Question

(a) Balance the following half-reactions by the ion-electron half-reaction method:

(i) (acid solution) NO3 – (aq) → N2O (g)

(ii) (acid solution) FeO4 2– (aq) → Fe3+ (aq)

(iii) (base solution) Fe2O3 (s) → Fe(OH)2 (s)

(iv) (base solution) Cu2O (s) → Cu(OH)2 (s)

(b) Which of the above equations are oxidation half-reactions?

(c) Give the formulas of the reactants that become oxidized in the course of the reaction.

Answer 1

+5 +1

(i) (acid solution) NO3– (aq) → N2O (g)

Balance N atoms : 2NO3– (aq) → N2O (g)

Balance O atoms : 2NO3– (aq) → N2O (g) + 5H2O(l)

Balance H atoms : 2NO3– (aq) +10H+(aq) → N2O (g) + 5H2O(l)

Balance charge : 2NO3– (aq) +10H+(aq)+8e- → N2O (g) + 5H2O(l)

Here the oxidation state of N decreases from +5 to +1 so it is a reduction reaction.

+6 +3

(ii) (acid solution) FeO42– (aq) → Fe3+ (aq)

Balance O atoms : FeO42– (aq) → Fe3+ (aq) + 4H2O(l)

Balance H atoms :  FeO42– (aq)+8H+(aq) → Fe3+ (aq) + 4H2O(l)

Balance charge :   FeO42– (aq)+8H+(aq) +5e- → Fe3+ (aq) + 4H2O(l)

Here the oxidation state of Fe decreases from +6 to +3 so it is a reduction reaction.

+3 +2

(iii) (base solution) Fe2O3 (s) → Fe(OH)2 (s)

Balance Fe atoms : Fe2O3 (s) → 2Fe(OH)2 (s)

Balance O atoms : Fe2O3 (s)+ H2O(l) → 2Fe(OH)2 (s)

Balance H atoms : Fe2O3 (s)+ H2O(l) + 2H2O(l) → 2Fe(OH)2 (s) +2OH-(aq)

Balance charge :  Fe2O3 (s)+ 3H2O(l)+2e- → 2Fe(OH)2 (s) +2OH-(aq)

Here the oxidation state of Fe decreases from +3 to +2 so it is a reduction reaction.

+1 +2

(iv) (base solution) Cu2O (s) → Cu(OH)2 (s)

Balance Cu atoms : Cu2O (s) → 2Cu(OH)2 (s)

Balance O atoms : Cu2O (s)+3H2O(l) → 2Cu(OH)2 (s)

Balance H atoms : Cu2O (s) + 3H2O(l) +6OH-(aq) → 2Cu(OH)2 (s) + 6H2O(l)

Balnec charge : Cu2O (s) +6OH-(aq) → 2Cu(OH)2 (s) + 3H2O(l)+6e-

Here the oxidation state of Cu increases from +1 to +2 so it is an oxidation reaction.

So Cu2O (s) is oxidized.