This is reduced to chromium(III) ions, Cr3+. Add 5 electrons to the left-hand side to reduce the 7+ to 2+. The sequence is usually: The two half-equations we've produced are: You have to multiply the equations so that the same number of electrons are involved in both.
In the example above, we've got at the electron-half-equations by starting from the ionic equation and extracting the individual half-reactions from it. Reactions done under alkaline conditions. What is an electron-half-equation? The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. You will often find that hydrogen ions or water molecules appear on both sides of the ionic equation in complicated cases built up in this way. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions. Now that all the atoms are balanced, all you need to do is balance the charges. Which balanced equation represents a redox reaction called. Write this down: The atoms balance, but the charges don't. You need to reduce the number of positive charges on the right-hand side. What about the hydrogen? But this time, you haven't quite finished. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero.
You start by writing down what you know for each of the half-reactions. Example 2: The reaction between hydrogen peroxide and manganate(VII) ions. By doing this, we've introduced some hydrogens. The simplest way of working this out is to find the smallest number of electrons which both 4 and 6 will divide into - in this case, 12. What we know is: The oxygen is already balanced.
This technique can be used just as well in examples involving organic chemicals. Add 6 electrons to the left-hand side to give a net 6+ on each side. Check that everything balances - atoms and charges. It is a fairly slow process even with experience. Manganate(VII) ions, MnO4 -, oxidise hydrogen peroxide, H2O2, to oxygen gas. In the chlorine case, you know that chlorine (as molecules) turns into chloride ions: The first thing to do is to balance the atoms that you have got as far as you possibly can: ALWAYS check that you have the existing atoms balanced before you do anything else. Which balanced equation represents a redox reaction cycles. Allow for that, and then add the two half-equations together. These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing! Now you need to practice so that you can do this reasonably quickly and very accurately! Any redox reaction is made up of two half-reactions: in one of them electrons are being lost (an oxidation process) and in the other one those electrons are being gained (a reduction process). It would be worthwhile checking your syllabus and past papers before you start worrying about these!
Using the same stages as before, start by writing down what you know: Balance the oxygens by adding a water molecule to the left-hand side: Add hydrogen ions to the right-hand side to balance the hydrogens: And finally balance the charges by adding 4 electrons to the right-hand side to give an overall zero charge on each side: The dichromate(VI) half-equation contains a trap which lots of people fall into! All you are allowed to add to this equation are water, hydrogen ions and electrons. Which balanced equation represents a redox reaction below. There are 3 positive charges on the right-hand side, but only 2 on the left. Now you have to add things to the half-equation in order to make it balance completely. Practice getting the equations right, and then add the state symbols in afterwards if your examiners are likely to want them. This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction.
You know (or are told) that they are oxidised to iron(III) ions. You are less likely to be asked to do this at this level (UK A level and its equivalents), and for that reason I've covered these on a separate page (link below). This topic is awkward enough anyway without having to worry about state symbols as well as everything else. At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Add two hydrogen ions to the right-hand side. If you think about it, there are bound to be the same number on each side of the final equation, and so they will cancel out. When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is: Note: I am going to leave out state symbols in all the equations on this page. WRITING IONIC EQUATIONS FOR REDOX REACTIONS.
The oxidising agent is the dichromate(VI) ion, Cr2O7 2-. All you are allowed to add are: In the chlorine case, all that is wrong with the existing equation that we've produced so far is that the charges don't balance. Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH. If you want a few more examples, and the opportunity to practice with answers available, you might be interested in looking in chapter 1 of my book on Chemistry Calculations. That's easily done by adding an electron to that side: Combining the half-reactions to make the ionic equation for the reaction. That's doing everything entirely the wrong way round! Example 3: The oxidation of ethanol by acidified potassium dichromate(VI). In this case, everything would work out well if you transferred 10 electrons. If you don't do that, you are doomed to getting the wrong answer at the end of the process! The technique works just as well for more complicated (and perhaps unfamiliar) chemistry. During the checking of the balancing, you should notice that there are hydrogen ions on both sides of the equation: You can simplify this down by subtracting 10 hydrogen ions from both sides to leave the final version of the ionic equation - but don't forget to check the balancing of the atoms and charges! You can split the ionic equation into two parts, and look at it from the point of view of the magnesium and of the copper(II) ions separately. What we have so far is: What are the multiplying factors for the equations this time? When you come to balance the charges you will have to write in the wrong number of electrons - which means that your multiplying factors will be wrong when you come to add the half-equations... A complete waste of time!
If you forget to do this, everything else that you do afterwards is a complete waste of time! This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them. These can only come from water - that's the only oxygen-containing thing you are allowed to write into one of these equations in acid conditions. Let's start with the hydrogen peroxide half-equation. This is the typical sort of half-equation which you will have to be able to work out. So the final ionic equation is: You will notice that I haven't bothered to include the electrons in the added-up version.
The best way is to look at their mark schemes. You can simplify this to give the final equation: 3CH3CH2OH + 2Cr2O7 2- + 16H+ 3CH3COOH + 4Cr3+ + 11H2O. The multiplication and addition looks like this: Now you will find that there are water molecules and hydrogen ions occurring on both sides of the ionic equation. That means that you can multiply one equation by 3 and the other by 2. You should be able to get these from your examiners' website. To balance these, you will need 8 hydrogen ions on the left-hand side. Chlorine gas oxidises iron(II) ions to iron(III) ions.
The manganese balances, but you need four oxygens on the right-hand side. That's easily put right by adding two electrons to the left-hand side. You would have to know this, or be told it by an examiner. Note: Don't worry too much if you get this wrong and choose to transfer 24 electrons instead. The first example was a simple bit of chemistry which you may well have come across. All that will happen is that your final equation will end up with everything multiplied by 2. Don't worry if it seems to take you a long time in the early stages. Note: You have now seen a cross-section of the sort of equations which you could be asked to work out. But don't stop there!! If you add water to supply the extra hydrogen atoms needed on the right-hand side, you will mess up the oxygens again - that's obviously wrong! Example 1: The reaction between chlorine and iron(II) ions. Working out electron-half-equations and using them to build ionic equations. Note: If you aren't happy about redox reactions in terms of electron transfer, you MUST read the introductory page on redox reactions before you go on.
The final version of the half-reaction is: Now you repeat this for the iron(II) ions. © Jim Clark 2002 (last modified November 2021). In the process, the chlorine is reduced to chloride ions. Now for the manganate(VII) half-equation: You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. In reality, you almost always start from the electron-half-equations and use them to build the ionic equation. It is very easy to make small mistakes, especially if you are trying to multiply and add up more complicated equations. In building equations, there is quite a lot that you can work out as you go along, but you have to have somewhere to start from! Working out half-equations for reactions in alkaline solution is decidedly more tricky than those above.
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