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Balance each of the following redox reactions in basic solution. (a) \(\mathrm{Cl}_{2}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{ClO}_{3}^{-}(\mathrm{aq})\) (b) \(\cdot \mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s})\) (c) \(\mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{CN}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{CNO}^{-}(\mathrm{aq})\)

Short Answer

Expert verified
(a) \(\mathrm{Cl}_{2}+\mathrm{OH}^{-}\rightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}\) (b) \(\mathrm{MnO}_{4}^{-}+\mathrm{I}^{-}+\mathrm{OH}^{-}\rightarrow \mathrm{IO}_{3}^{-}+\mathrm{MnO}_{2}\) (c) \(\mathrm{ClO}_{3}^{-}+\mathrm{CN}^{-}\rightarrow \mathrm{Cl}^{-}+\mathrm{CNO}^{-}\)

Step by step solution

01

Write Unbalanced Half-Reactions for Reaction (a)

For the reaction \(\mathrm{Cl}_{2} \rightarrow \mathrm{Cl}^{-} + \mathrm{ClO}_{3}^{-}\), split into two half-reactions:1. \(\mathrm{Cl}_{2} \rightarrow \mathrm{Cl}^{-}\)2. \(\mathrm{Cl}_{2} \rightarrow \mathrm{ClO}_{3}^{-}\).
02

Balance Atoms Other Than O and H in Reaction (a)

Since the \(\mathrm{Cl}_{2} \rightarrow \mathrm{Cl}^{-}\) half-reaction already has balanced chlorine atoms, focus on oxygen:- Add 3 \(\mathrm{H}_2\mathrm{O}\) to the right to balance oxygens in \(\mathrm{Cl}_{2} \rightarrow \mathrm{ClO}_{3}^{-}\): \(\mathrm{Cl}_{2} + 3\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{ClO}_{3}^{-}\).
03

Balance Oxygen with Water and Hydrogen with Hydroxide in Reaction (a)

For \(\mathrm{Cl}_{2} + 3\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{ClO}_{3}^{-}\):- Add 6 \(\mathrm{OH}^{-}\) to the left to balance the hydrogen, forming water: \(\mathrm{Cl}_{2} + 6\mathrm{OH}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O}\).- For \(\mathrm{ClO}_{3}^{-} \rightarrow\): Add 5 \(\mathrm{OH}^{-}\) to the left to form balance with waters formed on the right.
04

Balance Charges with Electrons in Reaction (a)

For \(\mathrm{Cl}_{2} + 6\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_{2}\mathrm{O}\),- Add 6 electrons to the right: \(\mathrm{Cl}_{2} + 6\mathrm{OH}^{-} + 6e^- \rightarrow 2\mathrm{Cl}^{-} + 3\mathrm{H}_{2}\mathrm{O}\).- For \(\mathrm{Cl}_{2} + 5\mathrm{OH}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O}\), add electrons so: \(6e^- + \mathrm{Cl}_{2} + 5\mathrm{OH}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O}+ e^- \).
05

Add the Half-Reactions for Reaction (a)

Combine the two balanced half-reactions and cancel the electrons: \[\mathrm{Cl}_{2} + 6\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O}\]This is the balanced reaction for (a).
06

Separate and Write Unbalanced Half-Reactions for Reaction (b)

For the reaction \(\mathrm{MnO}_{4}^{-} + \mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-} + \mathrm{MnO}_{2}\):1. \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{MnO}_{2}\)2. \(\mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-}\).
07

Balance Atoms in Reaction (b)

- For \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{MnO}_{2}\): Add 2 \(\mathrm{H}_2\mathrm{O}\) to the right and 4 \(\mathrm{OH}^{-}\) to the left.- For \(\mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-}\) add 3 \(\mathrm{H}_2\mathrm{O}\) to the left and add 6 \(\mathrm{OH}^{-}\) to the right.
08

Balance Charges with Electrons in Reaction (b)

- For \(\mathrm{MnO}_{4}^{-} + 4\mathrm{e}^- + 4\mathrm{OH}^{-} \rightarrow \mathrm{MnO}_{2} + 2\mathrm{H}_2\mathrm{O}\).- For \(\mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-}\), add 6 electrons to the left: \(\mathrm{I}^{-} + 6\mathrm{OH}^{-} \rightarrow \mathrm{IO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O}+ 6e^- \).
09

Add the Half-Reactions for Reaction (b)

Combine the two balanced half-reactions and cancel the electrons:\[\mathrm{MnO}_{4}^{-} + \mathrm{I}^{-} + 2\mathrm{OH}^{-} \rightarrow \mathrm{IO}_{3}^{-} + \mathrm{MnO}_{2} + \mathrm{H}_{2}\mathrm{O}\]This is the balanced reaction for (b).
10

Separate and Write Unbalanced Half-Reactions for Reaction (c)

For the reaction \(\mathrm{ClO}_{3}^{-} + \mathrm{CN}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{CNO}^{-}\):1. \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\)2. \(\mathrm{CN}^{-} \rightarrow \mathrm{CNO}^{-}\).
11

Balance Atoms in Reaction (c)

