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Half Equations

In Redox, we learned that a redox reaction features two separate reactions: oxidation and reduction. We can show the overall process with a redox equation. But this equation makes it hard to see the individual oxidation and reduction processes and the movement of electrons. Instead, we can break the reaction down into two separate half equations.This article is about half…

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Half Equations

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Jetzt kostenlos anmeldenIn Redox, we learned that a **redox reaction** features two separate reactions: **oxidation** and **reduction**. We can show the overall process with a **redox equation**. But this equation makes it hard to see the individual oxidation and reduction processes and the movement of electrons. Instead, we can break the reaction down into two separate **half equations**.

- This article is about
**half equations**in physical chemistry. - We'll first
**define half equation.** - We'll then learn how to
**write half equations**, using**oxidation states**to help us. - After that, we'll
**combine half equations**to create an overall**redox equation**. - You'll be able to practice your skills with our
**worked examples**.

**Half equations** are equations that show **one-half of a redox reaction**.

Redox reactions consist of two processes: **oxidation** and **reduction**. Half equations show each separate process in terms of electron movement. One half equation shows the oxidation process, which is the **loss of electrons**, whilst the other shows the reduction process, which is the **gain of electrons**.

Check out Redox for a more in-depth look at oxidation and reduction.

To write half equations, we consider the oxidation and reduction processes involved in the redox equation separately. We pick a species that is oxidised or reduced and form a new equation to show how it changes. We add in electrons to show the processes of oxidation and reduction, and potentially add in water molecules or hydrogen ions to balance the equation.

These steps are a handy guide:

- Pick out the reactants and products involving a particular species from the overall redox equation. This is the start of your first half equation.
- Balance the elements in this new equation, apart from oxygen and hydrogen. Half equations have to be balanced - you must have the same number of moles of each element on both sides of the equation.
- Next, add in water molecules (H
_{2}O) to balance the number of oxygen atoms on each side of the equation. - Then add in hydrogen ions (H
^{+}) to balance the number of hydrogen atoms on each side of the equation. - Finally, add in electrons (e
^{-}) to balance the charges on each side of the equation. - Repeat the process with the other species involved in the reaction to create your second half equation.

The only three things you can add to half equations, besides more of the reactant or product, are water (H_{2}O), hydrogen ions (H^{+}), and electrons (e^{-}). You can't sneak in oxygen gas (O_{2}), for example. You should also note that some redox reactions might involve more than two half equations - the name can be a bit misleading! However, you are unlikely to come across these in your exams.

Ready to give it a go? Here are some examples.

Let's now practice writing half equations for real-life reactions, using the method we learned above. We'll start with a simple redox reaction between bromine gas and iodide ions.

**Write half equations for the displacement reaction between bromine and iodide ions. The unbalanced equation is given below: **

\(Br_2+I^-\rightarrow I_2+Br^- \)

First, let’s pick a reactant. We'll start with bromine. Bromine (Br_{2}) reacts to form bromide ions (Br^{-}):

\(Br_2\rightarrow Br^- \)

This equation isn’t balanced. There are two Br on the left, but only one on the right. To balance it, we need to double the number of bromide ions:

\(Br_2\rightarrow 2Br^- \)

There aren’t any oxygen or hydrogen atoms involved in the equation, so we don’t need to add in any water molecules or hydrogen ions to balance it. However, we\(Br_2+2e^-\rightarrow 2Br^- \)

Both sides now have the same number of each element *and* the same charge. The half equation is balanced. But we’re not done. We now need to write a second half equation for iodine.

In this reaction, iodide ions react to form iodine:

\(I^-\rightarrow I_2\)

Balancing the equation in terms of I gives us the following:

\(2I^-\rightarrow I_2\)

This time, the charge on the left-hand side of the equation is 2(-1) = -2, and the charge on the right-hand side is +0. This means that we need to add two negative electrons to the right-hand side:

\(2I^-\rightarrow I_2+2e^- \)

Once again, this equation is now balanced in terms of elements and charges. Your two half equations are complete.

Can you tell which species has been oxidised, and which has been reduced? Here, bromine gains electrons and is reduced, whilst iodide ions lose electrons and so are oxidised.

