Identify the oxidized and reduced compound in the given reaction CuO (s) + H2(g) → Cu(s) + H2O(l) Copper oxide + Hydrogen gas → Copper + Water

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Oxidized compounds are those whose oxidation number increases, or they gain oxygen in a process, or they lose electrons or electronegative species combine with them in a reaction. Reduced substances are those whose oxidation number lowers, meaning they lose oxygen or gain electrons, or electropositive species join with them in a reaction. The oxidation number of copper changes from +2 to 0. Also, the oxygen atom is removed from copper oxide. Therefore copper oxide is reduced here.

Oxidized and Reduced Substances

CuO (s) → Cu (s)

Copper oxide → Copper

+2 → 0

Hydrogen gas is oxidized here as an oxygen atom is added to it. Also, the oxidation number of hydrogen changes from 0 to +1.

Uses of Copper Oxide

  1. Used in antifouling coatings for boat and ship floors. It is effective corrosion control.
  2. Used for coating glass and porcelain.
  3. Used as a p-type semiconductor material used in the manufacture of photocells and rectifiers for light meters.
  4. Used as fungicide and seed dressing.

Therefore, in the given reaction oxidized substance is hydrogen and the reduced substance is copper oxide.


Identify the oxidized and reduced compound in the given reaction: CuO (s) + H2(g) → Cu(s) + H2O(l)

In the above reaction, copper oxide is the reduced substance and hydrogen is the oxidized substance. Copper(I) Oxide is known as cuprous oxide which is an inorganic compound with Cu2O as a chemical formula.

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