Big Idea #4: Redox equations - Online A Level Chemistry & Physics Tutor

Many reactions in Chemistry are redox reactions. This means that reduction and oxidation reactions are occuring at the same time, with one substance gaining electrons (reduction) and the other losing electrons (oxidation).

You need to be able to work out the balanced symbol equation for a redox reaction, usually starting from with a few bits of information:

Which reactant is reduced?: e.g., MnO₄^– –> Mn²⁺
Which reactant is oxidised? e.g., Fe²⁺ -> Fe³⁺
Are the conditions acidic / neutral or alkaline?

There is a method you can use to work out each half-equation in acidic or neutral conditions:

Balance the central atom (i.e., anything but O or H)
Balance the O with H₂O
Balance the H with H⁺
Balance the charge with electrons

Once we have the two half-equations, we can then combine them to make the overall balanced symbol equation. To do this, we must multiply the equations so that the number of electrons cancels out completely when we add them.

(In some cases, where they appear in both equations, cancelling of the H⁺ and H₂O is also necessary.)

Example 1: MnO₄^– and Fe²⁺ react in acidic conditions to form Mn²⁺ and Fe³⁺. Determine the overall balanced symbol equation.

If we apply the method, we get:

Reduction:
8 H⁺ + 5 e^– + MnO₄^– –> Mn²⁺ + 4 H₂O

Oxidation:
Fe²⁺ -> Fe³⁺ + e^–

Notice that the reduction half-equation transfers 5 electrons and the oxidation half-equation transfers 1 electron only. To combine these two half-equations into an overall redox equation, we need to multiply the oxidation reaction by 5, and then add both reactions.

Reduction:
8 H⁺ + 5 e^– + MnO₄^– –> Mn²⁺ + 4 H₂O

Oxidation:
5 Fe²⁺ -> 5 Fe³⁺ + 5 e^–

Overall: 8 H⁺ + MnO₄^– + 5 Fe²⁺–> Mn²⁺ + 4 H₂O + 5 Fe³⁺

You will notice that the 5 e^– on each side cancel and so do not appear in the overall equation. H⁺ and H₂O appear only in the reduction equation so no additional cancelling is needed.

Here is another, more complicated, example:

Example 2: VO²⁺ and Cr₂O₇^2- react in acidic conditions to form VO₂⁺ and Cr³⁺. Determine the overall balanced symbol equation.

If we apply the method, we get:

Reduction:

14 H⁺ + 6 e^– + Cr₂O₇^2- –> 2 Cr³⁺ + 7 H₂O

Oxidation:

H₂O + VO²⁺ -> VO₂⁺ + 2 H⁺ + e-

Notice that the reduction half-equation transfers 6 electrons and the oxidation half-equation transfer 1 electron only. To combine these two half-equations into an overall redox equation, we need to multiply the oxidation reaction by 6, and then add both reactions.

Reduction:

14 H⁺ + 6 e^– + Cr₂O₇^2- –> 2 Cr³⁺ + 7 H₂O

Oxidation:

6 H₂O + 6 VO²⁺ -> 6 VO₂⁺ + 12 H⁺ + 6 e-

Overall:

2 H⁺ + Cr₂O₇^2- + 6 VO²⁺-> 2 Cr³⁺ + H₂O + 6 VO₂⁺

You will notice that the 6 e^– on each side cancel and so do not appear in the overall equation. Since H⁺ and H₂O appear in both reduction and oxidation equations they also need to be cancelled.

So… how does all this link to calculations?

The most common type of question where you’ll need to balance a redox equation is a redox titration. Most simply – if you know the overall balanced symbol equation between two reactants, and the change in moles of one reactant, you can work out the change in moles of the other reactant. This is the starting point to a very wide range of calculations in analytical chemistry to determine e.g., moles, concentration, purity, calculating M_r, the water of crystallisation etc! The range of possible question types here is huge. We’ll do all of these in later posts.

Another interesting type of calculation for now is where only one half-equation is known and the other is not known:

Example 3: 0.20 mol MnO₄^– and 0.50 mol V²⁺ react in acidic conditions to form Mn²⁺ and another ion of vanadium. Determine the final oxidation state of V.

We know from the reduction half-equation that 0.2 mol of MnO₄^– would transfer 1.0 mol of electrons, since each MnO₄^– gains 5 electrons.

Reduction:

8 H⁺ + 5 e^– + MnO₄^– –> Mn²⁺ + 4 H₂O

This means that 0.50 mol of V²⁺would also transfer 1.0 mol of electrons. This means that each V²⁺ loses 2 electrons, so its final oxidation state is +4.

Example 4: 0.10 mol Cr₂O₇^2- and 0.20 mol of Fe react in acidic conditions to form Cr³⁺ and an ion of Fe. Determine the final oxidation state of Fe.

We know from the reduction half-equation that 0.10 mol of Cr₂O₇^2- would transfer 0.60 mol of electrons.

Reduction:

14 H⁺ + 6 e^– + Cr₂O₇^2- –> 2 Cr³⁺ + 7 H₂O

This means that 0.20 mol of Fewould also transfer 0.60 mol of electrons. This means that each Fe loses 3 electrons, so its final oxidation state is +3.

All the examples we have looked at so far involve acidic or neutral conditions. If you are required to balance redox equations in alkaline conditions, an extra step is needed:

Find the overall balanced symbol equation using the same method used for acidic and neutral conditions.
For each H⁺ in the equation, add a OH^– to both sides.

Example 5: H₂O₂ and Cl^– react in alkaline conditions to form Cl₂ and H₂O. Determine the overall balanced symbol equation.

We first work out the overall equation in acidic conditions:

Reduction:

2 H⁺ + 2 e^– + H₂O₂ –> 2 H₂O

Oxidation:

2 Cl^– -> Cl₂ + 2 e-

Overall (in acidic conditions):

2 H⁺ + 2 Cl^– + H₂O₂ –>Cl₂ + 2 H₂O

We then need to add 2 OH- to both sides to cancel the 2 H⁺. The 2 H⁺ on the left will be neutralised to form 2 H₂O. We then need to cancel the H₂O, which is now present on both sides.

Overall (in alkaline conditions):

2 Cl^– + H₂O₂ –>Cl₂ + 2 OH^–

Note – you are much less likely to encounter questions in alkaline conditions. This is because many transition metal hydroxides are insoluble. The halogens, in contrast, have quite an interesting set of reactions in alkaline conditions.