Titrations

Test your knowledge with free interactive questions on Seneca — used by over 10 million students.

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⁺