Reducing Power of Halide Ions | A Level Chemistry | OCR, AQA, Edexcel

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This week we are revising Reducing Power of Halide Ions

A-Level Chemistry can be tough but fortunately we’ve made this tutorial to help you score the A* you need for questions on everything to do with Reducing Power of Halide Ions.

As you move down the group, the reducing power of the halides increases. As a reducing agent, a halide ion loses an outermost electron and the ability to lose this electron depends on the shielding effect and the ionic radius. The greater the reducing power of a halide, the more easily it can lose electrons.

As you move down the group of halides, the attraction between the nucleus and the outer electrons decreases. The attraction between the outermost electrons the the positive nucleus weakens as the ionic radius increases. This means that the distance between the outer electrons and the nucleus increases. Additionally there is a greater shielding effect as more inner electron shells are present which also weakens the attraction.
The reaction of halides with sulphuric acid can be used to compare the reducing power of the halides. When halides react with concentrated sulphuric acid, a hydrogen halide is produced as an initial product. Other products can also be produced depending on the halides.

You should be learn the following different reactions of sodium halides with sulphuric acid:

1. Reaction of NaF or NaCl with H2SO4:
Hydrogen fluoride or hydrogen chloride gas is formed depending on which halide is used. You can identify these substances as misty fumes when the gas contacts the moisture in the air.
The reaction does not proceed after this as HF and HCl are weak reducing agents. They are not strong enough to reduce sulphuric acid.
This is not a redox reaction as the oxidation states of halide and sulphur do not change.

2. Reaction of NaBr with H2SO4:
Hydrogen bromide gas is formed in the first reaction as observed by the misty fumes.
Sulfur dioxide gas is produced as choking fumes as well as bromine gas as orange fumes.

3. Reaction of NaI with H2SO4:
The first two reactions are the same as before. Hydrogen iodide is formed and can be observed as misty fumes.
Hydrogen iodide then reduces sulphuric acid because it is a strong reducing agent.

Acidified silver nitrate solution can be used to identify and distinguish between halide ions. Silver nitrate solution must be acidified using dilute nitric acid first as this removes any excess ions present in the solution that might react and affect the results.

A precipitate of the silver halide will form when silver nitrate solution is added. The colour of the precipitate will help you identify the halide ion present. The table below summarises the different coloured precipitates and their associated halides. The precipitate of silver chloride forms the slowest and the precipitate of silver iodide forms the fastest.

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