often require special consideration as the presence of H+ ions (protons) may participate directly in the redox process. This involvement is particularly pertinent when the balance of oxygen or hydrogen atoms is necessary. In acidic solutions, H ions and water molecules are used to balance O and H atoms respectively, which is not the case in neutral or basic solutions.

To illustrate, if a reaction is taking place in an acidic solution and oxygen needs to be balanced, water molecules are added to the side deficient in oxygen, and H ions are added to the other side to balance the added hydrogen from water. This technique stabilizes the half-reactions by counterbalancing the charges that oxygen and hydrogen contribute. When we balance the reaction of arsenic trioxide (As2O3) with nitrate ion (NO3-) to form arsenic acid (H3AsO4) and nitric oxide (NO) within an acidic medium, we invoke such principles to maintain stoichiometric and charge balance amid the overall redox transformation.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It encompasses the calculation of masses, volumes, and concentrations of substances involved. Moreover, stoichiometry is pivotal in ensuring that reactions are balanced not only for the atoms but also for the amount of matter.

In the context of redox reactions, stoichiometry ensures that the number of atoms of each element is the same on both sides of the equation and that the total charge is balanced. When half-reactions are combined, the coefficients of reactants and products must be multiplied by appropriate factors to balance the electrons. This careful adjustment ensures the overall reaction adheres to the Law of Conservation of Mass and Charge. For instance, in the reaction where methanol (CH3OH) is oxidized to formaldehyde (CH2O) and dichromate ion (Cr2O72-) is reduced to chromium ion (Cr3+), we apply stoichiometric principles to guarantee the reaction’s mass and charge are exactly balanced, exemplifying the harmonious interplay of matter.

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Most popular questions from this chapter

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. \(\mathrm{Ag}\) j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M\). Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{~kJ}\). What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

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