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Half Equations: The Foundation of Redox Chemistry
Dive into the world of half equations and master the art of analyzing redox reactions. Our comprehensive guide simplifies complex concepts, helping you excel in chemistry with confidence.

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  2. Examples0/5 watched
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Now Playing:Half equations – Example 0a
Intros
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  1. Using half-equations
  2. Using half-equations
    Balancing half equations: Example 1.
  3. Using half-equations
    Balance half equations in basic conditions: Example1.
Examples
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  1. Balance the half-equations in the conditions stated.
    Balance the half-equations below for the given conditions.
    1. SO42-   \, →  \, S2- using acidic conditions.

    2. Cr2O72-   \, →  \, Cr3+ using basic conditions.

    3. NO3-   \, →  \, NH4+ using acidic conditions.

    4. CrO42-   \, →  \, Cr3+ using basic conditions

    5. S2O32-   \, →  \, SO42- using acidic conditions.

Introduction to electrochemistry
Jump to:NotesConceptExampleFAQsPrerequisites
Notes

In this lesson, we will learn:

  • How to complete and balance half equations in basic and acidic conditions.
  • How to use half equations to balance full redox equations.

Notes:

  • Because redox reactions always involve both reduction and oxidation, to understand them more easily the whole reaction can be split into two half-reactions or half-equations. Half-equations are the two reduction and oxidation processes written separately, as if they were occurring alone.
    Redox exam questions normally ask you to complete a redox equation, but only give you the major elements (anything except O and H) to start. They may or may not already be split into half-equations. Either way, you need to add everything else to complete it.
    This lesson will show you how to balance half-equations so that they are ready to be combined for the full redox equation. There is one worked example for this here; more practice on this last part is in Balancing redox equations.

  • The most straightforward way to do this is to split a redox reaction into half-equations and complete these steps in order, using a worked example:

    MnO4- \enspace → \enspace Mn2+

    • First, balance the major elements. This is any atom that isnt oxygen or hydrogen. There is normally only one input reactant and output product for these, so just increase the number of moles to balance.
      In our example, manganese is already balanced so no change is necessary:

      MnO4- \enspace → \enspace Mn2+

    • Next, balance oxygen. Do this by adding H2O molecules (remember redox reactions nearly always occur in aqueous solution!).
      There are currently four oxygen atoms in the reactants and none in the products, so we add four H2O molecules there:

      MnO4- \enspace → \enspace Mn2+ + 4H2O

    • Next, balance hydrogen. This is done by adding H+ (aq) ions.
      There are currently eight hydrogens in the products and none in the reactants, so we add eight H+ there:

      MnO4- + 8H+ \enspace → \enspace Mn2+ + 4H2O

    • Finish balancing charge by adding electrons (e-).
      The reactants have an overall +7 charge while the products have an overall +2 charge, so 5 electrons (a total of 5- charge) needs to be added to the reactants to bring it down to +2:

      MnO4- + 8H+ + 5e- \enspace → \enspace Mn2+ + 4H2O

    This is the complete half equation and is accurate for acidic conditions. As always, here you can manually check the number of atoms and charge on each side to be sure.
    If the reaction is in basic conditions then you must make a change because H+ will not be present, OH- will be. will be. To balance a half equation in basic conditions from here, you need to use the autoionization of water equilibrium:

    H+ + OH- ⇌\enspace \rightleftharpoons \enspace H2O

    Multiply this equation by the number of H+ in your balanced half equation and insert it so that H+ from the half-equation and H+ from the autoionization of water equilibrium will be on opposite sides. This will cancel the terms out and leave you with just H2O and OH- in the half-equation, which is correct for basic conditions!
    Our example (the water equilibrium compounds are highlighted in red ) looks like this:

    MnO4- + 8H+ + 5e- + 8H2O \enspace → \enspace Mn2+ + 4H2O + 8H+ + 8OH-

    This leaves us with equal H+ and some H2O on both sides of the equation. Cancel it out:

    MnO4- + 8H+ + 5e- + 8 4H2O \enspace → \enspace Mn2+ + 4H2O + 8H+ + 8OH-

    This gives us the correct half equation in basic conditions:

    MnO4- + 5e- + 4H2O \enspace → \enspace Mn2+ + 8OH-


  • WORKED EXAMPLE 2: Balance in basic conditions.

    MnO2 \enspace → \enspace Mn2+

    • First, balance major atoms:

      MnO2 \enspace → \enspace Mn2+

    • Next, balance oxygen. 

      MnO2 \enspace → \enspace Mn2+ + 2H2O

    • Next, balance hydrogen. 

      MnO2 + 4H+\enspace → \enspace Mn2+ + 2H2O

    • Balance by adding e-.

      MnO2 + 4H+ + 2e-\enspace → \enspace Mn2+ + 2H2O

    • As this is in basic conditions, again we need to cancel out the H+ which will not be present we need to balance this using the dissociation of water equation.

