Redox chemistry

Oxidation numbers - the charge an element would have if the electrons in each bond of the molecule or ion belonged to the more electronegative element.

• Uncombined element - always 0
• H = +1 (except when bonded to a metal)
• O = -2 (except when bonded to F or in peroxides, e.g. Na₂O₂)

O I L R I G: OXIDATION is loss, REDUCTION is gain. 

Disproportionation - reaction in which the same substance is both oxidised and reduced.

Redox - involves both reduction and oxidation.

With iron(II) ions: self-indicating

• reduction: MnO₄^-(aq) + 8H⁺(aq) + 5e^- --> Mn²⁺(aq) + 4H₂O(l)
• oxidation: Fe²⁺(aq) --> Fe³⁺(aq) + e^-

MnO4-(aq) + 5Fe2+(aq) + 8H⁺(aq) --> Mn2+(aq) + 4H₂O(l) + 5Fe3+(aq)

With ethanedioic acid: autocatalysis

• reduction: MnO₄^-(aq) + 8H⁺(aq) + 5e^- ---> Mn²⁺(aq) + 4H₂O(l)
• oxidation: C₂O₄^2-(aq) --> 2CO₂(g) + 2e^-

MnO₄^-(aq) + 5C₂O₄^2-(aq) + 16H⁺(aq) + 5e^- --> Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

Iodometric

• 2S₂O₃^2-(aq)  → S₄O₆^2-(aq) + 2e^-
• I₂(aq) + 2e^- → 2I^-(aq)

I₂(aq)  +  2S₂O₃^2-(aq)  → 2I^-(aq) + S₄O₆^2-(aq)

Chlorine in bleach

ClO^-(aq) + 2H⁺(aq) + 2e^- → Cl^-(aq) + H₂O(l)
Cl: from +1 to -1 so it's the oxidising agent

Step 1: ClO^-(aq) + 2H⁺(aq) + 2I^-(aq) → Cl^-(aq) + I₂(aq)+ H₂O(l)
Step 2: back-titrating with standardised soln of sodium thiosulfate
Step 3: work backwards to find the conc of chlorate(I) in bleach.

Step 1: Add excess KI to known volume of the soln containing copper(II) ions. White ppt forms.

• 2Cu²⁺( aq) + 4I^-(aq) → 2CuI(s) + I₂(aq)

Step 2: Titrate the iodine solution with standardised S₂O₃^2- to find the conc of iodine

Step 3: Work backwards to find the copper(II) in the original sample.