Q.Write the two electron-transfer half-reactions for $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)$, label each as oxidation or reduction, and add them to recover the overall equation.

[!TLDR] $\text{Zn}\to\text{Zn}^{2+}+2e^{-}$ is oxidation (electrons lost); $\text{Cu}^{2+}+2e^{-}\to\text{Cu}$ is reduction (electrons gained); adding them cancels the electrons and recovers the overall equation. [!ANSWER] Oxidation: $\text{Zn}\to\text{Zn}^{2+}+2e^{-}$; Reduction: $\text{Cu}^{2+}+2e^{-}\to\text{Cu}$; sum $= \text{Zn}+\text{Cu}^{2+}\to\text{Zn}^{2+}+\text{Cu}$.

The overall reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)$ can be split into two half-reactions that show electron transfer explicitly. Zinc metal is converted to $\text{Zn}^{2+}$ ions, releasing two electrons: $\text{Zn} \to \text{Zn}^{2+} + 2e^{-}$. Since electrons appear as a product here, zinc is losing them — this is the oxidation half-reaction. Copper(II) ions are converted to metallic copper by accepting two electrons: $\text{Cu}^{2+} + 2e^{-} \to \text{Cu}$. Since electrons appear as a reactant here, copper is gaining them — this is the reduction half-reaction. Adding the two half-reactions: $\text{Zn} + \text{Cu}^{2+} + 2e^{-} \to \text{Zn}^{2+} + 2e^{-} + \text{Cu}$; the $2e^{-}$ appears identically on both sides and cancels, leaving exactly the observed overall equation $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$, confirming that the number of electrons released by zinc exactly equals the number captured by copper. [!ANSWER] Oxidation: $\text{Zn}\to\text{Zn}^{2+}+2e^{-}$; Reduction: $\text{Cu}^{2+}+2e^{-}\to\text{Cu}$; sum $= \text{Zn}+\text{Cu}^{2+}\to\text{Zn}^{2+}+\text{Cu}$.

Write each species as losing electrons (if it becomes more positively charged) or gaining electrons (if it becomes less positively charged), then add the two half-reactions and cancel the electrons.

Writing the electrons on the wrong side of a half-reaction (e.g. putting $2e^-$ as a reactant in the oxidation half instead of a product); forgetting to check that the electron counts in the two half-reactions are equal before adding.