      H+ + OH- ⇌\enspace \rightleftharpoons \enspace H2O

      There are four H+ in our current half-equation:

      MnO2 + 4H+ + 2e-\enspace → \enspace Mn2+ + 2H2O

      We therefore need to multiply the water equation four times and insert it in to our half-equation to cancel out the four H+ already present.

      MnO2 + 4H+ + 4H2O + 2e-\enspace → \enspace Mn2+ + 2H2O + 4H+ + 4OH-

      This allows us to cancel the H+ that was in our half equation already and the H+ from the water equation we just added.

      MnO2 + 4H+ + 4H2O + 2e-\enspace → \enspace Mn2+ + 2H2O + 4H+ + 4OH-

      This gives a final equation:

      MnO2 + 2H2O + 2e-\enspace → \enspace Mn2+ + 4OH-

  • WORKED EXAMPLE 3: With the ability to build correct half-equations, we can now combine two half-equations and make full equations. This is done by checking the number of electrons in each half-equation.
    Like mass, charge is conserved in a reaction, so the number of electrons in the reactants must equal the number in the products!
    To combine half-equations into a full equation, multiply the half equations until you have equal electrons on both sides, then combine them. Using an example, with two half equations:

    MnO4- + 8H+ + 5e-\enspace → \enspace Mn2+ + 4H2O
    \qquad \qquad \qquad \enspace \enspace Os + 4H2O \enspace → \enspace OsO4 + 8H+ + 8e- \quad (acidic solution)

    Remember that electrons lost equal electrons gained in a redox reaction, so we must have an equal amount on both sides. To get this, we need to multiply the equations until we get a common number in our example, we have to find a number that both 5 and 8 go into!

    \qquad MnO4- + 8H+ + 5e- \enspace → \enspace Mn2+ + 4H2O \qquad x8
    \qquad \quadOs + 4H2O \enspace → \enspace OsO4 + 8H+ + 8e- \qquad \enspace x5

    Multiplying by these gives us 40 electrons in each half equation these will now cancel when we combine the equation:

    8MnO4- + 64H+ + 40e- + 5Os + 20H2O \enspace → \enspace 8Mn2+ + 32H2O + 5OsO4 + 40H+ + 40e-

    Cancel the common species:

    8MnO4- + 64 24H+ + 40e- + 5Os + 20H2O \enspace → \enspace 8Mn2+ + 32 12H2O + 5OsO4 + 40H+ + 40e-

    This gives the final balanced equation:

    8MnO4- + 24H+ + 5Os + \enspace → \enspace 8Mn2+ + 12H2O + 5OsO4
Concept

Introduction to Half Equations

Welcome to the fascinating world of half equations! These essential components of redox reactions are crucial in understanding electron transfer in chemical processes. Half equations break down complex redox reactions into simpler, more manageable parts, making it easier to grasp the flow of electrons between species. Our introduction video serves as an excellent starting point, providing a clear and concise explanation of this fundamental concept. As we delve deeper into half equations, you'll discover their significance in various chemical reactions, from battery operations to corrosion processes. By mastering half equations, you'll gain a powerful tool for balancing redox reactions and predicting chemical behavior. Remember, understanding half equations is like unlocking a secret code in chemistry it opens up a whole new level of comprehension. So, let's embark on this exciting journey together and unravel the mysteries of half equations!

Example

Using half-equations
Balancing half equations: Example 1.

Step 1: Introduction to Half Equations

Half equations are a way to split a full redox equation into two separate parts, each representing either the oxidation or reduction process. This method helps in balancing complex redox reactions by focusing on one part of the reaction at a time. In this guide, we will learn how to balance half equations step-by-step, which is essential for understanding and solving full redox equations.

Step 2: Balancing Major Atoms

The first step in balancing a half equation is to balance the major atoms. Major atoms are any atoms that are not oxygen or hydrogen. For example, consider the half equation where manganate (MnO4-) is reduced to manganese ion (Mn2+). Here, manganese is the major atom. Since there is one manganese atom on both sides of the equation, it is already balanced in terms of major atoms.

Step 3: Balancing Oxygen Atoms

The next step is to balance the oxygen atoms. This is done by adding H2O molecules to the side that needs oxygen. In our example, MnO4- has four oxygen atoms, while Mn2+ has none. To balance the oxygen atoms, we add four H2O molecules to the side with Mn2+. This gives us four oxygen atoms on both sides of the equation.

Step 4: Balancing Hydrogen Atoms

Adding H2O molecules in the previous step introduces hydrogen atoms that need to be balanced. In our example, adding four H2O molecules introduces eight hydrogen atoms. To balance these hydrogen atoms, we add eight H+ ions to the side with MnO4-. This ensures that the number of hydrogen atoms is balanced on both sides of the equation.

Step 5: Balancing Charge

The final step is to balance the charge of the half equation by adding electrons (e-). In our example, the left side of the equation has a charge of -1 (from MnO4-) and +8 (from eight H+), totaling +7. The right side has a charge of +2 (from Mn2+). To balance the charges, we need to add five electrons to the left side, making the total charge on both sides equal.

