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! Example 2: The reaction between hydrogen peroxide and manganate(VII) ions. Which balanced equation represents a redox reaction called. You can simplify this to give the final equation: 3CH3CH2OH + 2Cr2O7 2- + 16H+ 3CH3COOH + 4Cr3+ + 11H2O. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions. So the final ionic equation is: You will notice that I haven't bothered to include the electrons in the added-up version. 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. 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!
The first example was a simple bit of chemistry which you may well have come across. 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. You should be able to get these from your examiners' website. 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. What we've got at the moment is this: It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Example 3: The oxidation of ethanol by acidified potassium dichromate(VI). There are links on the syllabuses page for students studying for UK-based exams. Which balanced equation represents a redox réaction de jean. In this case, everything would work out well if you transferred 10 electrons. Now you have to add things to the half-equation in order to make it balance completely. Allow for that, and then add the two half-equations together. Write this down: The atoms balance, but the charges don't.
If you don't do that, you are doomed to getting the wrong answer at the end of the process! That's easily done by adding an electron to that side: Combining the half-reactions to make the ionic equation for the reaction. Chlorine gas oxidises iron(II) ions to iron(III) ions. The best way is to look at their mark schemes. You need to reduce the number of positive charges on the right-hand side. This topic is awkward enough anyway without having to worry about state symbols as well as everything else. It is a fairly slow process even with experience.
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. How do you know whether your examiners will want you to include them? But this time, you haven't quite finished. Manganate(VII) ions, MnO4 -, oxidise hydrogen peroxide, H2O2, to oxygen gas. Reactions done under alkaline conditions. In reality, you almost always start from the electron-half-equations and use them to build the ionic equation. Practice getting the equations right, and then add the state symbols in afterwards if your examiners are likely to want them. 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. Always check, and then simplify where possible. Add 5 electrons to the left-hand side to reduce the 7+ to 2+. 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. The oxidising agent is the dichromate(VI) ion, Cr2O7 2-. The technique works just as well for more complicated (and perhaps unfamiliar) chemistry. This is reduced to chromium(III) ions, Cr3+. 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. To balance these, you will need 8 hydrogen ions on the left-hand side. 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. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero. Check that everything balances - atoms and charges. Now balance the oxygens by adding water molecules...... and the hydrogens by adding hydrogen ions: Now all that needs balancing is the charges. Now all you need to do is balance the charges. You start by writing down what you know for each of the half-reactions. Take your time and practise as much as you can.
There are 3 positive charges on the right-hand side, but only 2 on the left. What about the hydrogen? Note: Don't worry too much if you get this wrong and choose to transfer 24 electrons instead. Let's start with the hydrogen peroxide half-equation. Now you need to practice so that you can do this reasonably quickly and very accurately! 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! 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. 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! This technique can be used just as well in examples involving organic chemicals. 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. That means that you can multiply one equation by 3 and the other by 2.
That's doing everything entirely the wrong way round! We'll do the ethanol to ethanoic acid half-equation first. What is an electron-half-equation? Now that all the atoms are balanced, all you need to do is balance the charges. 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 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). In the process, the chlorine is reduced to chloride ions. The final version of the half-reaction is: Now you repeat this for the iron(II) ions. Electron-half-equations. Aim to get an averagely complicated example done in about 3 minutes. It would be worthwhile checking your syllabus and past papers before you start worrying about these!
Don't worry if it seems to take you a long time in the early stages. That's easily put right by adding two electrons to the left-hand side. 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. You know (or are told) that they are oxidised to iron(III) ions. This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them. All that will happen is that your final equation will end up with everything multiplied by 2. This is the typical sort of half-equation which you will have to be able to work out. Add two hydrogen ions to the right-hand side.
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