Half equations show electrons lost or gained

1. Many reactions involve electron transfer, in which one species loses electrons and another gains them.
2. A half equation shows one of these changes on its own, with each electron written as $e^{-}$.
3. When a species loses electrons, the electrons appear on the right-hand side: $$Na\to {Na}^{+}+e^{-}$$
4. When a species gains electrons, the electrons appear on the left-hand side: $${Cu}^{2+}+2e^{-}\to Cu$$
5. A correct half equation balances both the atoms and the total charge on each side.
   1. In the copper half equation, the left side has a total charge of $2+$ plus $2-$, which is zero, matching the uncharged copper atom on the right.

Writing half equations by balancing atoms then charge

1. Write the starting species and the species it becomes, such as ${Cl}^{-}\to {Cl}_2$.
2. Balance the atoms with balancing numbers: $$2{Cl}^{-}\to {Cl}_2$$
3. Work out the total charge on each side, which here is $2-$ on the left and $0$ on the right.
4. Add electrons to the side that is more positive until the charges match: $$2{Cl}^{-}\to {Cl}_2+2e^{-}$$
5. The same method gives ${Al}^{3+}+3e^{-}\to Al$ and $2H^{+}+2e^{-}\to H_2$.
   1. Hydrogen forms diatomic molecules, so two hydrogen ions are needed to make each $H_2$ molecule.

Ionic equations show only the particles that change

1. In solution, an ionic compound is split into separate ions, so $NaCl(aq)$ is present as ${Na}^{+}(aq)$ and ${Cl}^{-}(aq)$.
2. Some of these ions take part in the reaction, while others stay unchanged in the solution.
   1. The unchanged ions appear on both sides of the full equation, so they are left out of the ionic equation.
3. Solids, liquids, gases and covalent molecules such as water are kept as whole formulae, because they are not present as separate ions.
4. Every acid and alkali neutralisation shares one ionic equation: $$H^{+}(aq)+{OH}^{-}(aq)\to H_{2}O(l)$$
   1. The metal ion from the alkali and the negative ion from the acid are both spectator ions.

Building ionic equations by cancelling spectator ions

1. Start from the full balanced equation with state symbols.
2. Split every aqueous ionic compound into its ions, keeping the balancing numbers.
3. Cross out any ion that appears unchanged on both sides.
4. Write what is left, then check that the atoms and charges balance.
5. For magnesium reacting with hydrochloric acid, the chloride ions are spectators, leaving $Mg(s)+2H^{+}(aq)\to {Mg}^{2+}(aq)+H_2(g)$.
   1. The total charge is $2+$ on each side, so this ionic equation is balanced.

Two half equations combine into an ionic equation

1. In an electron-transfer reaction, every electron lost by one species is gained by another.
2. Multiply one or both half equations so that the numbers of electrons match.
3. Add the half equations and cancel the electrons, which now appear equally on both sides.
4. For magnesium displacing copper from copper sulfate solution, $Mg\to {Mg}^{2+}+2e^{-}$ and ${Cu}^{2+}+2e^{-}\to Cu$ already have two electrons each.
   1. Adding them gives $Mg(s)+{Cu}^{2+}(aq)\to {Mg}^{2+}(aq)+Cu(s)$, with the sulfate ions as spectators.
5. For aluminium displacing copper, each aluminium atom loses three electrons and each copper ion gains two, so the aluminium half equation is multiplied by $2$ and the copper half equation by $3$, giving six electrons each.
   1. Adding and cancelling gives $2Al(s)+3{Cu}^{2+}(aq)\to 2{Al}^{3+}(aq)+3Cu(s)$, with a total charge of $6+$ on each side.