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

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.

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The answer is 65.3312