/*! 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 82 Balance these redox equations by... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 82

Balance these redox equations by any method. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{P}+\mathrm{H}_{2} \mathrm{O}+\mathrm{HNO}_{3} \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{NO}} \\ {\text { b. } \mathrm{KClO}_{3}+\mathrm{HCl} \rightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{KCl}}\end{array} \end{equation}

Short Answer

Expert verified
The balanced equations are: (a) \(\text{P} + 4 \text{HNO}_3 + 3 \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 4 \text{NO}\); (b) \(2 \text{KClO}_3 + 6 \text{HCl} \rightarrow 3 Cl_2 + 2 ClO_2 + 3 \text{H}_2\text{O} + 2 KCl\).

Step by step solution

01

Break down the reaction into half-reactions (a)

For reaction (a), identify the oxidation and reduction parts. Phosphorus (P) is oxidized to \(\text{H}_3\text{PO}_4\), and \(\text{HNO}_3\) is reduced to \(\text{NO}\). Write these as half-reactions: \(\text{P} \rightarrow \text{H}_3\text{PO}_4\) and \(\text{HNO}_3 \rightarrow \text{NO}\).
02

Balance the atoms other than oxygen and hydrogen (a)

For phosphorus: \(\text{P} \rightarrow \text{H}_3\text{PO}_4\) includes 1 P atom. It's already balanced.For nitrogen: \(\text{HNO}_3 \rightarrow \text{NO}\) includes 1 N atom. It's balanced.
03

Balance oxygen atoms (a)

For the oxidation half-reaction: Add \(4\ \text{H}_2\text{O}\) to the right side to balance 4 O in \(\text{H}_3\text{PO}_4\): \(\text{P} + \text{4 H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4\). For the reduction: Each \(\text{HNO}_3\) has 3 O, so this part needs adjustment within the next steps.
04

Balance hydrogen atoms using H+ (a)

Now balance using \(\text{H}^+\). In the oxidation half-reaction: \(\text{P} + \text{4 H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 4\ \text{H}^+\). For reduction: Include \(2\ \text{H}^+\) on the left side of \(\text{HNO}_3\rightleftharpoons \text{NO}\).
05

Balance charges using electrons (a)

For the oxidation half-reaction: Electrons balance by 0 = 4 by providing \(4 \text{e}^-\) on the left. In reduction: add electrons \(\text{HNO}_3 + \text{3 e}^-\rightarrow\text{NO}+\text{2 H}^+\).
06

Combine and balance electrons across half-reactions (a)

Multiply the reduction by 4 to equalize electrons. Combine balanced equations:\(\text{4HNO}_3 + \text{P} + \text{4H}_2\text{O} \rightarrow \text{4NO} + \text{H}_3\text{PO}_4 + 4\ \text{H}^+\), which balances electrons, atoms, and charges.
07

Break down the reaction into half-reactions (b)

For reaction (b), identify half-reactions: \(\text{KClO}_3\rightarrow\text{ClO}_2\) and \(\text{HCl}\rightarrow\text{Cl}_2\).
08

Balance atoms other than O and H (b)

For potassium: 1 K in \(\text{KClO}_3\) and \(\text{KCl}\) are balanced. For chlorine: match 2 \(\text{Cl}\) in \(\text{HCl}\) leading to \(\text{Cl}_2\). This holds for the \(\text{ClO}_2\) product.
09

Balance oxygen atoms (b)

In oxidation, add \(2 \text{O}_2\) on the left in \(\text{KClO}_3 \rightarrow \text{ClO}_2+O_2\), doubling for 3 \(\text{O}\).
10

Balance H using H+ and e- (b)

For \(\text{HCl}\rightarrow \text{Cl}_2\), adjust using electrons, adding \(1\ \text{e}^-\). Add hydrogen ions to the oxidation half: \(\text{2HCl} +\text{2Kb}+\text{2ClO}_2+\text{Cl}_2+\text{H}_2\text{O}\rightarrow\text{H}^++\text{E}\).
11

Balance charges via electrons (b)

Balance charges through electrons: The KClO_3 and HCl reactions each release electrons in processes.
12

Combine and verify balanced equation (b)

Combine reactions: \(2 KClO_3 + 2 HCl \rightarrow 2 Cl_2 + 2 ClO_2 + 2 H_2 O + 2 KCl\). Ensure all atoms and charges equal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a crucial skill in understanding chemistry, especially redox reactions. It ensures that the number of atoms for each element is the same on both sides of the equation. This is based on the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Here鈥檚 how to approach balancing equations for redox reactions:

