Consider the following reaction equation: CuO(s) + H2(g) $\to$ Cu(s) + H2O(1) Which substance is oxidized?

Analyze the question: We need to find which substance loses electrons (or gains oxygen / loses oxidation number increase) in the reaction.

Concept: Oxidation is:

• Gain of oxygen
• Loss of hydrogen
• Loss of electrons
• Increase in oxidation number

Reduction is the opposite (OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons).

Solution:

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

Oxidation numbers:

• Cu in CuO: +2 → Cu in Cu: 0 (decrease, so Cu²⁺ is reduced)
• H in H₂: 0 → H in H₂O: +1 (increase, so hydrogen is oxidized)
• O stays at −2 throughout.

In terms of oxygen transfer: CuO loses oxygen (reduction), while H₂ gains oxygen to form H₂O (oxidation).

So CuO is the oxidizing agent (it is reduced), and H₂ is the reducing agent (it is oxidized).

Checking the options:

• A. Cu: this is a product, formed by reduction.
• B. CuO: this is reduced, not oxidized.
• C. H₂: this is oxidized (0 → +1). ✔
• D. H₂O: this is a product, not the substance that was oxidized.

Tip: The substance that is oxidized is the reducing agent. Remember "OIL RIG" and "the reducing agent gets oxidized." In reactions where a metal oxide is heated in hydrogen, the hydrogen is always the one oxidized, and the metal oxide is reduced to the metal.