Redox and Electrons

Oxidation and Reduction in Terms of Electrons

• In Higher Tier chemistry, oxidation and reduction are defined strictly by the movement of electrons, not just the gain or loss of oxygen.
• Oxidation is the loss of one or more electrons by a substance.
• Reduction is the gain of one or more electrons by a substance.
• You can remember these definitions using the standard mnemonic OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).
• A redox reaction is a reaction in which both oxidation and reduction occur simultaneously. They must happen together because any electrons lost by one species must be immediately gained by another.
• In these reactions, species take on specific roles based on the reactivity series:
  • An oxidising agent is a species that accepts electrons and becomes reduced.
  • A reducing agent is a species that donates electrons and becomes oxidised.

Analysing Displacement Reactions

Every displacement reaction is a redox reaction. Electrons are transferred from the more reactive atom (which acts as the reducing agent) to the less reactive ion (which acts as the oxidising agent).

To understand exactly what is happening, chemists use ionic equations.

• A spectator ion is an ion that exists in the exact same form (same formula, same charge, and usually aqueous) on both the reactant and product sides of a chemical equation. Because it does not take part in the reaction, it is "just watching".
• An ionic equation is a chemical equation that shows only the species (atoms or ions) that undergo a chemical change, completely excluding the spectator ions.
• Crucial Rule: Only substances with the aqueous state symbol (aq) are split into ions when deriving an ionic equation. Solids (s), liquids (l), and gases (g) must stay written as whole molecules or atoms.

Step-by-Step: Deriving Net Ionic Equations

To successfully write an ionic equation and identify the oxidised/reduced species, follow this step-by-step breakdown.

Example 1: Magnesium displacing Copper from Copper(II) Sulfate

Step 1: Write the full balanced equation. $Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)$

Step 2: Expand the aqueous substances into their ions. $Mg(s) + Cu^{2+}(aq) + SO_4^{2-}(aq) \rightarrow Mg^{2+}(aq) + SO_4^{2-}(aq) + Cu(s)$

Step 3: Identify and remove the spectator ions. The sulfate ion, $SO_4^{2-}$, is identical on both sides of the equation. We cross it out.

Step 4: Write the final net ionic equation. $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$

Step 5: Identify the oxidised and reduced species based on electron transfer.

• Magnesium goes from a neutral atom $Mg$ to a positively charged ion $Mg^{2+}$. It has lost electrons, so magnesium is oxidised.
• Copper goes from a positively charged ion $Cu^{2+}$ to a neutral atom $Cu$. It has gained electrons, so copper ions are reduced.

Example 2: Halogen Displacement (Chlorine and Potassium Bromide)

• Full Equation: $Cl_2(g) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(l)$
• Net Ionic Equation: $Cl_2(g) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(l)$ (The spectator $K^+$ ions are removed)
• Oxidised Species: The $2Br^-$ ions lose electrons to form $Br_2$.
• Reduced Species: The $Cl_2$ molecules gain electrons to form $2Cl^-$ ions.

Half Equations

A half equation describes either the oxidation or the reduction process separately, explicitly showing the exact number of electrons ($e^-$) transferred.

• When constructing a half equation, electrons are always added to the more positive side of the equation to balance the electrical charges.

From the Iron and Copper(II) Sulfate displacement:

• Net Ionic Equation: $Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$
• Oxidation Half Equation: $Fe \rightarrow Fe^{2+} + 2e^-$ (The neutral iron atom loses 2 electrons).
• Reduction Half Equation: $Cu^{2+} + 2e^- \rightarrow Cu$ (The copper ion gains 2 electrons).

From the Halogen Displacement example:

• Oxidation Half Equation: $2Br^- \rightarrow Br_2 + 2e^-$
• Reduction Half Equation: $Cl_2 + 2e^- \rightarrow 2Cl^-$