Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Balance the following oxidation-reduction reactio by the half-reaction method. (a) $$ \mathrm{CuCl}_{2}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{Cu}(s) $$ (b) $$ \mathrm{Cr}^{3+}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Cr}(s)+\mathrm{Zn}^{2+}(a q) $$

Short Answer

Expert verified
(a) \(2 \text{Al} + 3 \text{CuCl}_{2} \rightarrow 2 \text{AlCl}_{3} + 3 \text{Cu}\); (b) \(3 \text{Zn} + 2 \text{Cr}^{3+} \rightarrow 3 \text{Zn}^{2+} + 2 \text{Cr}\).

Step by step solution

01

Identify Oxidation and Reduction Half-Reactions in (a)

In the given reaction \( ext{CuCl}_{2} + ext{Al} \rightarrow ext{AlCl}_{3} + ext{Cu}\), determine which species are oxidized and reduced:- **Oxidation half-reaction**: \( ext{Al} \rightarrow ext{Al}^{3+} + 3e^{-}\) - Aluminum transforms from a pure element to AlCl3, gaining a positive charge.- **Reduction half-reaction**: \( ext{Cu}^{2+} + 2e^{-} \rightarrow ext{Cu}\) - Copper in \( ext{CuCl}_{2}\) is reduced, losing its positive charge to become elemental Cu.
02

Balance Each Half-Reaction in (a)

Adjust each half-reaction so that the number of electrons lost in oxidation equals the electrons gained in reduction.- **Oxidation half-reaction**: \[ ext{2 Al} \rightarrow 2 ext{Al}^{3+} + 6 e^{-}\]- **Reduction half-reaction**: \[3 ext{Cu}^{2+} + 6 e^{-} \rightarrow 3 ext{Cu}\]
03

Combine Balanced Half-Reactions for (a)

Add the balanced half-reactions together, ensuring electrons are canceled:\[2 ext{Al} + 3 ext{Cu}^{2+} \rightarrow 2 ext{Al}^{3+} + 3 ext{Cu}\]Thus, the balanced reaction is:\[2 ext{Al} + 3 ext{CuCl}_{2} \rightarrow 2 ext{AlCl}_{3} + 3 ext{Cu}\]
04

Identify Oxidation and Reduction Half-Reactions in (b)

For the reaction \( ext{Cr}^{3+} + ext{Zn} \rightarrow ext{Cr} + ext{Zn}^{2+}\), determine which species are oxidized and reduced:- **Oxidation half-reaction**: \( ext{Zn} \rightarrow ext{Zn}^{2+} + 2e^{-}\) - Zinc goes from elemental to a Zn^2+ ion.- **Reduction half-reaction**: \( ext{Cr}^{3+} + 3e^{-} \rightarrow ext{Cr}\) - Chromium ions gain electrons to become elemental Cr.
05

Balance Each Half-Reaction in (b)

Adjust the half-reactions to have equal numbers of electrons involved:- **Oxidation half-reaction**: \[3 ext{Zn} \rightarrow 3 ext{Zn}^{2+} + 6e^{-}\]- **Reduction half-reaction**: \[2 ext{Cr}^{3+} + 6e^{-} \rightarrow 2 ext{Cr}\]
06

Combine Balanced Half-Reactions for (b)

Add the balanced oxidation and reduction equations:\[3 ext{Zn} + 2 ext{Cr}^{3+} \rightarrow 3 ext{Zn}^{2+} + 2 ext{Cr}\]This gives the balanced overall reaction.

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 Vaia!

