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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.
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:
The only three things you can add to half equations, besides more of the reactant or product, are water (H2O), hydrogen ions (H+), and electrons (e-). You can't sneak in oxygen gas (O2), 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:
First, let’s pick a reactant. We'll start with bromine. Bromine (Br2) reacts to form bromide ions (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:
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 do need to balance the charges on both sides of the equation. The left-hand side has a charge of +0, whereas the right has a charge of 2(-1), equalling -2. To balance the charges in the equation, we need to add a charge of -2 to the left-hand side. This means we must add in two negative electrons:
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:
Balancing the equation in terms of I gives us the following:
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:
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 (Br2) reacts to form bromide ions (Br-). If we balance the number of bromines on each side, we end up with the following:
Now, look at the oxidation states of the two species. Br2 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 Br2 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 Br2, and so we need to add two electrons to the right-hand side:
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 (I2):
The oxidation state of I in I- is -1, whilst the oxidation state of I in I2 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:
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 tofind 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:
Let's start with iron. In this reaction, Fe2+ ions turn into Fe3+ ions:
To balance the charges, we need to add one electron to the right-hand side of the equation:
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:
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 (H2O), hydrogen ions (H+), and electrons (e-). So, to add more oxygen to the right-hand side, we need to include some H2O. We need 4 O, so we add in 4 H2O:
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:
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:
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:
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:
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:
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:
You'll see that the electrons cancel out. We can remove these, giving us our final answer:
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.
To write half equations, you take the following steps:
Writing half equations can seem a little complicated, but it is simple if you follow these steps:
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:
Br2 + 2e- → 2Br-
2I- → I2 + 2e-
These combine to give the overall redox equation:
Br2 + 2I- → I2 + 2Br-
A half equation is an equation that shows one-half of a redox reaction. It shows how a species is either oxidised or reduced and indicates the movement of electrons.
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 + H2O, is H+ + OH- → H2O. 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.
What is a half equation?
An equation that shows one-half of a redox reaction.
True or false? Half equations show the movement of protons in a redox reaction.
True or false? Half equations must have the same number of moles of each element on both sides of the equation.
True or false? Half equations must have the same number of free electrons on both sides of the equation.
True or false? Half equations must have the same charge on both sides of the equation.
Which of the following species can you add to half equations?
More of the products and reactants
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