/*! 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 23 Balance the following redox equa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 23

Balance the following redox equations. (a) \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}_{2}(g)\) (acidic) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) (basic)

Short Answer

Expert verified
Question: Balance the following redox reactions: (a) Cu(s) + NO3^-(aq) 鈫� Cu^2+(aq) + NO2(g) (acidic) (b) Cr(OH)3(s) + ClO^-(aq) 鈫� CrO4^2-(aq) + Cl^-(aq) (basic) Answer: (a) Cu(s) + 2NO3^-(aq) + 4H+(aq) 鈫� Cu^2+(aq) + 2NO2(g) + 2H2O(l) (b) 2Cr(OH)3(s) + 4ClO^-(aq) 鈫� 2CrO4^2-(aq) + 4Cl^-(aq) + 2H2O(l)

Step by step solution

01

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\)
02

Balance the atoms, except for oxygen and hydrogen

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)\) (already balanced) Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\) (already balanced)
03

Balance the oxygen atoms by adding water molecules

Oxidation: No need to balance oxygen atoms Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)
04

Balance hydrogen atoms by adding protons

Oxidation: No need to balance hydrogen atoms Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 2H^+(aq) \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)
05

Balance charges by adding electrons

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 2H^+(aq) + e^- \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)
06

Make the number of electrons equal and combine half-reactions

\(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) \(\mathrm{2NO}_{3}^{-}(a q) + 4H^+(aq) + 2e^- \longrightarrow 2\mathrm{NO}_{2}(g) + 2H_2O (l)\) Combine: \(\mathrm{Cu}(s) + \mathrm{2NO}_{3}^{-}(a q) + 4H^+(aq) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2\mathrm{NO}_{2}(g) + 2H_2O (l)\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) (basic)
07

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\)
08

Balance the atoms, except for oxygen and hydrogen

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (already balanced) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (already balanced)
09

Balance the oxygen atoms by adding water molecules

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)
10

Balance hydrogen atoms by adding hydroxide ions

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6OH^-(aq)\) Reduction: \(\mathrm{ClO}^{-}(a q) + 2OH^-(aq) \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)
11

Balance charges by adding electrons

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) + 2e^- \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6OH^-(aq)\) Reduction: \(\mathrm{ClO}^{-}(a q) + 2OH^-(aq) + e^- \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)
12

Make the number of electrons equal and combine half-reactions

\(\mathrm{2Cr}(\mathrm{OH})_{3}(s) + 6H_2O (l) + 4e^- \longrightarrow \mathrm{2CrO}_{4}^{2-}(a q) + 12OH^-(aq)\) \(\mathrm{4ClO}^{-}(a q) + 8OH^-(aq) + 4e^- \longrightarrow \mathrm{4Cl}^{-}(a q) + 4H_2O(l)\) Combine: \(\mathrm{2Cr}(\mathrm{OH})_{3}(s) + 4\mathrm{ClO}^{-}(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 4\mathrm{Cl}^{-}(a q) + 2H_2O (l)\)

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.

Half-Reaction Method
The half-reaction method is an essential tool in chemistry for balancing redox (reduction-oxidation) reactions. Redox reactions involve the transfer of electrons between different chemical species. Using half-reactions can simplify the balancing process:
  • In a redox reaction, two half-reactions are identified: one for oxidation and one for reduction.
  • In each half-reaction, we separately consider the oxidation and reduction processes.
  • The oxidation half-reaction shows electrons being lost, while the reduction half-reaction shows electrons being gained.
For example, in our exercise:
  • Oxidation: The transformation from Cu(s) to Cu2+(aq).
  • Reduction: The transformation from NO3-(aq) to NO2(g).
By focusing individually on these processes, we can systematically balance each half-reaction concerning atoms and charge, making the final combination straightforward.
Oxidation and Reduction
Understanding oxidation and reduction is pivotal in chemistry, especially when discussing redox reactions. These processes involve electron transfer and are two halves of a redox reaction.
  • Oxidation occurs when a substance loses electrons, leading to an increase in oxidation state. For instance, Cu in our first example becomes Cu2+, losing two electrons.
  • Reduction, on the other hand, is the gain of electrons, resulting in a decrease in oxidation state. In the same reaction, NO3- gains electrons to become NO2.
To track these changes, we often use oxidation numbers and compare them before and after the reaction. Overall, one substance's oxidation accompanies another's reduction, highlighting the interconnected nature of these processes.
Balancing Chemical Equations
Balancing chemical equations ensures that the same number of each type of atom is present on both sides of a reaction, adhering to the law of conservation of mass. In the context of redox reactions, balancing also involves ensuring charge balance.
  • Start by writing the unbalanced equation and then separate it into half-reactions as mentioned earlier.
  • Balance all atoms except hydrogen and oxygen first.
  • Balance oxygen atoms by adding water (H2O) molecules.
  • Next, balance hydrogen atoms by adding protons (H+) or hydroxide ions (OH-) depending on the solution鈥檚 acidity or basicity.
  • Finally, balance the charge by adding electrons (e-).
Each step brings us closer to a balanced chemical equation, ensuring both the mass and charge are aligned, as seen in our examples, whether the environment is acidic or basic.

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

Write a balanced equation to represent (a) the roasting of nickel(II) sulfide to form nickel(II) oxide. (b) the reduction of nickel(II) oxide by carbon monoxide.

Write a balanced equation for the reaction that occurs when (a) iron(III) oxide is reduced with carbon monoxide. (b) the excess carbon in pig iron is removed by the basic oxygen process.

A \(0.500-g\) sample of steel is analyzed for manganese. The sample is dissolved in acid and the manganese is oxidized to permanganate ion. A measured excess of \(\mathrm{Fe}^{2+}\) is added to reduce \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\). The excess \(\mathrm{Fe}^{2+}\) is determined by titration with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{27}\). If \(75.00 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{FeSO}_{4}\) is added and the excess requires \(13.50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to oxidize \(\mathrm{Fe}^{2+}\), calculate the percent by mass of \(\mathrm{Mn}\) in the sample.

Show by calculation which of the following metals will react with hydrochloric acid (standard concentrations). (a) \(\mathrm{Cd}\) (b) \(\overline{\mathrm{Cr}}\) (c) Co (d) \(\mathrm{Ag}\) (c) Au

Write a balanced redox equation for the reaction of mercury with aqua regia, assuming the products include \(\mathrm{HgCl}_{4^{2-}}\) and \(\mathrm{NO}_{2}(g)\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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.