Question

3SO2 + 2HNO3 + 2H2O → 3H2SO4 + 2NO Which statement is true about this chemical equation? H (hydrogen) was oxidized and S (sulfur) was reduced. H (hydrogen) was reduced and N (nitrogen) was oxidized. S (sulfur) was oxidized and N (nitrogen) was reduced. S (sulfur) was reduced and N (nitrogen) was oxidized.

Answer 1

Answer:

S (sulfur) was oxidized and N (nitrogen) was reduced.

Explanation:

In order to determine which element was oxidized or reduced in the equation, the change in the oxidation number of the element is observed.

Increase in oxidation number refers to oxidation and decrease in oxidation number refers to reduction

3SO2 + 2HNO3 + 2H2O → 3H2SO4 + 2NO

Sulphur;

Reactant side: +4

Product side: +6

This means there was a decrease in oxidation number. Sulfur was oxidized

Nitrogen;

Reactant side: +5

Product side: +2

This means there was a decrease in oxidation number. Nitrogen was reduced

Hydrogen and Oxygen both have a fixed oxidation number!

The correct option is;

S (sulfur) was oxidized and N (nitrogen) was reduced.