Chemistry Question

1.
Oxidation: Cu(s) → Cu(NO3)2(aq)
The O.N. (oxidation number) of Cu: 0 → +2
In oxidation, total change in O.N. = (+2) - 0 = +2

Reduction: HNO3(aq) → NO2(g)
The O.N. of N: +5 → +4
In reduction, total change in O.N. = (+4) - (+5) = -1

Hence, multiply Cu and Cu(NO3)2 by 1, and multiply HNO3 and NO2 by 2.
1Cu(s) + 2HNO3(aq) → 1Cu(NO3)2(aq) + 2NO2(g)

To balance N atoms, add 2HNO3 to the left hand side such that there are 4 N atoms on the both sides.
1Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g)

To balance H atoms, add 2H2O to the right hand side such that there are 4 H atoms on the both sides.
1Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Ans: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

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2.
Reduction: Cl2(aq) → 2Cl¯(aq)
Change in O.N. of Cl: 0 → (-1)
In reduction, total change in O.N. = 2[(-1) - 0] = -2

Oxidation: Cl2(aq) → 2ClO3¯(aq)
Change in O.N. of Cl: 0 → (+5)
In oxidation, total change in O.N. = 2[(+5) - (0)] = +10

Hence, multiply Cl2 and 2Cl¯ by 5, and multiply Cl2 and 2ClO3¯ by 1,
5Cl2(aq) + Cl2(aq) → 10Cl¯(aq) + 2ClO3¯(aq)
i.e. 6Cl2­(aq) → 10Cl¯(aq) + 2ClO3¯(aq)

Divide all coefficients by 2:
3Cl2(aq) → 5Cl¯(aq) + 1ClO3¯(aq)

To balance O atoms and H atoms, add 6OH¯ to the left hand side and 3H2O to the right hand side such that there are 6 H atoms and 6 O atoms on the both sides.
3Cl2(aq) + 3OH¯ → 5Cl¯(aq) + 1ClO3¯(aq) + 3H2O(l)

Ans: 3Cl2(aq) + 3OH¯ → 5Cl¯(aq) + ClO3¯(aq) + 3H2O(l)