In the following reaction, 451.4g of lead reacts with excess oxygen forming 356.7g of lead (II) oxide. Calculate the percent yie
1 answer:
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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction
<u>Explanation:</u>
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.
For the given chemical reaction:
<u>On the reactant side:</u>
Oxidation number of H = +1
Oxidation number of N = +5
Oxidation number of O = -2
Oxidation number of S = 0
<u>On the product side:</u>
Oxidation number of H = +1
Oxidation number of N = +2
Oxidation number of O = -2
Oxidation number of S = +6
As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.
The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.
Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction