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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
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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
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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
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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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