/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Balance the following oxidation-... [FREE SOLUTION] | 91影视

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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鈦 + ClO鈦 鈫 2I鈧冣伝 + Cl鈦 b. As鈧侽鈧 + 6NO鈧冣伝 鈫 2H鈧傾sO鈧 + 6NO c. 10Br鈦 + 2MnO鈧勨伝 鈫 5Br鈧 + 2Mn虏鈦 d. 3CH鈧僌H + Cr鈧侽鈧嚶测伝 鈫 3CH鈧侽 + 2Cr鲁鈦

Step by step solution

01

Identify half-reactions

: Oxidation: I鈦 鈫 I鈧冣伝 Reduction: ClO鈦 鈫 Cl鈦
02

Balance atoms

: Oxidation: 3I鈦 鈫 I鈧冣伝 Reduction: ClO鈦 鈫 Cl鈦 (no changes needed)
03

Balance charges

: Oxidation: Already balanced. Reduction: Add 2 electrons to the right side: ClO鈦 鈫 Cl鈦 + 2e鈦
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鈦 鈫 I鈧冣伝) + (ClO鈦 鈫 Cl鈦 + 2e鈦) 6I鈦 + ClO鈦 鈫 2I鈧冣伝 + Cl鈦 + 2e鈦
05

Simplify and Check

: 6I鈦 + ClO鈦 鈫 2I鈧冣伝 + Cl鈦 The reaction is balanced. b. As鈧侽鈧(s) + NO鈧勨伝(aq) 鈫 H鈧傾sO鈧(aq) + NO(g)
06

Identify half-reactions

: Oxidation: As鈧侽鈧 鈫 H鈧傾sO鈧 Reduction: NO鈧勨伝 鈫 NO
07

Balance atoms

: Oxidation: As鈧侽鈧 鈫 2H鈧傾sO鈧 Reduction: NO鈧冣伝 鈫 NO (no changes needed)
08

Balance charges

: Oxidation: Add 6 electrons to the right side: As鈧侽鈧 鈫 2H鈧傾sO鈧 + 6e鈦 Reduction: Add 2 electrons to the right side: NO鈧冣伝 鈫 NO + e鈦
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. (As鈧侽鈧 鈫 2H鈧傾sO鈧 + 6e鈦) + 6(NO鈧冣伝 鈫 NO + e鈦) As鈧侽鈧 + 6NO鈧冣伝 鈫 2H鈧傾sO鈧 + 6NO + 6e鈦
10

Simplify and Check

: As鈧侽鈧 + 6NO鈧冣伝 鈫 2H鈧傾sO鈧 + 6NO The reaction is balanced. c. Br鈦(aq) + MnO鈧勨伝(aq) 鈫 Br鈧(l) + Mn虏鈦(aq)
11

Identify half-reactions

: Oxidation: Br鈦 鈫 Br鈧 Reduction: MnO鈧勨伝 鈫 Mn虏鈦
12

Balance atoms

: Oxidation: 2Br鈦 鈫 Br鈧 Reduction: MnO鈧勨伝 鈫 Mn虏鈦 (no changes needed)
13

Balance charges

: Oxidation: Add 2 electrons to the right side: 2Br鈦 鈫 Br鈧 + 2e鈦 Reduction: Add 5 electrons to the left side: 5e鈦 + MnO鈧勨伝 鈫 Mn虏鈦
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鈦 鈫 Br鈧 + 2e鈦) + 2(5e鈦 + MnO鈧勨伝 鈫 Mn虏鈦) 10Br鈦 + 2MnO鈧勨伝 鈫 5Br鈧 + 2Mn虏鈦 + 10e鈦
15

Simplify and Check

: 10Br鈦 + 2MnO鈧勨伝 鈫 5Br鈧 + 2Mn虏鈦 The reaction is balanced. d. CH鈧僌H(aq) + Cr鈧侽鈧嚶测伝(aq) 鈫 CH鈧侽(aq) + Cr鲁鈦(aq)
16

Identify half-reactions

: Oxidation: CH鈧僌H 鈫 CH鈧侽 Reduction: Cr鈧侽鈧嚶测伝 鈫 Cr鲁鈦
17

Balance atoms

: Oxidation: CH鈧僌H 鈫 CH鈧侽 (no changes needed) Reduction: Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦
18

Balance charges

: Oxidation: Add 2 electrons to the right side: CH鈧僌H 鈫 CH鈧侽 + 2e鈦 Reduction: Add 6 electrons to the left side: 6e鈦 + Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦
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(CH鈧僌H 鈫 CH鈧侽 + 2e鈦) + (6e鈦 + Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦) 3CH鈧僌H + Cr鈧侽鈧嚶测伝 鈫 3CH鈧侽 + 2Cr鲁鈦 + 6e鈦
20

Simplify and Check

: 3CH鈧僌H + Cr鈧侽鈧嚶测伝 鈫 3CH鈧侽 + 2Cr鲁鈦 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鈥檚 mass and charge are exactly balanced, exemplifying the harmonious interplay of matter.

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Most popular questions from this chapter

Direct methanol fuel cells (DMFCs) have shown some promise as a viable option for providing "green" energy to small electrical devices. Calculate \(\mathscr{E}^{\circ}\) for the reaction that takes place in DMFCs: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+3 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Use values of \(\Delta G_{\mathrm{f}}^{\circ}\) from Appendix \(4 .\)

Gold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-}\). Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathrm{O}_{2}\) will be produced during the production of \(1.00 \mathrm{~mole}\) of gold?

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

Calculate the \(\mathrm{pH}\) of the cathode compartment for the following reaction given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\) when \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), and \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

You have a concentration cell with Cu electrodes and [Cu^{2+} ] \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) by the following equation: \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) \(K=1.0 \times 10^{13}\) Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

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