Question

Find out the amount of water that will be formed when 3.975 gm. of pure cupric oxide are reduced
by pure hydrogen. Also find loss of weight of the cupric oxide. [Cu = 63.5, H = 1, 0 = 16]​

Answer 1

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Explanation: CuO + H2 → Cu + H2O is a oxidation-reduction reaction (redox). One element gets oxidized and the other gets reduced. This occurs because the oxidation state of the elements changes as a result of the reaction. Cu in CuO has a oxidation state of 2+ and oxygen in CuO has a oxidation state of 2-. All elements in their elemental state have an oxidation state of 0. So elemental hydrogen and copper have oxidation states of 0. The hydrogen in water has an oxidation state of 1+ and oxygen again has a oxidation state of 2-. The Cu goes from an oxidation state of 2+ as a reactant to a state of 0 as a product. Hydrogen goes from an oxidation state of 0 as a reactant to a state of 1+ as a product. Copper has gained 2 electrons and each hydrogen atom has lost 1 electron. Oxidation is loss of electrons and reduction is gain of electrons (OILRIG), so copper has been reduced and hydrogen has been oxidized. The copper acted as an oxidizing agent for the hydrogen and hydrogen acted as a reducing agent for the copper.

Copper is the oxidizing agent.