4.2.6

Titrations (A2 Only)

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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.

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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.
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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.
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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.

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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-
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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.
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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.
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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
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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+

Jump to other topics

1Physical Chemistry

2Physical Chemistry 2 (A2 Only)

3Inorganic Chemistry

4Inorganic Chemistry 2 (A2 Only)

5Organic Chemistry 1

6Organic Chemistry 2 (A2 Only)

6.1Optical Isomerism (A2 Only)

6.2Aldehydes & Ketones (A2 Only)

6.3Carboxylic Acids & Esters (A2 Only)

6.4Aromatic Chemistry (A2 Only)

6.5Amines (A2 Only)

6.6Polymers (A2 Only)

6.7Biological Organic (A2 Only)

6.8Organic Synthesis (A2 Only)

6.9NMR Spectroscopy (A2 Only)

6.10Chromatography (A2 Only)

6.11A-A* (AO3/4) - Organic 2

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