Which half-reaction from the given notes represents oxidation?
A$Cu^{2+} + 2e^{-} \rightarrow Cu$
B$Zn^{2+} + 2e^{-} \rightarrow Zn$
C$Zn \rightarrow Zn^{2+} + 2e^{-}$
D$SO_4^{2-} \rightarrow SO_2 + 2e^{-}$
Answer & Solution
Correct answer: C. $Zn \rightarrow Zn^{2+} + 2e^{-}$
In oxidation, electrons appear on the product side because the species loses electrons.
In $Zn \rightarrow Zn^{2+} + 2e^{-}$, zinc loses two electrons, so this is oxidation.
**Option A is reduction because copper ions gain electrons.**
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