4.2.6

# Titrations (A2 Only)

Test yourself

## 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+