Zn (s) + CuSO 4(aq) → ZnSO 4(aq) + Cu (s) In the reaction above, the oxidation number of the reducing agent changes from
Question 1 Report
Zn (s) + CuSO 4(aq) → ZnSO 4(aq) + Cu (s)
In the reaction above, the oxidation number of the reducing agent changes from
Answer Details
In the given chemical reaction, Zn is the reducing agent because it causes the reduction of Cu²⁺ to Cu. Reduction is a process where an atom or ion gains electrons. In this reaction, Zn loses electrons and gets oxidized to Zn²⁺. The oxidation number of an atom or ion is the charge it would have if all its bonds were completely ionic.
Before the reaction, the oxidation state of Zn is 0 because it is in its elemental form. After the reaction, Zn is oxidized to Zn²⁺, which has an oxidation state of +2. Therefore, the oxidation number of the reducing agent (Zn) changes from 0 to +2.
So, the correct answer to the question is "0 to +2".
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