Group 7(17) Redox Reactions (Oxford AQA International A Level Chemistry)

Halogens as Oxidising Agents

• Halogens react with metals by accepting an electron from the metal atom to become an ion with 1- charge

• In the following redox reaction:

  Ca (s) + Cl₂ (g) → CaCl₂ (s) 

  • Calcium loses electrons

  • Chlorine gains electrons

• Halogens are therefore oxidising agents because they cause another substance to lose electrons

• Halogens are themselves reduced as they gain electrons

• The oxidising ability of the halogens decreases going down the group

• This ability can be demonstrated in displacement reactions

Displacement reactions

• Halogen displacement reactions involve a halogen displacing a less reactive halide in a metal halide

• For example, the addition of chlorine water to a solution of sodium bromide:

• Cl₂ (aq) + 2NaBr (aq) → 2NaCl (aq) + Br₂ (aq)

• The chlorine has displaced the bromide from solution as it is more reactive

• The ionic equation for the reaction is:

  Cl₂ (aq) + 2Br^- (aq) → 2Cl^- (aq) + Br₂ (aq)

• Bromine has been oxidised

  • The oxidation state of bromine has increased from -1 to 0

• The chlorine has been reduced

  • The oxidation state of chlorine has decreased from 0 to -1

  • Chlorine acts as an oxidising agent

The displacement of a halide by a halogen

Halide Ions as Reducing Agents

• Halide ions can act as reducing agents and donate electrons to another atom

• The halide ions themselves get oxidised and lose electrons

• The reducing ability of the halide ions increases going down the group

• This trend can be explained by looking at the relative sizes of the halide ions

The relative sizes of halide ions

• Going down the group, the reducing ability of halides increases because:

  • The halide ions become larger

  • The outer electrons get further away from the nucleus

  • The outer electrons also experience more shielding by inner electrons

  • The outer electrons are held less tightly to the positively charged nucleus and are more easily lost

Sodium halides with concentrated sulfuric acid

• The trend in reducing ability of the halides can be demonstrated by reacting solid sodium halides with concentrated sulfuric acid

• The ionic equation for these reactions is:

H₂SO₄(l) + X^-(aq) → HX(g) + HSO₄^-(aq)

Where X^- is the halide ion

Sodium chloride with conc. sulfuric acid

• Concentrated sulfuric acid is added drop by drop to sodium chloride crystals

• The reaction that takes place is:

H₂SO₄ (l) + NaCl (s) → HCl (g) + NaHSO₄ (s)      

• The HCl gas produces is seen as steamy fumes 

• Solid sodium hydrogensulfate is also observed

• A redox reaction does not take place as chlorine is too weak a reducing agent

• The oxidation state of each substance remains the same:

  • Cl = -1

  • Na = +1

  • S = +6

  • H = +1

  • O = -2

Sodium bromide with conc. sulfuric acid

• The reaction that takes place initially is:    

  H₂SO₄ (l) + NaBr (s) → HBr (g) + NaHSO₄ (s)      

  • The HBr gas produced is seen as steamy fumes 

  • Solid sodium hydrogensulfate is also observed

• Then, bromide ions reduce sulfuric acid to sulfur dioxide gas:

  2HBr (g) + H₂SO₄ (l) → Br₂ (g) + SO₂ (g) + 2H₂O (l)

  • Brown fumes of bromine are observed

  • Colourless sulfur dioxide is formed

• Sulfur has been reduced

  • The oxidation state of sulfur has decreased from +6 in H₂SO₄ to +4 in SO₂

• Bromine has been oxidised

  • The oxidation state of bromine has increased from -1 in HBr to 0 in Br₂

• A redox reaction occurred because bromide ions are stronger reducing agents than chloride ions

Sodium iodide with conc. sulfuric acid

• The reaction that takes place initially is the same as that of sodium chloride and sodium bromide:    

  H₂SO₄ (l) + NaI (s) → HI (g) + NaHSO₄ (s)      

  • The HI gas produced is seen as steamy fumes 

  • Solid sodium hydrogensulfate is also observed

• Iodide ions are much stronger reducing agents than chloride and bromide ions

• They are able to reduce the sulfur in H₂SO₄ (oxidation state = +6)

  • to sulfur dioxide (oxidation state = +4)

  • then to sulfur (oxidation state = 0)

  • and finally to hydrogen sulfide (oxidation state = -2)

• The equation for formation of hydrogen sulfide is:

  8HI (g) + H₂SO₄ (l) → 4I₂ (s) + H₂S (g) + 4H₂O (l)

  • Iodine is seen as a black solid

  • Hydrogen sulfide has a strong smell of bad eggs

  • Sulfur is seen as a yellow solid as the sulfur passes through the oxidation state 0