1. Home
  2. >Chemistry
  3. >Redox and Electrochemistry
Help

Half equations

Get the most by viewing this topic in your current grade. Pick your course now.

?
Intros
Lessons
  1. Using half-equations
  2. Balancing half equations: Example 1.
  3. Balance half equations in basic conditions: Example1.
  4. Balance half equations in basic conditions: Example 2.
  5. Combining half equations.
?
Examples
Lessons
  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.
Topic Notes
?

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!

Understanding Half Equations

What are Half Equations?

Half equations are a fundamental concept in chemistry, particularly in the study of redox reactions. These equations represent either the oxidation or reduction process occurring in a chemical reaction, but not both simultaneously. They are called "half" equations because they show only one part of the complete redox reaction.

The Purpose of Half Equations

Half equations serve several important purposes in chemistry:

  • They simplify complex redox reactions by breaking them down into manageable parts.
  • They help in balancing redox reactions accurately.
  • They provide a clear view of electron transfer in chemical processes.
  • They aid in understanding the separate oxidation and reduction processes.

Role in Oxidation and Reduction Processes

In redox reactions, oxidation and reduction always occur together. Half equations allow us to examine these processes separately:

  • Oxidation Half Equation: Shows the loss of electrons from a species.
  • Reduction Half Equation: Shows the gain of electrons by a species.

By combining these half equations, we can represent the complete redox reaction.

Writing Half Equations

To write a half equation, follow these steps:

  1. Write the formula of the reactant and product involved in the half-reaction.
  2. Balance all atoms except hydrogen and oxygen.
  3. Balance oxygen atoms by adding H2O to the appropriate side.
  4. Balance hydrogen atoms by adding H+ ions.
  5. Balance the charge by adding electrons to the more positive side.

Examples of Half Equations

Example 1: Oxidation of Iron
Fe Fe² + 2e
This half equation shows iron losing two electrons to form an iron(II) ion.

Example 2: Reduction of Oxygen
O + 4H + 4e 2HO
This half equation represents oxygen gaining four electrons and combining with hydrogen ions to form water.

Importance in Electrochemistry

Half equations are crucial in electrochemistry, particularly in understanding:

  • Electrode processes in galvanic and electrolytic cells
  • Standard electrode potentials
  • The direction of electron flow in electrochemical reactions

Balancing Redox Reactions Using Half Equations

To balance a complete redox reaction:

  1. Write separate half equations for oxidation and reduction.
  2. Balance each half equation independently.
  3. Multiply each half equation by a factor to equalize the number of electrons transferred.
  4. Add the two half equations together, canceling out common terms.

Conclusion

Half equations are an essential tool in chemistry for understanding and analyzing redox reactions. They provide a clear representation of electron transfer processes, simplify complex reactions, and are invaluable in electrochemistry and balancing redox equations. By mastering the concept of half equations, chemists can better comprehend and predict the behavior of oxidation-reduction reactions in various chemical systems.

Balancing Half Equations: Step-by-Step Guide

Welcome to our comprehensive guide on balancing half equations! This essential skill in chemistry can seem daunting at first, but with our step-by-step approach, you'll master it in no time. Let's dive in and explore how to balance atoms, charges, and electrons in half equations.

Step 1: Write the unbalanced half equation
Let's start with a simple example: the reduction of copper(II) ions to copper metal. Our unbalanced equation looks like this: Cu² Cu

Step 2: Balance the atoms (except H and O)
In our example, we already have one copper atom on each side, so we're good to go. For more complex equations, make sure the number of atoms (excluding hydrogen and oxygen) is the same on both sides.

Step 3: Balance oxygen atoms by adding HO
If your equation involves oxygen, add water molecules to the side that needs more oxygen. For instance, if we were balancing MnO Mn², we'd add 4 HO to the right side: MnO Mn² + 4HO

Step 4: Balance hydrogen atoms by adding H
To balance the hydrogen atoms introduced by water, add H ions to the opposite side. In our MnO example: MnO + 8H Mn² + 4HO

Step 5: Balance the charges by adding electrons
Now, let's focus on balancing the charges. Count the total charge on each side and add electrons to the more positive side to balance. In our copper example: Cu² + 2e Cu

For the MnO example: MnO + 8H + 5e Mn² + 4HO

Step 6: Check your work
Always double-check that your final equation has balanced atoms and charges on both sides. This step is crucial to ensure accuracy in your work.

Let's practice with another example: the oxidation of iron(II) to iron(III).
Step 1: Write the unbalanced equation: Fe² Fe³
Step 2: Atoms are already balanced
Step 3 & 4: No oxygen or hydrogen to balance
Step 5: Balance charges: Fe² Fe³ + e
Step 6: Check - all balanced!

Remember, in reduction half equations, electrons are added to the reactant side. In oxidation half equations, electrons appear on the product side. This is a key distinction to keep in mind as you practice.

