Question

Given the reaction that occurs in an electrochemical cell: Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s) During this reaction, the oxidation number of Zn changes from

Answer 1

Zn⁰(s) + CuSO4(aq) --> Zn²⁺SO4²⁻(aq) + Cu(s)

Zn⁰ --- in free elements oxidation number is 0.
Zn²⁺SO4²⁻ ---- ionic compound , it has ion Zn²⁺, so oxidation number here is +2.
The oxidation number of Zn changes from 0 to +2.