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Chapter 19: Problem 79

Balance these ionic redox equations by any method. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Sb}^{3+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{SbO}_{4}^{3-}+\mathrm{Mn}^{2+}(\text { in acid solution })} \\ {\text { b. } \mathrm{N}_{2} \mathrm{O}+\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{NO}_{2}^{-} \text { (in basic solution) }}\end{array} \end{equation}

Short Answer

Expert verified
a) \[5\mathrm{Sb}^{3+} + 3\mathrm{MnO}_4^- + 4\mathrm{H}_2\mathrm{O} \rightarrow 5\mathrm{SbO}_4^{3-} + 3\mathrm{Mn}^{2+} + 8\text{H}^+\]. b) \[2\mathrm{N}_2\mathrm{O} + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NO}_2^- + \mathrm{Cl}^- + 2\mathrm{OH}^-\].

Step by step solution

01

Identify Oxidation and Reduction Parts

For reaction (a): \[ \mathrm{Sb}^{3+} \rightarrow \mathrm{SbO}_4^{3-} \] is the oxidation half-reaction, and \[ \mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+} \] is the reduction half-reaction. For reaction (b): \[ \mathrm{N}_2\mathrm{O} \rightarrow \mathrm{NO}_2^{-} \] is the oxidation half-reaction, and \[ \mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-} \] is the reduction half-reaction.
02

Balance Atoms Other Than O and H

For reaction (a), as Sb and Mn atoms are already balanced, no action is needed here. For reaction (b), N and Cl atoms are balanced in their half-reactions.
03

Balance Oxygen by Adding H2O

For reaction (a), add \( 4\text{ H}_2\text{O} \) to the right side of the oxidation reaction. For reaction (b), add \( \text{H}_2\text{O} \) to the left side of the oxidation reaction.
04

Balance Hydrogen by Adding H+

For reaction (a), add \( 8\text{ H}^+\) ions to the left side of the oxidation reaction.For reaction (b), add \( 2\text{ H}_2\text{O} \) to the right side of the reduction half-reaction to balance hydrogen created from adding water to the left side.
05

Balance Charge by Adding Electrons

For reaction (a): - Add \( 3\text{ e}^鈭� \) to the left of the oxidation equation: \[ \mathrm{Sb}^{3+} + 4\text{ H}_2\text{O} \rightarrow \mathrm{SbO}_4^{3-} + 8\text{ H}^+ + 3\text{ e}^- \] - Add \( 5\text{ e}^- \) to the right of the reduction: \[ \mathrm{MnO}_4^- + 8\text{ H}^+ + 5\text{ e}^- \rightarrow \mathrm{Mn}^{2+} + 4\text{ H}_2\text{O} \] For reaction (b): - Add \( 4\text{ e}^{-} \) to the right of the oxidation equation: \[\mathrm{N}_2\mathrm{O}+\text{H}_2\text{O} \rightarrow \mathrm{NO}_2^{-} + 4\text{ e}^{-} + 2\text{ H}^+\]- Add \( 2\text{ e}^- \) to the left of the reduction: \[ \mathrm{ClO}^- + 2\text{ e}^- + \text{H}_2\text{O} \rightarrow \mathrm{Cl}^- + 2\text{ OH}^- \]
06

Equalize Electrons and Combine Half-Reactions

For reaction (a):- Multiply the oxidation half-reaction by \( 5 \) and the reduction half-reaction by \( 3 \) to equalize electrons:\[5(\mathrm{Sb}^{3+} + 4\text{ H}_2\text{O} \rightarrow \mathrm{SbO}_4^{3-} + 8\text{ H}^+ + 3\text{ e}^-)\]\[3(\mathrm{MnO}_4^- + 8\text{ H}^+ + 5\text{ e}^- \rightarrow \mathrm{Mn}^{2+} + 4\text{ H}_2\text{O})\]- Combine to get the overall balanced equation.For reaction (b):-Multiply the oxidation half-reaction by \(2\) and the reduction half-reaction by \(1\) to equalize electrons:\[2(\mathrm{N}_2\mathrm{O}+\text{H}_2\text{O} \rightarrow \mathrm{NO}_2^{-} + 4\text{ e}^{-} + 2\text{ H}^+ )\]\[1(\mathrm{ClO}^- + 2\text{ e}^- + \text{H}_2\text{O} \rightarrow \mathrm{Cl}^- + 2\text{ OH}^-) \]- Combine to get the overall balanced equation.
07

Write the Balanced Equations

For reaction (a), the balanced equation in acidic solution is:\[5\mathrm{Sb}^{3+} + 3\mathrm{MnO}_4^- + 4H_2O \rightarrow 5\mathrm{SbO}_4^{3-} + 3\mathrm{Mn}^{2+} + 8\text{H}^+\]For reaction (b), the balanced equation in basic solution is:\[2\mathrm{N}_2\mathrm{O} + \mathrm{ClO}^- + \text{H}_2\text{O} \rightarrow 2\mathrm{NO}_2^{-} + \mathrm{Cl}^- + 2\text{OH}^-\]

