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11.3 Reactions of halide ions
understand the trend in reducing ability of the halide ions
know the different products formed by reaction of NaX and H2SO4
Halide ions can act as reducing agents and in these reactions the halide ions lose electrons and
become halogen molecules.
The larger the ion, the more easily it loses an electron as the electron is lost from the outer shell the
outer shell gets further from the nucleus as the ion gets larger.
Ion F- Cl- Br- I-
Ionic radius 0.133nm 0.180nm 0.195nm 0.215nm
This trend can be seen in the reactions of solid sodium halides with concentrated sulphuric acid.
Solid sodium halides all react with concentrated sulphuric acid.
Sodium chloride + concentrated sulphuric acid hydrogen chloride + sodium hydrogen sulphate
NaCl(s) + H2SO4(l) NaHSO4(s) +HCl(g)
This is not a redox reaction because no oxidation state has changed. It is an acid-base reaction.
Sodium bromide in this case we see steamy fumes of hydrogen bromide and brown fumes of
NaBr(s) + H2SO4(l) NaHSO4(s) + HBr(g)
Two reactions occur:
Firstly sodium hydrogen sulphate and hydrogen bromide are produced.
However, bromide ions are strong enough reducing agents to reduce the sulphuric acid to sulphur
2H+ + 2Br- + H2SO4(l) SO2(g) + 2H2O(l) +Br2(l)
This is a redox reaction.
*Reactions take place between solid halide salts and concentrated sulphuric acid*
Sodium iodide add concentrated sulphuric acid
Several reactions occur; Hydrogen iodide is produced in an acid-base reaction.
NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)
Iodide ions reduce the sulphur in sulphuric acid even further so that sulphur dioxide, sulphur and hydrogen
sulphide gas are produced.
8H+ + 8I- + H2SO4 (l) H2S(g) + 4H2O(l) +4I2(s)
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