Zn _(s) + CuSO ₄(aq) → ZnSO ₄(aq) + Cu _(s)
In the reaction above, the oxidation number of the reducing agent changes from

ibukun246

8 years ago

Zn in its free state is 0,
From ZnSO4 ,oxidation no of S = 6,oxidation no of O = -2 .
Hence, Zn + 6 + -2(4) = 0
Zn + 6 - 8 = 0
Zn = -6 + 8
Zn = +2,
Therefore, oxidation no of Zn moved from 0 to +2.