Halide ion reactions

​​In a nutshell

Halides can be used as reducing agents, and this is highlighted when they're  reacted with sulfuric acid $$(H_2SO_4)$$. The reducing power of halides increases as you go down the group. Silver ions can be used to identify halides as a precipitate is formed.​

Reduction power of halides

Halide ions can act as reducing agents, by losing an electron from their outer shell. How readily the outermost electron is lost is dependent on the attraction between the nucleus and outer electrons. As you go down the group the attraction weakens, due to a larger atomic radius and additional inner shells causing a greater shielding effect.

Reactions with sulfuric acid 

​​Reaction of $$F^-$$ or $$Cl^-$$ with sulfuric acid $$(H_2SO_4)$$​​

​$$HF$$ or $$HCl$$ is formed in the reaction with sulfuric acid. $$F^-$$ and $$Cl^-$$ are not strong enough reducing agents and therefore they do not reduce the sulfuric acid, they just simply react with the sulfuric acid. This is not a redox reaction as the oxidation numbers of the halide ($$-1$$​) and sulfur ($$+6$$​) stay the same.

$$KF(s)\ +\ H_2SO_4(l)\ \rightarrow\ KHSO_4(s)\ +\ HF(g)\\KCl(s)\ +\ H_2SO_4(l)\ \rightarrow\ KHSO_4(s)\ +\ HCl(g)$$​​

Reaction of $$Br^-$$ with sulfuric acid $$(H_2SO_4)$$​​

The reaction between sulfuric acid $$(H_2SO_4)$$ and potassium bromide $$(KBr)$$ produces hydrogen bromide $$(HBr)$$. Hydrogen bromide is a stronger reducing agent than fluoride and chloride and can reduce sulfuric acid, where bromine $$(Br_2)$$​, sulfur dioxide $$(SO_2)$$​ and water $$(H_2O)$$ are produced. This is a redox reaction as the oxidation numbers of the bromine and sulfur change. 

​$$KBr(s)\ + \ H_2SO_4(l)\ \rightarrow\ KHSO_4(s)\ +HBr(g) \\2HBr(aq)\ +\ H_2SO_4(l)\ \rightarrow\ Br_2(g)\ +\ SO_2(g)\ +\ 2H_2O(l)$$​​

Reaction of $$(I^-)$$ with sulfuric acid $$(H_2SO_4)$$​

The first reaction produces hydrogen iodide $$(HI)$$​. The iodide ions $$(I^-)$$​ then reduce the sulfuric acid $$(H_2SO_4)$$. Since iodide ions are a stronger reducing agent than bromide, the iodide ions further reduce the sulfur dioxide $$(SO_2)$$​ to hydrogen sulfide $$(H_2S)$$. When hydrogen sulfide is formed it'll be clear as a very foul smell of rotten eggs is released.  

​$$KI(s)\ + \ H_2SO_4(l)\ \rightarrow\ KHSO_4(s)\ +HI(g) \\2HI(aq)\ +\ H_2SO_4(l)\ \rightarrow\ I_2(g)\ +\ SO_2(g)\ +\ 2H_2O(l)\\6HI(g)\ +\ SO_2(g)\ \rightarrow\ H_2S(g)\ +3I_2(s)\ +\ 2H_2O(l)$$​​

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Hydrogen halides 

Hydrogen halides are colourless acidic gases. Hydrogen halides dissolve in water. Hydrogen chloride forms hydrochloric acid, hydrogen bromide forms hydrobromic acid and hydrogen iodide forms hydroiodic acid. As you go down the group the weaker the bond is between the hydrogen and the halide.

Example

​$$HCl(g)\ \rightarrow\ H^+(aq)\ +\ Cl^-(aq)\\HBr(g)\ \rightarrow\ H^+(aq)\ +\ Br^-(aq)\\HI(g)\ \rightarrow\ H^+(aq)\ +\ I^-(aq)\\$$​​

Hydrogen halides also react with ammonia gas to produce white fumes.

​$$NH_3(g)\ +\ HCl(g)\ \rightarrow\ NH_4Cl(g)$$

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Silver ions and halides

Silver ions are used as a test for halides. Firstly, add dilute nitric acid to remove ions which may interfere with the reaction. Then add silver nitrate solution $$(AgNO_3)$$ and a precipitate of a silver halide is formed. 

$$Ag^+(aq)\ +\ X^-(aq)\ \rightarrow\ AgX(s)$$

Ammonia solution can be used to confirm the results of a silver halide test.