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Which substance is reduced in this redox reaction? CuO(s) + H₂(g) → Cu(s) + H₂O(l)

User Pezhvak
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Final answer:

In the redox reaction CuO(s) + H₂(g) → Cu(s) + H₂O(l), copper(II) oxide (CuO) is the substance that is reduced as it gains electrons going from an oxidation state of +2 to 0.

Step-by-step explanation:

The substance that is reduced in the redox reaction CuO(s) + H₂(g) → Cu(s) + H₂O(l) is copper(II) oxide (CuO). In a redox reaction, reduction refers to the gain of electrons. You can remember it with the mnemonic 'oil rig', which stands for oxidation is loss, reduction is gain. We determine the substance that is reduced by analyzing the changes in oxidation states of the reactants and products. Copper's oxidation state changes from +2 in the CuO to 0 in the metallic copper (Cu), indicating it has gained electrons. Therefore, CuO is reduced to Cu in this reaction. Hydrogen, on the other hand, goes from an oxidation state of 0 in H₂ to +1 in H₂O, which means hydrogen is oxidized.

The substance that is reduced in the redox reaction CuO(s) + H₂(g) → Cu(s) + H₂O(l) is copper(II) oxide (CuO). In this reaction, copper(II) oxide is being reduced to copper (Cu). Reduction is the process of gaining electrons, and CuO is the species that undergoes this change. On the other hand, hydrogen gas (H₂) is being oxidized to form water (H₂O). Oxidation is the process of losing electrons, and H₂ is the species that undergoes this change.

User Krekto
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