You can also write half equations by considering the species' change in oxidation state. We'll walk you through this method as well.

**Write half equations for the same displacement reaction between bromine and iodide ions using changes in oxidation state.**

Once again, we'll start by considering bromine. In this reaction, bromine (Br_{2}) reacts to form bromide ions (Br^{-}). If we balance the number of bromines on each side, we end up with the following:

\(Br_2\rightarrow 2Br^-\)

** **

Now, look at the oxidation states of the two species. Br_{2} is an uncombined element, and so each Br within it has an oxidation state of 0. On the other hand, Br^{-} is an ion with a charge of -1, and so in this species, Br has an oxidation state of -1. Consider how Br's oxidation state has changed from the left-hand side of the equation to the right-hand side: it has decreased by 1. Each Br in Br_{2} must gain an electron to become Br^{-}. We can add this spare electron to the left-hand side of the equation. But note that there are two Br in Br_{2}, and so we need to add *two *electrons to the right-hand side:

\(Br_2+2e^ \rightarrow 2Br^- \)

This is our first half equation. We can also apply the same process to the second half equation, involving iodine. We start with iodide ions (I^{-}) reacting to form iodine (I_{2}):

\(2I^-\rightarrow I_2\)

The oxidation state of I in I^{-} is -1, whilst the oxidation state of I in I_{2} is 0. In this reaction, I's oxidation state increases by 1. This means that each I loses an electron; we can add this spare electron to the right-hand side of the equation. But once again, note that we start with 2 I^{-}, and so we need to add *two* electrons to the right-hand side:

\(2I^-\rightarrow I_2+2e^- \)

This is our final answer.

No matter which method you choose, you should end up with the same half equations for one particular redox reaction. Don't be afraid to try both techniques to find out which one works best for you.

Here's another example. This one is a little trickier. Give it a go and then check your answer against our worked solution.

**Write half equations for the reaction between manganate(VII) ions and iron(II) ions to form manganese(II) ions and iron(III) ions. The unbalanced equation is given below:**

\(MnO_4^- + Fe^{2+} + H^+ \rightarrow Mn^{2+} + Fe^{3+} + H_2O\)

Let's start with iron. In this reaction, FeTo balance the charges, we need to add one electron to the right-hand side of the equation:

\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)

Both elements and charges are balanced; that's our first half equation done. We'll now consider the manganate ion. Here's our starting equation:

\(MnO_4^- \rightarrow Mn^{2+}\)

There are the same number of Mn on each side of the equation, so we don’t need to worry about them. However, there are four oxygens (O) on the left-hand side but none on the right-hand side. We need to balance the equation by adding more O to the right-hand side. Remember that the only substances we can add to half equations are water (H_{2}O), hydrogen ions (H^{+}), and electrons (e^{-}). So, to add more oxygen to the right-hand side, we need to include some H_{2}O. We need 4 O, so we add in 4 H_{2}O:

\(MnO_4^- \rightarrow Mn^{2+} + 4H_2O\)

We've now encountered another issue: there are 8 hydrogens (H) on the right-hand side of the equation, but none on the left. Luckily for us, hydrogen ions (H^{+}) are one of the species that we are allowed to add to half equations. Therefore, we add 8 H^{+} to the left-hand side:

\(MnO_4^- + 8H^+\rightarrow Mn^{2+} + 4H_2O\)

We're almost done. However, the charges aren't balanced: There is an overall charge of +7 on the left-hand side, but only +2 on the right. To make the charges equal, we add 5 negative electrons to the left-hand side:

\(MnO_4^- + 8H^+ + 5e^-\rightarrow Mn^{2+} + 4H_2O\)

Give the equation a final check to make sure that the number of moles of each element and the overall charge is balanced on each side of the equation. In this case, everything looks good. Well done - we've written our two half equations.

In your exam, you might also be asked to **combine two half equations** to make one overall redox equation. This is a lot simpler than writing half equations. You'll notice that in redox equations, we don't see any electrons. To form an overall redox equation, we combine multiples of the two half equations so that the electrons cancel out.

Here are the steps that you should follow:

- Multiply each of the half equations by a constant so that they both feature the same number of electrons.
- Add the half equations together to create an overall redox equation.
- Cancel the electrons and any other species that appear on both sides of the equation.

