7.1 Oxidation-reduction reactions and half-reactions

Oxidation-reduction (redox) reactions are all about electron transfer between chemical species. They drive processes ranging from corrosion to battery operation, and they form the foundation for the electrochemistry you'll study throughout this unit. Getting comfortable with half-reactions and oxidation numbers here will pay off when you move into galvanic cells and cell potentials.

Oxidation-Reduction Reactions

Oxidation and reduction definitions

Oxidation is the loss of electrons, which causes a species to increase in oxidation state. Reduction is the gain of electrons, which causes a species to decrease in oxidation state. The two always occur together: you can't have one without the other.

• When $\text{Fe}^{2+}$ loses an electron to become $\text{Fe}^{3+}$, that's oxidation. The oxidation state went up from +2 to +3.
• When $\text{Cl}_2$ gains electrons to become $\text{Cl}^-$, that's reduction. The oxidation state dropped from 0 to -1.

A useful mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain). For electrochemical cells specifically, oxidation occurs at the anode and reduction occurs at the cathode.

Oxidizing and reducing agents

This is where students often get tripped up, because the names seem backwards at first.

• An oxidizing agent accepts electrons from another species, causing that other species to be oxidized. The oxidizing agent itself gets reduced in the process. Common examples: $\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$, $\text{H}_2\text{O}_2$.
• A reducing agent donates electrons to another species, causing that other species to be reduced. The reducing agent itself gets oxidized. Common examples: $\text{H}_2$, $\text{Fe}^{2+}$, $\text{I}^-$.

Think of it this way: the agent is named for what it does to the other species, not what happens to itself.

Relative strengths: Among the halogens, oxidizing strength follows $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. Among the alkali metals, reducing strength follows $\text{Li} > \text{Na} > \text{K} > \text{Rb} > \text{Cs}$. You'll see these trends connect directly to standard reduction potentials later in this unit.

Balancing Redox Reactions and Oxidation Numbers

Balancing redox reactions

The half-reaction method keeps atom and charge bookkeeping manageable. Follow these steps in order:

1. Split the overall reaction into two half-reactions (one for oxidation, one for reduction).
2. Balance all atoms except H and O in each half-reaction.
3. Balance O by adding $\text{H}_2\text{O}$ to whichever side needs oxygen.
4. Balance H by adding $\text{H}^+$ to whichever side needs hydrogen.
5. Balance charge by adding electrons ($e^-$) to the more positive side of each half-reaction.
6. Equalize electrons by multiplying each half-reaction by the appropriate integer so both transfer the same number of electrons.
7. Add the half-reactions together and cancel species that appear on both sides.
8. Verify that atoms and charge both balance in the final equation.

For basic solutions, complete steps 1–7 in acidic conditions first. Then, for every $\text{H}^+$ in the final equation, add an equal number of $\text{OH}^-$ to both sides. Each $\text{H}^+ + \text{OH}^-$ pair becomes $\text{H}_2\text{O}$. Simplify any water molecules that appear on both sides.

Oxidation numbers in compounds

Oxidation numbers are a bookkeeping tool that tracks how electrons are distributed in a compound. Assigning them correctly is how you identify which species are oxidized and which are reduced.

Core rules (apply in this priority order):

• Free elements have an oxidation number of 0. This applies to any uncombined element: $\text{Na}$, $\text{O}_2$, $\text{P}_4$.
• Monatomic ions have an oxidation number equal to their charge.
• Fluorine is always -1 in compounds (it's the most electronegative element).
• Hydrogen is +1 in most compounds ($\text{HCl}$, $\text{H}_2\text{O}$), but -1 in metal hydrides like $\text{NaH}$ and $\text{CaH}_2$.
• Oxygen is -2 in most compounds ($\text{H}_2\text{O}$, $\text{CO}_2$), but -1 in peroxides ($\text{H}_2\text{O}_2$, $\text{Na}_2\text{O}_2$), and positive in compounds with fluorine ($\text{OF}_2$).

Sum rules:

• In a neutral compound, oxidation numbers must sum to 0. For example, in $\text{CH}_4$: carbon is -4 and each hydrogen is +1, giving $-4 + 4(+1) = 0$.
• In a polyatomic ion, oxidation numbers must sum to the ion's charge. For example, in $\text{SO}_4^{2-}$: sulfur is +6 and each oxygen is -2, giving $+6 + 4(-2) = -2$.

When you compare oxidation numbers before and after a reaction, any species whose oxidation number increased was oxidized, and any species whose oxidation number decreased was reduced. This is the most reliable way to identify the redox changes in a complex reaction.