Titrations (A2 Only)

Chemistry: AQA A Level, topic 4.2.6

Redox Titrations

Redox titrations use a redox reaction to measure the concentrations of unknown solutions. Like acid/base titrations, they require an indicator and use the same technique.

Common reagents

A redox titration is a good way of determining the concentration of Fe²⁺ in a solution.
Fe²⁺ is a reducing agent, so you can titrate it with an oxidising agent.
A common oxidising agent to use is potassium permanganate, KMnO₄.
KMnO₄ can only be reduced in acidic conditions, so you need to add an excess of sulfuric acid.

Doing a titration

First, fill the burette with KMnO₄.
Measure out a known volume of Fe²⁺ solution into a conical flask. Add an excess of sulfuric acid.
Titrate the Fe²⁺ solution with the KMnO₄ solution. When the colour turns slightly pink, your titration is complete.
This is because the KMnO₄ goes colourless when reduced, so when you see a colour change, all the reducing agent has reacted.

Concordant titrations

There are a lot of places to make errors in titrations, so we need to be sure that our result is accurate.
One way of doing this is to carry out titrations until we get the same result.
Concordant titrations are titrations that give the same result, no more than 0.10cm³ apart.
In practice, you repeat your titration until you get two values this far apart.

Balancing Half-Equations

To do some calculations with your titration result, you need to know the equation for the reaction. We use redox half-equations to develop the full equation.

Oxidising Fe²⁺

We know that if we’re oxidising Fe²⁺, it’s gonna go to Fe³⁺ (we can’t take any more electrons off than that!)
So the half-equation for the oxidation of Fe²⁺ is:
Fe²⁺ → Fe³⁺ + e^-

Reducing KMnO₄

We know that the KMnO₄ goes colourless, so which ion of manganese is colourless?
Mn²⁺ is technically a pale pink, but unless it’s very concentrated, you can’t actually see that. This is what we’re reducing it to.
So part of our half equation is: MnO₄^- + 5e^- → Mn²⁺
We now need to balance the number of oxygen atoms in this half equation.

Balancing oxygen and hydrogen

We’re in solution, so things that can get involved in the reaction are: H⁺, OH^- and water.
We added acid to the reaction flask, so we can’t have any OH^- (it would react with the acid and give water).
So what’s involved is acid on one side, and water on the other.
This is a general rule - you can add acid or hydroxide to one side, and water to the other.

Applying this to KMnO₄

For our half equation, we need to react four oxygen atoms from the KMnO₄, so here we’re going to use 8H⁺ ions to make four water molecules.
The eventual half-equation is:
MnO₄^- + 5e^- + 8H⁺ → Mn²⁺ + 4H₂O

Working out the overall equation

As earlier, our two half equations are:
MnO₄^- + 5e^- + 8H⁺ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e^-
To combine them, just multiply one of the equations to make the number of electrons match.
So we need 5Fe²⁺ → 5Fe³⁺ + 5e^- to balance the electrons in the other equation.
The overall equation is:
MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Test yourself on Titrations (A2 Only)

What's the end purpose of a redox titration?
Performing a Titration
Which equation does the oxidation of iron (II) always follow?
When adding the half equations of iron and potassium permanganate, what is the integer we must multiply the iron half equation through by?

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