Now, let's tackle a more complex example: the reduction of dichromate ions to chromium(III) ions.
Step 1: CrO² Cr³
Step 2: Balance chromium: CrO² 2Cr³
Step 3: Balance oxygen: CrO² 2Cr³ + 7HO
Step 4: Balance hydrogen: CrO² + 14H 2Cr³ + 7HO
Step 5: Balance charges: CrO² + 14H + 6e 2Cr³ + 7HO

Half Equations in Acidic and Basic Conditions

Half equations are essential tools in understanding redox reactions, but their representation can differ significantly between acidic and basic conditions. This difference primarily stems from the role of H+ ions in acidic environments and OH- ions in basic environments. Understanding these distinctions is crucial for chemistry students to accurately balance redox equations and predict reaction outcomes.

In acidic conditions, H+ ions play a central role in half equations. These protons are readily available and participate in the reduction of oxygen-containing species. For example, the reduction of permanganate (MnO4-) in acidic conditions can be represented as:

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

Here, H+ ions are explicitly included in the half equation, reflecting their abundance in acidic solutions. The presence of these protons facilitates the formation of water molecules as oxygen is reduced.

Conversely, in basic conditions, OH- ions take center stage. These hydroxide ions are abundant in alkaline solutions and participate in the oxidation of hydrogen-containing species. For instance, the oxidation of chromite (CrO2-) in basic conditions can be written as:

CrO2- + 4OH- CrO42- + 2H2O + 3e-

In this case, OH- ions are explicitly included, demonstrating their role in forming water molecules as hydrogen is oxidized.

The key difference between acidic and basic half equations lies in how they handle protons and hydroxide ions. Acidic half equations often involve the consumption of H+ ions, while basic half equations frequently show the consumption of OH- ions or the production of H2O.

Converting between acidic and basic half equations is a valuable skill for chemistry students. To convert an acidic half equation to its basic counterpart, follow these steps:

1. Add OH- to both sides of the equation to neutralize H+ ions.

2. Combine H+ and OH- to form H2O on the appropriate side.

3. Cancel out any excess water molecules.

For example, let's convert the acidic half equation for the reduction of dichromate:

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O (acidic)

To convert this to basic conditions:

1. Add 14OH- to both sides:

Cr2O72- + 14H+ + 14OH- + 6e- 2Cr3+ + 7H2O + 14OH-

2. Combine H+ and OH- to form H2O:

Cr2O72- + 14H2O + 6e- 2Cr3+ + 7H2O + 14OH-

3. Cancel out excess water:

Cr2O72- + 7H2O + 6e- 2Cr3+ + 14OH-

This process yields the basic half equation for dichromate reduction.

Understanding the differences between acidic and basic half equations is crucial for accurately representing redox reactions in various environments. H+ ions dominate in acidic conditions, while OH- ions prevail in basic conditions. By mastering the conversion between these forms, chemistry students can confidently approach redox problems in any pH environment, enhancing their ability to balance complex equations and predict reaction outcomes accurately.

Combining Half Equations to Form Full Redox Equations

Hey there, chemistry enthusiasts! Today, we're going to dive into the exciting world of redox reactions and learn how to combine half equations to form full redox equations. Don't worry if it sounds complicated we'll break it down step by step, and you'll be a pro in no time!

First things first, let's understand what half equations are. In redox reactions, we have two separate processes happening simultaneously: oxidation (where electrons are lost) and reduction (where electrons are gained). Each of these processes is represented by a half equation.

Now, the magic happens when we combine these half equations to create a full redox equation. But here's the catch we need to make sure our electrons are balanced. This is where balancing redox equations comes into play, and it's crucial for writing accurate full redox equations.

So, how do we achieve this electron balance? It's all about finding common multiples! We need to make sure that the number of electrons lost in the oxidation half equation matches the number of electrons gained in the reduction half equation. This might sound tricky, but with a little practice, you'll get the hang of it.

Let's walk through a step-by-step example to see how this works in action:

Step 1: Write out your two half equations.
Oxidation: Zn Zn² + 2e
Reduction: Cu² + 2e Cu

Step 2: Check the number of electrons in each half equation.
In this case, we're lucky! Both half equations involve 2 electrons, so we don't need to find a common multiple.

Step 3: Combine the half equations, canceling out the electrons.
Zn + Cu² Zn² + Cu

And there you have it your full redox equation!

But what if the number of electrons doesn't match up so neatly? Don't worry, we've got you covered. Let's look at another example:

Oxidation: Fe² Fe³ + e
Reduction: MnO + 8H + 5e Mn² + 4HO

In this case, we need to find a common multiple of 1 and 5. The least common multiple is 5, so we'll multiply the oxidation half equation by 5:

5(Fe² Fe³ + e)
MnO + 8H + 5e Mn² + 4HO

Now we can combine these balanced half equations:

5Fe² + MnO + 8H 5Fe³ + Mn² + 4HO

See? It's not so scary after all! The key is to take it step by step and always make sure your electrons are balanced.

Remember, practice makes perfect when it comes to balancing redox equations. Don't get discouraged if it takes a few tries to get it right even experienced chemists sometimes need to double-check their work.

As you continue your chemistry journey, you'll find that mastering the art of combining half equations to form full redox equations is an essential skill. It's not just about getting the right answer on a test understanding this process helps you grasp the fundamental principles of electron transfer in chemical reactions.