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 technique used to balance oxidation-reduction (redox) reactions. It involves splitting the overall reaction into two distinct parts: oxidation and reduction half-reactions. Each part is handled separately and then combined to achieve a balanced equation.
  • Oxidation Half-Reaction: Represents loss of electrons.
  • Reduction Half-Reaction: Represents gain of electrons.
In this method, you deal with the electron transfer directly, ensuring that the electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction. This focus on electron balance sets it apart from other methods of balancing equations, making it ideal for redox reactions.
Balancing Chemical Equations
Balancing chemical equations is essential for accurately representing chemical reactions. In the context of redox reactions, using the half-reaction method ensures that you account for both the mass and charge balance. The principle involves adjusting coefficients so that the number of atoms for each element in the reactants equals those in the products.
For half-reactions:
  • Balance atoms other than O and H.
  • Balance O atoms by adding water molecules.
  • Balance H atoms using hydrogen ions.
  • Balance charge by adding electrons.
After balancing each half-reaction separately, they are added together, ensuring electrons are canceled out. This results in a balanced overall equation for the redox reaction.
Redox Reactions
Redox reactions are chemical processes involving the transfer of electrons between two substances. The term "redox" is derived from two concepts: reduction and oxidation. These reactions are central to many biological and industrial processes, such as energy production, corrosion, and metabolism. In a redox reaction, one substance undergoes oxidation (loses electrons) while the other undergoes reduction (gains electrons).
Identifying redox reactions involves:
  • Determining changes in oxidation states of elements.
  • Recognizing electron transfer between reactants.
  • Using the half-reaction method to simplify balancing.
Redox reactions are balanced for the electrons transferred, making the half-reaction method particularly useful.
Oxidation Reactions
Oxidation reactions are a type of redox reaction where a substance loses electrons, increasing its oxidation state. The substance undergoing oxidation donates these electrons to the reduction half-reaction. In the oxidation process:
  • The oxidized species turns into a positively charged ion or a less negative state.
  • Commonly involves metals like zinc and aluminum changing from a neutral atom to a charged ion.
An example is the oxidation of zinc metal to zinc ions (\( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \)). Recognizing oxidation is crucial in accurately balancing reactions and contributes to understanding energy transfer mechanisms in chemistry.
Reduction Reactions
Reduction reactions occur when a substance gains electrons and decreases its oxidation state. This is the second half of a redox reaction, balancing the process initiated by oxidation. The species that gains electrons is said to be "reduced," as its charge becomes more negative or less positive.
Some notable points about reduction reactions include:
  • Reduction often occurs with nonmetals, such as copper ions converting to copper metal (\( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \)).
  • It is essential for processes like battery operations, where reduction occurs at the cathode.
  • Balancing reduction reactions involves ensuring electron uptake matches the electrons released during oxidation.
Recognizing and balancing reduction helps understand redox reactions and ties into various chemical applications.

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 the molecular equation and the net ionic equation for the reaction of solid barium carbonate and hydrogen bromide in aqueous solution. Add phase labels.

A 10.0-mL sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to produce silver iodide crystals, which were filtered from the solution. $$ \mathrm{KI}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{KNO}_{3}(a q)+\mathrm{AgI}(s) $$ If \(2.183 \mathrm{~g}\) of silver iodide was obtained, what was the molarity of the original KI solution?

Potassium sulfate solution reacts with barium bromide solution to produce a precipitate of barium sulfate and a solution of potassium bromide. Write the molecular equation for this reaction. Then write the corresponding net ionic equation.

Complete and balance each of the following molecular equations, including phase labels, if a reaction occurs. Then write the net ionic equation. If no reaction occurs, write \(N R\) after the arrow. (a) \(\mathrm{HClO}_{4}+\mathrm{BaCO}_{3} \longrightarrow\) (b) \(\mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{Sr}(\mathrm{OH})_{2} \longrightarrow\) (c) \(\mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{MgCl}_{2} \longrightarrow\) (d) \(\mathrm{FeSO}_{4}+\mathrm{MgCl}_{2} \longrightarrow\)

When using \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in aqueous reactions, many problems are presented or solved by assuming both of the hydrogen ions to be strongly ionized, as if they were both strong acid ionization processes. At the same time, the second \(\mathrm{H}^{+}\) only weakly ionizes in aqueous solution. 1\. Write molecular, complete ionic, and net ionic equations for the reaction between sulfuric acid and barium hydroxide (a strong base) using the following assumptions: (a) Both hydrogen ions of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are assumed to ionize \(100 \%\) like strong acids. (b) Only the first hydrogen ion of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(100 \%\) ionized, and the second is weakly ionized. 2\. Write molecular, complete ionic, and net ionic equations for the reaction between sulfuric acid and sodium acetate, a weakly basic salt, using the following assumptions: Only the first hydrogen ion of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(100 \%\) ionized, and the second is weakly ionized.

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free