5.2.3(a)-(c) - Redox agents, half-equations and predictions

What oxidising and reducing agents do

An oxidising agent and a reducing agent are named by what they make happen to another species. That is why the words can feel backwards at first.

For example:

Zinc changes from $\ce{Zn}$ to $\ce{Zn^2+}$. It loses two electrons:

So zinc is oxidised. Because zinc gives electrons to $\ce{Cu^2+}$ and causes $\ce{Cu^2+}$ to be reduced, zinc is the reducing agent.

Copper(II) ions change from $\ce{Cu^2+}$ to $\ce{Cu}$. They gain two electrons:

So $\ce{Cu^2+}$ is reduced. Because $\ce{Cu^2+}$ removes electrons from zinc and causes zinc to be oxidised, $\ce{Cu^2+}$ is the oxidising agent.

Using oxidation numbers as evidence

In many redox questions, you cannot see the electrons directly. Oxidation numbers let you infer the electron transfer.

If an oxidation number increases, oxidation has occurred. If an oxidation number decreases, reduction has occurred.

Be precise about the species. The agent is usually a whole ion, molecule or atom in the equation, not just the element whose oxidation number changes.

For a neutral species, oxidation numbers sum to zero; for an ion they sum to its charge. Oxygen is normally −2, but in a peroxide such as $\ce{H2O2}$ it is −1. Hydrogen is normally +1 in these examples. Oxidation numbers are bookkeeping values and must not be mistaken for charges on the whole molecule.

Worked example: silver nitrate and phosphine

In an unfamiliar reaction, phosphine reacts with silver nitrate:

Look only at the elements whose oxidation numbers change.

Silver:

• in $\ce{AgNO3}$, silver is $\ce{Ag+}$, so $\ce{Ag}$ has oxidation number +1
• in $\ce{Ag(s)}$, $\ce{Ag}$ has oxidation number 0
• +1 to 0 is a decrease, so silver ions are reduced

Phosphorus:

• in $\ce{PH3}$, hydrogen is +1, so phosphorus is -3
• in $\ce{H3PO3}$, hydrogen is +1 and oxygen is -2, so phosphorus is +3
• -3 to +3 is an increase, so phosphorus is oxidised

Therefore $\ce{AgNO3}$, more exactly the $\ce{Ag+}$ in $\ce{AgNO3}$, provides the oxidising species. $\ce{PH3}$ is the reducing agent because it is oxidised.

The ratio also makes sense. Each $\ce{Ag+}$ gains 1 electron, and phosphorus loses 6 electrons per $\ce{PH3}$ molecule, so 6$\ce{Ag+}$ ions are needed for each $\ce{PH3}$ molecule. That is M0.2 ratio reasoning in a redox equation.

Writing half-equations

A half-equation shows one side of the redox process: either oxidation or reduction. It includes electrons so that both atoms and charge balance.

The electron position tells you the redox process.

• Oxidation: electrons are on the product side, because electrons are lost.
• Reduction: electrons are on the reactant side, because electrons are gained.

Worked example 1: iron(II) to iron(III)

Iron(II) ions are oxidised to iron(III) ions:

Charge check:

• left side: +2
• right side: +3 plus -1 = +2

The charge balances, so the half-equation is correct.

A common error is to start from Fe(s). Do not do that unless the question actually says iron metal is reacting. If the question says iron(II) ions are oxidised, the half-equation must start with $\ce{Fe^2+}$.

Worked example 2: chlorine to chloride

Chlorine molecules are reduced to chloride ions:

The 2 before $\ce{Cl^-}$ is needed because chlorine exists as $\ce{Cl2}$ here. The 2 electrons are needed because the two chlorine atoms each gain one electron.

Combining half-equations into redox equations

To construct a full redox equation from two half-equations, the electrons must cancel. The number of electrons lost in oxidation must equal the number of electrons gained in reduction.

The method is:

1. Write the oxidation half-equation.
2. Write the reduction half-equation.
3. Multiply one or both half-equations so the electron numbers match.
4. Add the half-equations.
5. Cancel electrons and any other species that appear unchanged on both sides.
6. Check atoms and total charge.