So keep at it, and before you know it, you'll be balancing redox equations like a pro! And who knows? You might even start to find it fun. After all, there's something satisfying

Applications of Half Equations in Chemistry

Half equations play a crucial role in understanding and solving various chemical processes, particularly in the fields of electrochemistry, corrosion studies, and battery technology. These simplified representations of oxidation and reduction reactions have far-reaching practical applications that impact our daily lives and industrial processes.

In electrochemistry, half equations are the building blocks for comprehending complex redox reactions. They allow scientists and engineers to break down intricate chemical processes into manageable components, making it easier to analyze and predict outcomes. For instance, in the production of aluminum through electrolysis, half equations help us understand the separate processes occurring at the anode and cathode, enabling optimization of the entire system.

Corrosion studies heavily rely on half equations to explain and mitigate the degradation of materials. By understanding the oxidation half-equation of a metal and the reduction half-equation of its environment, researchers can develop effective strategies to prevent or slow down corrosion. This knowledge is invaluable in industries such as construction, automotive, and aerospace, where protecting materials from corrosion is paramount for safety and longevity.

The field of battery technology has been revolutionized by the application of half equations. In designing and improving batteries, scientists use these equations to understand the chemical reactions occurring at each electrode. This knowledge allows for the development of more efficient and longer-lasting batteries, crucial for the advancement of electric vehicles and renewable energy storage systems. For example, the lithium-ion battery, a cornerstone of modern portable electronics, owes much of its development to the careful analysis of half equations governing lithium intercalation and de-intercalation processes.

Understanding half equations can help solve real-world problems in numerous ways. In environmental science, they aid in developing water treatment processes by elucidating the reactions involved in removing contaminants. In forensic science, half equations assist in analyzing evidence from electrochemical processes. Moreover, in the field of nanotechnology, these equations are instrumental in designing and optimizing nanomaterials for various applications, from catalysis to sensing devices.

The relevance of half equations extends beyond the laboratory. They are essential tools for engineers working on fuel cells, helping to improve energy conversion efficiency. In the medical field, half equations contribute to the development of biosensors and drug delivery systems. Even in the food industry, understanding these equations aids in preserving food quality and extending shelf life through controlled oxidation processes.

As we continue to face global challenges such as energy sustainability and environmental protection, the importance of half equations in chemistry only grows. By mastering these fundamental concepts, scientists and engineers are better equipped to innovate and solve complex problems across various disciplines, ultimately contributing to technological advancements and improvements in our quality of life.

Common Mistakes and Tips for Working with Half Equations

Working with half equations can be challenging for many students, but with the right approach and understanding, you can master this essential chemistry skill. Let's address some common mistakes and provide helpful tips to boost your confidence in handling half equations.

One frequent error is forgetting to balance the charges on both sides of the equation. Remember, the total charge must be equal on both sides. To avoid this, always double-check your charges before considering the equation complete. Another mistake is overlooking the role of water in aqueous solutions. In acidic or basic conditions, H+ or OH- ions are often involved, so be sure to include them when necessary.

Students sometimes struggle with identifying the correct species to include in half equations. To overcome this, carefully analyze the given information and consider which elements are changing oxidation states. Focus on the key reactants and products directly involved in the electron transfer.

Balancing strategies can help you tackle half equations systematically. Start by writing the unbalanced equation with the key species. Then, balance all atoms except hydrogen and oxygen. Next, add H2O to balance oxygen atoms, and H+ to balance hydrogen atoms. Finally, add electrons to balance the charges. Remember the mnemonic "OILRIG" - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons) - to help you determine which side of the equation needs electrons.

When combining half equations, a common pitfall is forgetting to multiply one or both equations to ensure the number of electrons transferred matches. To avoid this, clearly identify the number of electrons in each half equation and find the least common multiple. Multiply each equation as needed before combining them.

Building confidence in working with half equations comes with practice. Don't be discouraged if you make mistakes initially - they're part of the learning process. Create a step-by-step checklist to follow each time you approach a half equation problem. This will help reinforce the correct procedure and reduce errors.

Remember, mastering half equations is a valuable skill that will serve you well in your chemistry studies. With patience and consistent practice, you'll find yourself becoming more proficient and confident in handling these equations. Keep at it, and don't hesitate to seek help from your teachers or peers when you need clarification. You've got this!

Conclusion

In this article, we've explored the essential concept of half equations in chemistry. We've covered the key points of identifying oxidation and reduction processes, balancing electrons and charges, and combining half equations to form complete redox reactions. The introduction video provided a crucial foundation for understanding these concepts visually. Remember, mastering half equations is vital for comprehending more complex redox reactions and electrochemistry. We encourage you to practice working with various half equations to solidify your understanding. Try creating your own examples, balancing different reactions, and identifying oxidizing and reducing agents. The more you engage with this topic, the more comfortable you'll become with these fundamental principles. Don't hesitate to revisit the video and article content as needed. For further exploration, consider researching real-world applications of redox reactions in batteries, corrosion prevention, and industrial processes. Keep challenging yourself to deepen your knowledge of half equations and their significance in chemistry!

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.

Prerequisite Topics for Understanding Half Equations

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.

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