  • Identify reactants and products: Start by writing down the correct formulas for all reactants and products involved. This is the basic skeleton of your chemical equation.
  • List each type of atom and count them on both sides of the equation: Often, it鈥檚 helpful to start with the atoms that appear in a single reactant or product.
  • Adjust coefficients: Use whole numbers as coefficients to balance each type of atom on both sides. Remember, you only change coefficients, not subscripts, to balance the equation.
  • Check your work: Once you believe the equation is balanced, recount all the atoms for each element to ensure they're equal across both sides. Also, double-check that the charges are balanced if applicable.
Balancing chemical equations requires practice and careful attention to detail, especially the charges and number of electrons in redox reactions.
Oxidation-Reduction
Oxidation and reduction reactions, collectively known as redox reactions, are processes that involve the transfer of electrons between species. Understanding these processes is key to mastering redox balancing. Here's what you need to know:

  • Oxidation involves the loss of electrons. When a species is oxidized, it loses electrons, which results in an increase in oxidation state.
  • Reduction involves the gain of electrons. When a species is reduced, it gains electrons, causing its oxidation state to decrease.
  • Redox reactions are coupled: Oxidation and reduction occur simultaneously. One species loses electrons (oxidized), while another species gains those electrons (reduced).
  • Identifying redox pairs: Look at changes in oxidation states of elements in the chemical reaction to determine which are oxidized and which are reduced.
Understanding the changes in oxidation states helps you to break down complex reactions into simpler internal processes. This breakdown is essential for balancing redox reactions effectively.
Half-Reactions
Half-reactions are a critical part of balancing redox equations. By focusing on each part of the reaction separately, they simplify the balancing process.

  • Definition: A half-reaction is either the oxidation or reduction reaction component of a redox process.
  • Oxidation half-reaction: It illustrates how electrons are lost in the reaction. For instance, phosphorus oxidizing to phosphoric acid can be shown through a half-reaction.
  • Reduction half-reaction: It demonstrates the gain of electrons. An example is nitrate reducing to nitrogen oxide within the reaction.
  • Balancing steps: For both parts, ensure that separate steps balance elements other than oxygen and hydrogen first, then balance oxygen through water and hydrogen with hydrogen ions.
  • Combine and balance: After writing and balancing each half-reaction, you combine them, ensuring that the number of electrons lost equals the number of electrons gained.
Using half-reactions, students can focus solely on one process of electron transfer at a time, making it clearer to understand overall changes in the reaction.
Electrons Transfer
The transfer of electrons is at the heart of redox reactions, defining their nature and driving the chemical changes.

  • Electron transfer defines redox reactions: These involve movement of electrons from the reducing agent (which loses electrons) to the oxidizing agent (which gains electrons).
  • Significance of electron balance: While balancing redox equations, ensuring that electrons lost equals electrons gained is vital. This balance preserves charge neutrality.
  • Visualizing electron flow: Imagining electrons moving from one species to another can simplify understanding of redox reactions. Drawing diagrams or using models might help visualize these transfers.
  • Role of electrons in oxidation state changes: As electrons are transferred, oxidation states change. Monitoring these changes ensures correct balancing of the reaction.
Mastering electron transfer gives students the tools to predict and balance redox reactions with confidence. Observing how electrons move enables a deeper grasp of chemical behaviors and transformations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Use the half-reaction method to balance these equations. Add water molecules and hydrogen ions (in acid solutions) or hydroxide ions (in basic solutions) as needed. Keep balanced equations in net ionic form. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})} \\ {\text { (in acid solution) }} \\ {\text { b. } \mathrm{IO}_{3}-(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(1)+\mathrm{IBr}(\mathrm{s}) \\ {\text { (in acid solution) }}} \\ {\text { c. } \mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq})} \\\ {(\text { in acid solution })} \end{array} \end{equation}

Balance these net ionic equations for redox reactions. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Au}^{3+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{Au}(\mathrm{s})+\mathrm{I}_{2}(\mathrm{s})} \\ {\text { b. } \mathrm{Ce}^{4+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Ce}^{3+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq})}\end{array} \end{equation}

Describe why it is important to know the conditions under which an aqueous oxidation-reducation reaction takes place in order to balance the ionic equation for the reaction.

Use the oxidation-number method to balance the following net ionic redox equations. $$ \text {Challenge} \quad \mathrm{I}-(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{s})+\mathrm{MnO}_{2}(\mathrm{s}) \quad \text {(in basic solution)} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.