Oxidation and Reduction Concepts

Two definitions, one idea. The classical definition says oxidation is gain of oxygen or loss of hydrogen, and reduction is the reverse — useful for $2Mg + O_2 \rightarrow 2MgO$, where magnesium is oxidised. But it cannot explain $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$, since no oxygen appears anywhere. The general, electron-transfer definition covers both: oxidation is loss of electrons, reduction is gain of electrons. Zinc loses two electrons (oxidised), copper ion gains them (reduced). The species that gets oxidised is called the reducing agent because it donates electrons to reduce something else; the species that gets reduced is the oxidising agent because it accepts electrons to oxidise something else — a naming direction students frequently flip.

Oxidation number as a tracking tool. Since electron transfer is often shared or partial in covalent bonds, chemists assign each atom an oxidation number: the charge it would carry if every bond in the molecule were fully ionic, with shared electron pairs assigned to the more electronegative atom. This turns "who gained/lost electrons" into simple addition. The assignment rules, applied strictly in this priority order, are: Rule 1: any atom in its free elemental form (uncombined) has oxidation number $0$, such as in $Cl_2$, $Fe$, $P_4$. Rule 2: for a monatomic ion, the oxidation number equals the ionic charge, so $Na^+ = +1$ and $O^{2-} = -2$. Rule 3: hydrogen is usually $+1$, except $-1$ in metal hydrides like $NaH$; oxygen is usually $-2$, except $-1$ in peroxides like $H_2O_2$, and $+2$ in $OF_2$. Rule 4: alkali metals are always $+1$, alkaline earth metals always $+2$, and fluorine is always $-1$ in every compound it forms. Rule 5, the closing constraint that solves for unknowns, is the balance equation:

where: the sum runs over every atom in the formula unit; for a neutral molecule the overall charge is $0$, and for a polyatomic ion it equals the ion's charge (e.g., $-2$ for $SO_4^{2-}$). A reaction qualifies as redox only if at least one element's oxidation number rises (that element is oxidised) while another's falls (that element is reduced) somewhere between reactants and products. Reactions where no oxidation number changes at all — most acid-base neutralisations, for instance — are not redox.

Four reaction patterns. Combination reactions join two species where at least one reactant starts in elemental form, e.g. $C(s) + O_2(g) \rightarrow CO_2(g)$. Decomposition reactions break a compound apart into at least one elemental product, e.g. electrolysis of water, $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$. Displacement reactions swap one element for another inside a compound, e.g. $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$. Disproportionation is the odd one out: a single element in one oxidation state ends up in two different oxidation states among the products — simultaneously oxidised and reduced. In $2H_2O_2 \rightarrow 2H_2O + O_2$, oxygen starts at $-1$ and ends up split between $-2$ (in water) and $0$ (in $O_2$).

Question. Assign oxidation numbers to every element in potassium dichromate, $K_2Cr_2O_7$, and state the oxidation number of chromium — the element responsible for its strength as an oxidising agent.

Step 1. Apply the fixed rules first. Potassium is an alkali metal, so by Rule 4 it is always $+1$. Oxygen here is not in a peroxide or bonded to fluorine, so by Rule 3 it takes its usual value of $-2$.

Step 2. Let chromium's oxidation number be $x$. The compound is a neutral molecule, so by Rule 5 the overall charge $q = 0$, and the balance equation from the derivation above becomes

Step 3. Solve the linear equation:

Step 4. State and interpret the answer. In $K_2Cr_2O_7$, potassium is $+1$, oxygen is $-2$, and chromium is $+6$. This is chromium's highest common oxidation state, which is exactly why dichromate is such a powerful oxidising agent in reactions: it has "room" to fall all the way down to $Cr^{3+}$ ($+3$), gaining three electrons per chromium atom while oxidising whatever it reacts with.

Competitive stretch

Now check the thermite reaction, $Fe_2O_3(s) + 2Al(s) \rightarrow Al_2O_3(s) + 2Fe(l)$, for redox character. In $Fe_2O_3$, oxygen is $-2$, so $2x + 3(-2) = 0$ gives iron at $+3$; aluminium starts free at $0$. In the products, aluminium in $Al_2O_3$ is $+3$ by the same balance, while iron as free liquid metal is $0$. Iron falls from $+3$ to $0$ (reduced) and aluminium rises from $0$ to $+3$ (oxidised), so this is a redox displacement reaction, with $Fe_2O_3$ as the oxidising agent and $Al$ as the reducing agent.

• Swapping the agent names: the oxidising agent is the species that gets reduced (it oxidises something else by taking electrons); the species that gets oxidised is the reducing agent. Students often assign these backwards.
• Forgetting the peroxide and metal-hydride exceptions to the oxygen and hydrogen rules, which silently breaks the neutrality-balance equation and gives the wrong unknown.
• Assuming any reaction that forms a new compound must be redox. Combination and decomposition reactions are redox only when an element appears in free elemental form on one side; acid-base neutralisation, for example, is not redox at all.
• Missing disproportionation, where a single element shows up in two different oxidation states among the products of the same reaction — students often look only at whether one element rose and a different element fell.

Ask the tutor: Want to practice assigning oxidation numbers in a trickier ion like $Cr_2O_7^{2-}$ or classify a few more reactions as redox or not — ask the AI tutor to walk through extra examples with you.