Practice getting the equations right, and then add the state symbols in afterwards if your examiners are likely to want them. Now for the manganate(VII) half-equation: You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH. What about the hydrogen? Which balanced equation, represents a redox reaction?. The best way is to look at their mark schemes. How do you know whether your examiners will want you to include them?
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. Your examiners might well allow that. Add 5 electrons to the left-hand side to reduce the 7+ to 2+. By doing this, we've introduced some hydrogens. If you don't do that, you are doomed to getting the wrong answer at the end of the process! Note: Don't worry too much if you get this wrong and choose to transfer 24 electrons instead. Check that everything balances - atoms and charges. Which balanced equation represents a redox reaction equation. You know (or are told) that they are oxidised to iron(III) ions. 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.
Allow for that, and then add the two half-equations together. Now you have to add things to the half-equation in order to make it balance completely. 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. Which balanced equation represents a redox réaction chimique. You need to reduce the number of positive charges on the right-hand side. 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! Take your time and practise as much as you can. Working out electron-half-equations and using them to build ionic equations.
Example 1: The reaction between chlorine and iron(II) ions. Aim to get an averagely complicated example done in about 3 minutes. What is an electron-half-equation? Write this down: The atoms balance, but the charges don't. To balance these, you will need 8 hydrogen ions on the left-hand side. This is an important skill in inorganic chemistry. Now all you need to do is balance the charges. This technique can be used just as well in examples involving organic chemicals.
That's easily put right by adding two electrons to the left-hand side. Example 2: The reaction between hydrogen peroxide and manganate(VII) ions. 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! What we have so far is: What are the multiplying factors for the equations this time? 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. That's easily done by adding an electron to that side: Combining the half-reactions to make the ionic equation for the reaction. Example 3: The oxidation of ethanol by acidified potassium dichromate(VI). That means that you can multiply one equation by 3 and the other by 2. 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. The technique works just as well for more complicated (and perhaps unfamiliar) chemistry. But don't stop there!! If you forget to do this, everything else that you do afterwards is a complete waste of time! It would be worthwhile checking your syllabus and past papers before you start worrying about these! 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.
This is the typical sort of half-equation which you will have to be able to work out. 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. 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. Note: You have now seen a cross-section of the sort of equations which you could be asked to work out. 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). At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Don't worry if it seems to take you a long time in the early stages. 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. You start by writing down what you know for each of the half-reactions. Start by writing down what you know: What people often forget to do at this stage is to balance the chromiums. In the process, the chlorine is reduced to chloride ions. 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.
Manganate(VII) ions, MnO4 -, oxidise hydrogen peroxide, H2O2, to oxygen gas. 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! It is a fairly slow process even with experience. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero. 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. You should be able to get these from your examiners' website. There are 3 positive charges on the right-hand side, but only 2 on the left. It is very easy to make small mistakes, especially if you are trying to multiply and add up more complicated equations. 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! What we know is: The oxygen is already balanced. We'll do the ethanol to ethanoic acid half-equation first. But this time, you haven't quite finished. 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!
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. In this case, everything would work out well if you transferred 10 electrons. This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them. Let's start with the hydrogen peroxide half-equation.
Always check, and then simplify where possible. Working out half-equations for reactions in alkaline solution is decidedly more tricky than those above. The first example was a simple bit of chemistry which you may well have come across. Chlorine gas oxidises iron(II) ions to iron(III) ions. 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. There are links on the syllabuses page for students studying for UK-based exams. All you are allowed to add to this equation are water, hydrogen ions and electrons. In reality, you almost always start from the electron-half-equations and use them to build the ionic equation. This topic is awkward enough anyway without having to worry about state symbols as well as everything else.
The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. You can simplify this to give the final equation: 3CH3CH2OH + 2Cr2O7 2- + 16H+ 3CH3COOH + 4Cr3+ + 11H2O. All that will happen is that your final equation will end up with everything multiplied by 2. 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. WRITING IONIC EQUATIONS FOR REDOX REACTIONS. This is reduced to chromium(III) ions, Cr3+. Now that all the atoms are balanced, all you need to do is balance the charges. The final version of the half-reaction is: Now you repeat this for the iron(II) ions. Now balance the oxygens by adding water molecules...... and the hydrogens by adding hydrogen ions: Now all that needs balancing is the charges. Reactions done under alkaline conditions. 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). Now you need to practice so that you can do this reasonably quickly and very accurately!
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