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The given reaction is

`CuO(s) + H_2(g) -> Cu(s) + H2O(l)`

In this reaction, copper (II) oxide reacts with hydrogen to generate copper metal and water. This is an oxidation-reduction reaction, in which some species are oxidized and some reduced. 

The oxidation number of copper goes from +2 (in CuO)...

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The given reaction is

`CuO(s) + H_2(g) -> Cu(s) + H2O(l)`

In this reaction, copper (II) oxide reacts with hydrogen to generate copper metal and water. This is an oxidation-reduction reaction, in which some species are oxidized and some reduced. 

The oxidation number of copper goes from +2 (in CuO) to 0 (in Cu), while hydrogen's oxidation number goes from 0 (in H2) to +1 (in water). The oxidation number of oxygen stays the same and is equal to -2.

In oxidation, the oxidation number increases as the species lose electron(s). Since the species gain one or more electrons in reduction, the oxidation number decreases. 

Thus, hydrogen is oxidized while copper is reduced. We can also say hydrogen is the reducing agent, while the copper (II) oxide is the oxidizing agent. 

Hope this helps. 

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