Ready to give it a go? Let's use the two half equations we wrote above for the reaction between manganate(VII) and iron(II) ions.

**Combine the following two half equations to create one overall redox equation:**

\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)

\(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)

Take a look at the two half equations. The first has just one electron on the right-hand side, whilst the second has five on the left-hand side. We need to multiply each of the half equations by a constant so that they both feature the same number of electrons. The easiest way to do this is by multiplying the first equation by 5. This way, both equations will feature five electrons:

\(5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-\)

We now add the two reactions together to form one overall redox equation. Add all the reactants from both half equations to the left-hand side of this new redox equation, and add all the products to the right:

\(5Fe^{2+} + MnO_4^- + 8H^+ + 5e^- \rightarrow 5Fe^{3+} + 5e^-+ Mn^{2+} + 4H_2O\)

You'll see that the electrons cancel out. We can remove these, giving us our final answer:

\(5Fe^{2+} + MnO_4^- + 8H^+ +\cancel {5e^-}\rightarrow 5Fe^{3+} + \cancel {5e^-}+ Mn^{2+} + 4H_2O\)

\(5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\)

It is always a good idea to check that your overall redox equation is balanced, too. Is the number of moles of each element and the overall charge the same on both sides of the equation? If not, something has gone wrong - take another look and try again.

**Half equations**are equations that show**one-half of a redox reaction**. Each half equation shows either the individual oxidation or reduction process in terms of the**movement of electrons**.- To write half equations, you take the following steps:
- Pick out all the reactants and products involving a certain species from the original redox equation.
- Balance the elements in this new equation by adding more of the reactants or products to either side of the equation. You can also add water (H
_{2}O) and hydrogen ions (H^{+}). - Balance the charges in the equation by adding electrons (e
^{-}). - Repeat the process with the other species involved in the reaction to create your second half equation.

- When writing half equations, you must make sure that both the
**number of moles of each element**and the**overall charge**are balanced on each side of the equation. - The only species you can add to half equations are
**more of the reactants or products**,**water (H**_{2}**O)**,**hydrogen ions (H**^{+}**)**, and**electrons (e**^{-}**)**. - To combine half equations, you take the following steps:
- Multiply each of the half equations by a constant so that they both feature the same number of electrons.
- Add the half equations together to create an overall redox equation.
- Cancel the electrons and any other species that appear on both sides of the equation.

To write half equations, you take the following steps:

- Pick out all the reactants and products involving a certain species from the original redox equation.
- Balance the elements in this new equation by adding more of the reactants or products to either side of the equation. You can also add water (H
_{2}O) and hydrogen ions (H^{+}). - Balance the charges in the equation by adding electrons (e
^{-}). - Repeat the process with the other species involved in the reaction to create your second half equation.

Writing half equations can seem a little complicated, but it is simple if you follow these steps:

- Pick out all the reactants and products involving a certain species from the original redox equation.
- Balance the elements in this new equation by adding more of the reactants or products to either side of the equation. You can also add water (H
_{2}O) and hydrogen ions (H^{+}). - Balance the charges in the equation by adding electrons (e
^{-}).

A half equation is an equation that shows one-half of a redox reaction. Each half equation represents either the process of oxidation or reduction in terms of electron movement. For example, the reaction between bromine and iodide ions has the following two half equations:

Br_{2} + 2e^{-} → 2Br^{-}

2I^{-} → I_{2} + 2e^{-}

These combine to give the overall redox equation:

Br_{2} + 2I^{-} → I_{2} + 2Br^{-}

Ionic equations show what happens to ions in a chemical reaction. They don't necessarily show redox reactions, and they ignore any ions that are present in the system but don't actually take part in the reaction. For example, the ionic equation for the reaction HCl + NaOH → NaCl + H_{2}O, is H^{+} + OH^{-} → H_{2}O. We ignore the Na^{+} and Cl^{-} ions because they are present on both sides of the equation.

However, half equations show what happens to electrons in redox reactions. They show the individual oxidation and reduction processes, as well as the changes in oxidation states of the species involved. The reaction we looked at above doesn't have any half equations, because it isn't a redox reaction - there is no transfer of electrons and none of the species change oxidation states.

More about Half Equations

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