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7. Electrochemistry eLearn.Punjab

Example 7.2

Calculate the oxidation number of sulphur in H2SO4, when O. N. of
H = + l and O.N. of O = -2.

Solution

Applying the formula for H2SO4,
2[O.N. of H] + [O.N. of S] + 4[O.N. of O] = 0

Putting the values in above formula

2[+1] + [O.N. of S] + 4[-2] = 0

2 + [O.N. of S] -8 = 0

or O.N. of S = 8-2 = +6

Example 7.3

Find out the oxidation number of chlorine in KCIO3,
As O.N. of K = +1 and O.N. of O = -2

Solution

Putting the values in formula, we get

[+1] + [O.N. of Cl] + 3[-2] = 0

[+1] + [O.N. of Cl] + [-6] = 0

O.N. of Cl = 6-1

= +5

1. Find out the oxidation numbers of the following elements marked in bold in the

formulae:

Ba3(PO4)2, CaSO4, Cu(NO3)2, Al2(SO4)3
2. In a compound MX3, find out the oxidation number of M and X.
3. Why the oxidation number of oxygen in OF2 is +2
4. In H2S, SO2 and H2SO4 the sulphur atom has different oxidation number. Find out

the oxidation number of sulphur in each compound.

5. An element X has oxidation state 0. What will be its oxidation state when it gains

three electrons?

6. An element in oxidation state +7 gains electrons to be reduced to oxidation state

+2. How many electrons did it accept?

7. If the oxidation state of an element changes from +5 to -3. Has it been reduced or

oxidized? How many electrons are involved in this process?

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