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Key Concepts

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

Ionic Equations
Ionic equations are a way to represent chemical reactions that occur in aqueous solutions. Instead of showing the compounds as whole, ionic equations break down the reactants and products into their constituent ions. This is particularly useful for reactions that involve electrolytes. To write an ionic equation, follow these steps:
  • Identify the strong electrolytes and dissociate them into their ions. Weak electrolytes, non-electrolytes, and solids remain undissociated.
  • Write the balanced molecular equation first, then separate the dissociated species into their ions.
  • Cancel out the spectator ions which do not participate in the chemical change, leaving only the ions involved in the reaction. This results in the net ionic equation.
Ionic equations help students get an insight into what is actually happening on the molecular level during a reaction, focusing only on the components that are altered.
Balancing Equations
Balancing chemical equations is crucial for the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance an equation, ensure the same number and types of atoms are on both sides. Here鈥檚 how to do it effectively:
  • Start by balancing the atoms of metals first, then move to non-metals.
  • Balance hydrogen and oxygen atoms last, as they often appear in multiple compounds.
  • Use coefficients to balance the number of molecules, not by altering subscripts in chemical formulas.
  • In redox reactions, ensure the charge is balanced across the equation as well.
Balancing ensures that the equation accurately represents the substances and their amounts used and produced in the reaction.
Oxidation-Reduction
Oxidation-reduction (redox) reactions are processes in which electrons are transferred between chemical species. One species gains electrons (reduction) and the other loses electrons (oxidation). Understanding these reactions involves recognizing the change in oxidation states:
  • Oxidation occurs when a substance loses electrons, increasing its oxidation state (e.g., \( ext{Fe}^{2+} \rightarrow ext{Fe}^{3+} + ext{e}^- \)).
  • Reduction occurs when a substance gains electrons, decreasing its oxidation state (e.g., \( ext{Cu}^{2+} + 2 ext{e}^- \rightarrow ext{Cu} \)).
  • An oxidizing agent facilitates oxidation by accepting electrons, while a reducing agent facilitates reduction by donating electrons.
  • The sum of the oxidation states in a compound or molecule must equal its charge.
Recognizing the species being oxidized and reduced is key for understanding how electrons are shifted in a reaction.
Half-Reaction Method
The half-reaction method is a systematic way to balance redox reactions, especially in complex equations. This method breaks down the overall reaction into two separate equations鈥攁 reduction half and an oxidation half鈥攁llowing for straightforward balancing. Here's how to use the half-reaction method:
  • Identify and write the separate oxidation and reduction half-reactions.
  • Balance all atoms except hydrogen and oxygen first.
  • Balance oxygen by adding water molecules and hydrogen using \( ext{H}^+ \) ions or \( ext{OH}^- \) in basic solutions.
  • Balance the charge by adding electrons to either side of the half-reaction.
  • Multiply the half-reactions by appropriate coefficients so that the electrons lost and gained are equal.
  • Add the balanced half-reactions together, ensuring that electrons cancel out, resulting in the net balanced equation.
The half-reaction method allows tackling the complex nature of redox reactions by dealing with smaller, simpler parts that lead up to the overall balanced equation.

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Most popular questions from this chapter

Determine the oxidation number of nitrogen in each of these molecules or ions. $$ \text { a. }\mathrm{NO}_{3} \quad \text { b. } \mathrm{N}_{2} \mathrm{O} \quad \text { c. } \mathrm{NF}_{3} $$

Label each half-reaction as reduction or oxidation. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-}} \\ {\text { b. } \mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. } 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}} \\ {\text { d. } \mathrm{F}_{2} \rightarrow 2 \mathrm{F}^{-}+2 \mathrm{e}^{-}}\end{array} \end{equation}

Copper was a useful metal even before iron, silver, and gold metals were extracted and used from their ores and used as tools, utensils, jewelry, and artwork. Copper was smelted by heating copper ores with charcoal to high temperatures as early as 8000 years ago. Thousands of pieces of scrap copper have been unearthed in Virginia, where in the 1600 s the colonists might have traded this material for food. Compare and contrast the processing and use of copper in those older civilizations with today.

The following equations show redox reactions that are sometimes used in the laboratory to generate pure nitrogen gas and pure dinitrogen monoxide gas (nitrous oxide, \(\left.\mathrm{N}_{2} \mathrm{O}\right)\). \(\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{~s}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) a. Determine the oxidation number of each element in the two equations, and then make diagrams showing the changes in oxidation numbers that occur in each reaction. b. Identify the atom that is oxidized and the atom that is reduced in each of the two reactions. c. Identify the oxidizing and reducing agents in each of the two reactions. d. Write a sentence telling how the electron transfer taking place in these two reactions differs from that taking place here. \(2 \mathrm{AgNO}_{3}+\mathrm{Zn} \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\)

In terms of electrons, what happens when an atom is oxidized? When an atom is reduced?
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