Step 6: Final Balanced Half Equation

After following the steps above, the balanced half equation for the reduction of manganate to manganese ion is:
MnO4- + 8H+ + 5e- Mn2+ + 4H2O
This equation shows the reduction process, as indicated by the gain of electrons on the reactant side.

Step 7: Understanding Reduction and Oxidation

In half equations, the placement of electrons indicates whether a reduction or oxidation reaction is occurring. Electrons on the reactant side signify a reduction reaction, as the substance gains electrons. Conversely, electrons on the product side indicate an oxidation reaction, where the substance loses electrons. In our example, the presence of electrons on the reactant side confirms that it is a reduction reaction.

Step 8: Summary of Steps

To summarize, balancing a half equation involves the following steps:

  1. Balance the major atoms (excluding oxygen and hydrogen).
  2. Balance the oxygen atoms by adding H2O molecules.
  3. Balance the hydrogen atoms by adding H+ ions.
  4. Balance the charge by adding electrons (e-).
Following these steps ensures that both mass and charge are balanced in the half equation, providing a clear and accurate representation of the redox process.

FAQs

1. How do you write a half-reaction?
To write a half-reaction, start by identifying the species involved in either oxidation or reduction. Write the reactant and product, then balance all atoms except hydrogen and oxygen. Add H2O to balance oxygen atoms, H+ to balance hydrogen atoms, and finally, add electrons to balance the charge.

2. How do you find the balanced half equation?
To balance a half equation, follow these steps: 1) Write the unbalanced equation with key species. 2) Balance all atoms except H and O. 3) Add H2O to balance O atoms. 4) Add H+ to balance H atoms. 5) Add electrons to balance the charge. Always check that both atoms and charges are balanced in the final equation.

3. What is the difference between a half equation and an ionic equation?
A half equation specifically shows either the oxidation or reduction process in a redox reaction, including electron transfer. An ionic equation shows the complete reaction with all ions involved but doesn't necessarily focus on electron transfer. Half equations are used to analyze redox reactions, while ionic equations represent general solution reactions.

4. What is the half equation for chlorine?
The half equation for the reduction of chlorine to chloride ions is:
Cl2 + 2e- 2Cl-
This equation shows chlorine gas (Cl2) gaining two electrons to form two chloride ions (Cl-).

5. Why are half equations important in electrochemistry?
Half equations are crucial in electrochemistry because they help understand the separate processes occurring at each electrode in an electrochemical cell. They allow for the calculation of standard electrode potentials, prediction of spontaneous reactions, and analysis of complex redox processes in batteries, corrosion, and electrolysis.

Prerequisites

Understanding half equations is crucial in chemistry, particularly when dealing with redox reactions and electrochemistry. To fully grasp this concept, it's essential to have a solid foundation in several prerequisite topics. Let's explore how these topics relate to half equations and why they're so important.

First and foremost, calculating cell potential in voltaic cells is closely linked to half equations. This knowledge helps you understand how individual half-reactions contribute to the overall redox process. By mastering cell potential calculations, you'll be better equipped to analyze and predict the behavior of half equations in electrochemical systems.

Another critical prerequisite is balancing redox equations. Half equations are the building blocks of complete redox reactions, and being able to balance these equations is essential. This skill allows you to ensure that electron transfer is accounted for correctly, which is at the heart of understanding half equations.

Before diving into redox-specific balancing, it's crucial to have a strong grasp of balancing chemical equations in general. This fundamental skill provides the basis for more complex balancing techniques used in half equations and redox reactions. It ensures you can maintain atom and charge conservation, which is vital when working with half equations.

Interestingly, graphing from slope-intercept form, while seemingly unrelated, can be valuable when studying half equations. This mathematical skill helps in visualizing and interpreting data related to redox potentials and reaction rates, which are often represented graphically in electrochemistry studies involving half equations.

Understanding these prerequisite topics is not just about memorizing facts; it's about building a comprehensive framework for tackling half equations. Each topic contributes uniquely to your overall understanding. For instance, knowing how to calculate cell potentials allows you to predict the spontaneity of reactions described by half equations. Balancing skills ensure you can correctly represent the electron transfer in these equations, which is crucial for understanding their role in redox reactions.

Moreover, the ability to balance general chemical equations provides the foundational logic needed when dealing with the more complex half equations. This skill helps you maintain the conservation of mass and charge, which is paramount in electrochemistry. Lastly, graphing skills can aid in visualizing trends and relationships in redox potentials, enhancing your ability to interpret and apply half equations in various contexts.

By mastering these prerequisite topics, you'll find that understanding half equations becomes much more intuitive. You'll be able to approach problems with confidence, seeing how different aspects of chemistry and mathematics come together in the study of redox reactions and electrochemistry. Remember, each of these topics builds upon the others, creating a robust knowledge base that will serve you well in your chemistry studies and beyond.