WAEC Chemistry 2016 Paper

Concept:In a cell reaction, the total number of electrons lost must equal the total number of electrons gained.Explanation:The oxidation half-reaction is:$\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^-}$The reduction half-reaction is:$\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$Electrons lost are $3$ per Al atom, and electrons gained are $2$ per Cu ion.Their lowest common multiple is $6$.Multiply the aluminium half-reaction by $2$:$2\mathrm{Al(s) \rightarrow 2Al^{3+}(aq) + 6e^-}$Multiply the copper half-reaction by $3$:$3\mathrm{Cu^{2+}(aq) + 6e^- \rightarrow 3Cu(s)}$Add the two balanced half-reactions and cancel $6e^-$:$2\mathrm{Al(s)} + 3\mathrm{Cu^{2+}(aq)} \rightarrow 3\mathrm{Cu(s)} + 2\mathrm{Al^{3+}(aq)}$Answer:Option C: $2\mathrm{Al(s)} + 3\mathrm{Cu^{2+}(aq)} \rightarrow 3\mathrm{Cu(s)} + 2\mathrm{Al^{3+}(aq)}$