- For \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\): Add 3 \(\mathrm{H}_2\mathrm{O}\) to right and 6 \(\mathrm{OH}^{-}\) to left.- For \(\mathrm{CN}^{-} \rightarrow \mathrm{CNO}^{-}\), add a \(\mathrm{H}_2\mathrm{O}\) molecule to the left and \(\mathrm{OH}^-\) to the right.
12

Balance Charges with Electrons in Reaction (c)

- For \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\), add 6 electrons to left: \(\mathrm{ClO}_{3}^{-} + 6e^- + 3\mathrm{H}_2\mathrm{O}+ 6\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + 6\mathrm{OH}^{-}\).- For \(\mathrm{CN}^{-} \rightarrow \mathrm{CNO}^{-}\), no additional electrons are needed, as charge is already balanced.
13

Add the Half-Reactions for Reaction (c)

Combine the two balanced half-reactions and cancel the electrons:\[\mathrm{ClO}_{3}^{-} + \mathrm{CN}^{-} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{CNO}^{-} + \mathrm{H}_2\mathrm{O} \]This is the balanced reaction for (c).

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Key Concepts

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

Chemical Balancing
Chemical balancing is a vital process in chemistry that ensures the law of conservation of mass is followed in reactions. This means that the number of atoms of each element must be the same on both sides of a chemical equation. During balancing redox reactions, specific steps must be followed to ensure everything is correctly accounted for. Initially, all the reactants and products are written, then each element except oxygen and hydrogen is balanced. Oxygen atoms are typically balanced by adding water molecules, and hydrogen atoms are balanced by adding protons or hydroxide ions, depending on the solution's conditions (acidic or basic).

Balancing these atoms often requires changing coefficients in a reaction without altering the fundamental identity of any substance. Once the atoms are balanced, the charges on both sides of the equation are balanced separately, usually using electrons. This method ensures the reaction is in compliance with both mass and charge conservation, making chemical balancing essential for precise chemical reactions and stoichiometry.
Basic Solution
In a basic solution, the pH is above 7, indicating a higher concentration of hydroxide ions (\(\text{OH}^-\)). This environment affects how we balance redox reactions differently than in acidic solutions. When balancing in basic conditions, after splitting the reaction into half-reactions, any hydrogen atoms that need to be balanced are done so using hydroxide ions rather than protons.

For example, if you encounter additional hydrogen atoms during the process, you would add hydroxide ions to one side of the equation to maintain the balance, which will then usually form water with any existing protons, simplifying the equation. Maintaining the correct ratio of hydroxide ions is crucial for reactions in a basic medium as they directly affect the overall charge and balance of the reaction.
Half-Reactions
The half-reaction method is a strategy used to balance redox reactions. It involves dividing the original reaction into two separate reactions—one represents oxidation, and the other represents reduction. The purpose of this method is to simplify the balancing of complex reactions by concentrating on one process at a time.

Through half-reactions, you can see how electrons are transferred from one reactant to another, identifying which species loses electrons (oxidation) and which gains electrons (reduction). After writing the half-reactions, each is balanced for mass and charge separately. Often, this requires adding water molecules and hydroxide ions in basic solutions to account for oxygen and hydrogen atoms. Ultimately, the half-reactions are recombined to form a balanced overall equation, with electrons canceling out during the combination step.
Oxidation States
Oxidation states are numbers assigned to elements in chemical compounds that describe their degree of oxidation. They help in identifying which atoms in a reaction are undergoing oxidation or reduction, making them crucial in balancing redox reactions. An increase in oxidation state indicates oxidation, whereas a decrease suggests reduction.

For each element in a compound, a unique oxidation state is determined based on a set of rules, such as the oxidation state of a free element being zero, or oxygen generally having an oxidation state of -2 in compounds. Steps in a redox reaction may include determining the oxidation states of all involved elements to identify which are oxidized and which are reduced. Understanding oxidation states helps predict and balance redox reactions more easily, improving accuracy in chemical analysis and synthesis.

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Most popular questions from this chapter

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In the analytical technique called electrogravimetry, electrolysis is used to separate the analyte from a solution by depositing it on an inert electrode. The electrode is weighed before and after the experiment to find the mass of analyte deposited. A \(0.122-\mathrm{g}\) sample of a copper-zinc alloy was treated with concentrated sulfuric acid to produce a solution containing copper(II) and zinc(II) sulfates. The platinum cathode used in the electrolysis of this solution increased in mass by \(0.073 \mathrm{~g}\) after exhaustive electrolysis. (a) Which metal was deposited on the cathode during the electrolysis? Write the balanced equation for the electrolysis reaction. (b) What was the mass percentages of copper and zinc in the alloy sample?

The electrochemical processes that occur in the corrosion of iron are represented by the half-reactions $$ \begin{array}{l} \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \\ \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{array} $$ (a) Write the overall reaction. (b) What is the standard potential for the overall chemical reaction? (c) Natural water has a pH of about \(5.9,\) and air is 0.21 mol fraction oxygen. If the concentration of iron(II) in the water is \(5 \times 10^{-5} M\), what is the potential of the corrosion reaction in the presence of air and natural water at 1 atm pressure and \(298 \mathrm{~K} ?\)

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