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Chemguide: Support for CIE A level Chemistry Learning outcome 11.3 Group 17 (previously Group 7) Some reactions of the halide ions Learning outcome 11.3.2 This statement looks at two reactions of halide ions - with silver ions and with concentrated sulfuric acid. Before you go on, you should find and read the statement in your copy of the syllabus. Statement 11.3.2(a) This is about the reactions between the halide ions and silver ions from, for example, silver nitrate solution, and is the basis for the tests for halide ions. You will find this discussed on the page about testing for halide ions. All you need to know is what happens when you add silver nitrate solution followed by ammonia of different concentrations to solutions of the halides. Surprisingly, according to the syllabus, you don't need to know the formula of the complex silver ion formed. Learn the descriptions! Make sure you can write the ionic equations for the formation of the precipitates as well. Statement 11.3.2(b) This is about the reactions of halides such as sodium chloride or potassium iodide with concentrated sulfuric acid. You will find this discussed on the page about halide ions and sulfuric acid. You do not need to be able to explain this in detail. Ignore the section "Explaining the trend" entirely. The syllabus uses the word "explain", but the teacher support material doesn't suggest any more explanation than an ability to work out the equations for the bromide and iodide reactions. The questions that have been asked about this tend to just want you to know that concentrated sulfuric acid isn't a strong enough oxidising agent to oxidise either fluoride or chloride ions, but can oxidise bromide or iodide ions. | |
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Note: You should also be comfortable about reversing this and saying that only bromide ions and iodide ions are strong enough reducing agents to reduce the sulfuric acid. Reducing ability of the ions increases down the Group. | |
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Don't try to learn these equations involving the formation of SO2 or H2S. The more likely question would be to give you the unbalanced equation and ask you to balance it using oxidation states. On the Chemguide page, I have worked out the equations using electron-half-equations. However, the teacher support material suggests using oxidation states. Here is how to do it for the oxidation of iodide ions using concentrated sulfuric acid. The unbalanced equation is:
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Note: You might think that this is the equation for the reaction with HI, but it isn't. There are other spectator ions involved which are not included - sulfate and sodium. | |
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Look at the oxidation state change of the sulfur. In the sulfuric acid it is +6; in the hydrogen sulfide it is -2. The oxidation state of the sulfur has fallen by 8. That means that the oxidation state of the iodide ions (the only other thing whose oxidation state is changing) must increase by a total of 8. If you add up everything in a complete equation, overall oxidation state gains and losses exactly balance out. Each iodide ion goes from -1 in the iodide ion to 0 in the iodine molecules. To make an overall gain of 8 oxidation states, there must be 8 iodide ions for every 1 sulfuric acid molecule. That gives you enough information to start to balance the equation.
Now you can sort out the iodines and the oxygens, and finally the hydrogens to give:
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Note: If you aren't happy about oxidation states, you must follow this link before you go on. Use the BACK button on your browser to return to this page. | |
© Jim Clark 2020 |
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