Question

1. (a) Balance the following half-reactions by the ion-electron half-reaction method: (i) (acid solution) NO2(g) → NO,- (aq) (ii) (acid solution) H SeO(aq) → Se (s) (iii) (base solution) P (8) PH, (g) (iv) (base solution) Co(OH)(s) C002 () (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

+4 +5

(a) (i) NO2 (g) ----> NO3- (aq)

Balance O atoms : NO2 (g) +H2O(l) ----> NO3- (aq)

Balance H atoms : NO2 (g) +H2O(l) ----> NO3- (aq) + 2H+(aq)

Balance charge : NO2 (g) +H2O(l) ----> NO3- (aq) + 2H+(aq) +e-

Here the oxidation state of N increases from +4 to +5 so it is an oxidation reaction.

Here NO2 is oxidized

+4 0

(ii) H2SeO3(aq) ----> Se(s)

Balance O atoms : H2SeO3(aq) ----> Se(s) +3H2O(l)

Balance H atoms : H2SeO3(aq) + 4H+(aq) ----> Se(s) +3H2O(l)

Balance charge :  H2SeO3(aq) + 4H+(aq) + 4e- ----> Se(s) +3H2O(l)

Here the oxidation state of Se decreases from +4 to 0 so it is a reduction reaction.

Here H2SeO3 is reduced

0 -3

(iii) P4(s) ------> PH3(g)

Balance P atoms : P4(s) ------> 4PH3(g)

Balance H atoms : P4(s) + 12H2O(l) ------> 4PH3(g) +12OH-(aq)

Balance charge : PP4(s) + 12H2O(l) +12e- ------> 4PH3(g) + 12OH-(aq)

Here the oxidation state of P decreases from 0 to -3 so it is a reduction reaction.

Here P4 is reduced.

+3 +4

(iv) Co(OH)3 (s) ----> CoO2(s)

Balance O atoms : Co(OH)3 (s) ----> CoO2(s) +H2O(l)

Balance H atoms : Co(OH)3 (s) + OH-(aq) ----> CoO2(s) +H2O(l) + H2O(l)

Balance charge : Co(OH)3 (s) + OH-(aq) ----> CoO2(s) + 2H2O(l) + e-

Here the oxidation state of Co increases from +3 to +4 so it is an oxidation reaction.

Here Co(OH)3 is oxidized