Understanding why reactions happen, how to control them, and how to predict products.
Metals are arranged in order of decreasing reactivity. The higher a metal is in the reactivity series, the more easily it loses electrons to form positive ions.
A more reactive metal displaces a less reactive metal from its compound.
Example: $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$
Zinc is higher in the reactivity series, so it displaces Copper from its compound. If you tried $Cu + ZnSO_4$, there would be no reaction.
Q1: Why does Magnesium react with Copper Sulfate but Copper does not react with Magnesium Sulfate?
Magnesium is higher in the reactivity series (more reactive) than Copper, so it can displace it. Copper is less reactive, so it cannot displace Magnesium.
$$ \text{Metal} + \text{Acid} \rightarrow \text{Salt} + \text{Hydrogen} $$
Example: $Mg + 2HCl \rightarrow MgCl_2 + H_2$
Hydrogen test is a lit splint producing a squeaky pop. Never test Hydrogen with a "glowing splint"—that is used only for Oxygen.
In modern chemistry, redox is about electron transfer. Oxidation and reduction always occur together in a reaction.
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)
Q1: If an atom loses an electron, has it been oxidised or reduced?
It has been oxidised (Loss Is Oxidation).
Choosing the correct method depends on the reactivity of the reactants.
| Method | Reactants | Example |
|---|---|---|
| Metal + Acid | Reactive metal + Acid | $Mg + HCl$ |
| Insoluble Base | Acid + Metal Oxide/Carbonate | Making $CuSO_4$ from $CuO$ |
| Neutralisation | Acid + Alkali (Titration) | Making $NaCl$ from $NaOH$ |
| Precipitation | Two soluble salts | Making insoluble $AgCl$ |
Q1: Why do we add "excess" Copper Oxide when making Copper Sulfate?
To ensure all of the acid is neutralised so the final salt is pure.
Using electricity to move and break down ions in an ionic compound. Remember the mnemonic PANIC:
Q1: In the electrolysis of aqueous $NaCl$, why does Hydrogen form at the cathode instead of Sodium?
Because Sodium is more reactive than Hydrogen, so Hydrogen is released instead.
Half equations show what happens at each electrode. The total charge on both sides of a half-equation must be equal.
At Cathode (Reduction): $Cu^{2+} + 2e^- \rightarrow Cu$
At Anode (Oxidation): $2Cl^- \rightarrow Cl_2 + 2e^-$
Q1: Balance this half equation: $Al^{3+} + ?e^- \rightarrow Al$
$$ Al^{3+} + 3e^- \rightarrow Al $$
1. Don't say a metal "pushes out" another; use "displaces from its compound".
2. Don't forget that Calcium reacts slowly with cold water compared to Sodium.
3. Don't assume metals form in aqueous electrolysis; Hydrogen usually wins.
4. Don't forget that oxidation is at the anode and reduction is at the cathode.