Oxidation numbers and balancing redox equations

When a piece of zinc is dropped into hydrochloric acid, the gas that fizzes off is hydrogen. Each zinc atom hands over two electrons, and those electrons end up on hydrogen ions. In a reaction with many atoms, counting electrons one by one would be exhausting. Oxidation numbers give you a bookkeeping system to track them, and half-equations let you balance the equation step by step. By the end of this page you will be able to balance equations like these in the Leaving Cert style.

Oxidation numbers

Put a clean strip of copper into colourless silver nitrate solution. The liquid slowly turns blue and shiny crystals of silver grow on the metal. No wires, no sparks, yet electrons have moved. How can we track the electrons gained or lost by each element in any compound?

Rules for assigning oxidation numbers, in order of priority

An uncombined element has oxidation number $0$. A simple (monatomic) ion has an oxidation number equal to its charge.
Combined oxygen is $-2$, except in peroxides ($-1$, because of the O–O bond) and when combined with fluorine (positive, because F is more electronegative).
Combined hydrogen is $+1$, except in metal hydrides, where it is $-1$ because the metal is less electronegative than hydrogen.
The oxidation numbers of all the atoms add up to the overall charge on the species (zero for a neutral compound).

Why? Oxygen and fluorine are the most electronegative common elements, so in a bond they take the shared electrons, which is why oxygen is usually $-2$ and positive only when bonded to fluorine. Hydrogen is less electronegative than most non-metals ($+1$) but more electronegative than metals ($-1$ in hydrides). In a peroxide the two O atoms share their O–O pair equally, so oxygen is $-1$. Two identical atoms share electrons equally, so an element has oxidation number $0$. Since the total charge on a compound or ion is conserved, the oxidation numbers must add up to exactly that total.

When naming compounds, the oxidation number is written in Roman numerals, as in iron(II) sulfate or potassium manganate(VII). When writing it on an atom, put the sign first: $+2$, $-1$.

Oxidation, reduction and agents

Practice★☆☆

When a sheet of metallic zinc is placed in an aqueous solution of hydrochloric acid, vigorous bubbling of hydrogen gas occurs according to the following process:…

When a sheet of metallic zinc is placed in an aqueous solution of hydrochloric acid, vigorous bubbling of hydrogen gas occurs according to the following process:

$$\mathrm{Zn(s) + 2\,HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$$

a) Calculate the oxidation number of all the atoms in the reactants and products.
b) Identify the species that is oxidised, the one that is reduced, the oxidising agent and the reducing agent, justifying your answer in terms of electron transfer.

Solve

Balancing using oxidation numbers

The Leaving Cert method is to balance redox equations using the changes in oxidation numbers. The idea is simple: the electrons lost by the species oxidised must equal the electrons gained by the species reduced.

A subscript in the formula: $\mathrm{Cr_2O_7^{2-}}$ and $\mathrm{Fe^{2+}}$

Skeleton: $\mathrm{Cr_2O_7^{2-} + Fe^{2+} \rightarrow Cr^{3+} + Fe^{3+}}$. Cr goes from $+6$ to $+3$, a decrease of $3$ per atom. There are $2$ Cr atoms in the formula, so the total decrease is $2 \times 3 = 6$. Fe goes from $+2$ to $+3$, an increase of $1$.

To make the totals equal you need $6$ Fe for each dichromate: $\mathrm{Cr_2O_7^{2-} + 6\,Fe^{2+} \rightarrow 2\,Cr^{3+} + 6\,Fe^{3+}}$. Balance the Cr atoms too: $2$ on each side. ✓

Balance the oxygen with $\mathrm{H_2O}$, then the hydrogen with $\mathrm{H^+}$, and check the charges.

Seven O on the left need $7\,\mathrm{H_2O}$ on the right, which needs $14\,\mathrm{H^+}$ on the left: $\mathrm{Cr_2O_7^{2-} + 6\,Fe^{2+} + 14\,H^+ \rightarrow 2\,Cr^{3+} + 6\,Fe^{3+} + 7\,H_2O}$. Charge: left $-2 + 12 + 14 = +24$, right $+6 + 18 = +24$. ✓

The ion-electron method in acidic solution

Add purple manganate(VII) to iron(II) sulfate in sulfuric acid and the purple disappears instantly. The ion-electron method splits the change into two half-equations. Oxidation numbers are your main method for balancing whole equations. You need the ion-electron method for the half-equations at the electrodes in cells and electrolysis, so it is worth mastering.

Practice★☆☆

In acidic aqueous medium, the nitrate ion can be converted into nitrogen monoxide, while the tin(II) ion can change to tin(IV). a) Write and balance the reduction…

In acidic aqueous medium, the nitrate ion can be converted into nitrogen monoxide, while the tin(II) ion can change to tin(IV).

a) Write and balance the reduction half-equation of the nitrate ion, $\mathrm{NO_3^-(aq)}$, to nitrogen monoxide, $\mathrm{NO(g)}$, in acidic medium, stating the number of electrons transferred.
b) Write the oxidation half-equation of the cation $\mathrm{Sn^{2+}(aq)}$ to $\mathrm{Sn^{4+}(aq)}$ and check that both half-equations balance simultaneously in mass and electric charge.

Solve

Practice★★☆

The diagram below illustrates the flow of electrons that occurs spontaneously when two redox half-cells interact through a conducting circuit. Observe the arrangement of…

The diagram below illustrates the flow of electrons that occurs spontaneously when two redox half-cells interact through a conducting circuit. Observe the arrangement of the chemical species and the direction of the electron flow:

a) Write the balanced oxidation and reduction half-equations, deducing the numerical value that must replace the question mark in the number of electrons transferred, and write the overall net ionic equation.
b) Assign, with justification, the roles of oxidising agent and reducing agent from the changes in oxidation numbers.

Solve

Basic solution

Practice★★☆

In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidise the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide,…

In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidise the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide, $\mathrm{MnO_2(s)}$, and the sulfate anion, $\mathrm{SO_4^{2-}}$.

a) Write and balance, using the ion-electron method in basic medium, the oxidation and reduction half-equations, detailing the balancing of atoms using $\mathrm{H_2O}$ and $\mathrm{OH^-}$, as well as the balancing of charges using electrons.
b) Obtain the overall net ionic equation and check explicitly that the total charge of the reactants equals that of the products.

Solve

From the ionic to the full equation

On the lab shelf there are no bottles labelled "manganate(VII) ions" or "iron(II) ions". There are real salts such as $\mathrm{KMnO_4}$ and $\mathrm{FeSO_4}$. How do we turn an ionic equation into one with the real formulae?

Summary

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★★☆ The diagram below illustrates the flow of electrons that occurs spontaneously when two…★★☆ In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidise the…

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