Redox titrations
Why redox titrations suit this standard
- The mole ratios are almost never 1 : 1, so the stoichiometry does real work in the calculation.
- Several are self-indicating, so the endpoint chemistry can itself be explained and justified.
- The reactions are genuine Level 3 chemistry, with half equations that can be linked to the procedure.
The three common systems
Permanganate titrations
- Self-indicating — purple MnO4− becomes colourless Mn2+, so the endpoint is the first permanent faint pink.
- Must be acidified with sulfuric acid, never hydrochloric: chloride is a reductant and would be oxidised to chlorine in a competing reaction.
- Used for iron(II), oxalate, hydrogen peroxide.
Against iron(II), the overall equation gives the 1 : 5 ratio:
Iodine–thiosulfate titrations
- Starch is the indicator, giving a sharp blue-black to colourless endpoint.
- Add the starch only near the endpoint, when the solution has faded to pale straw. Added early, starch binds iodine so strongly that the endpoint is sluggish and reads high.
- Used for ascorbic acid, hypochlorite in bleach, copper(II), and for iodine liberated from iodate.
Dichromate titrations
- Not self-indicating — the orange-to-green change is gradual, so a redox indicator is needed.
- Dichromate is stable enough to be used as a primary standard, which is its main advantage.
- It can be used with hydrochloric acid, unlike permanganate, because its potential is not high enough to oxidise chloride.
Choosing which to use
| Analyte | Titrant | Ratio | Endpoint |
|---|---|---|---|
| Iron(II) in a supplement | acidified MnO4− | 1 : 5 | first permanent pink |
| Ascorbic acid (vitamin C) | I2 directly, or liberated I2 with S2O32− | 1 : 1 with I2, then 1 : 2 with thiosulfate | blue-black → colourless |
| Hypochlorite in bleach | excess I−, then S2O32− | 1 : 2 | blue-black → colourless |
| Hydrogen peroxide | acidified MnO4− | 2 : 5 | first permanent pink |
| Oxalate | acidified MnO4−, warmed | 2 : 5 | first permanent pink |
Justifying the procedure — the Excellence criterion
Excellence asks you to justify the steps used in the procedure in relation to the reactions occurring and to the nature of the samples. Every step in a redox titration has a chemical reason:
| Step | The reaction-based justification |
|---|---|
| Acidify with sulfuric acid | the permanganate half equation consumes 8H+ per MnO4−; chloride from HCl would be oxidised in competition |
| Add a large excess of acid | 8 mol of H+ per mole of permanganate is needed, and a shortfall gives brown MnO2 instead of colourless Mn2+ |
| Warm an oxalate titration | high activation energy makes the reaction too slow at room temperature |
| Add starch late in an iodine titration | starch–iodine binding is strong, so early addition makes the endpoint sluggish and high |
| Add excess iodide in an iodometric determination | ensures all the oxidant is converted to an equivalent amount of I2 |
| Titrate promptly after liberating iodine | iodine is volatile and is lost by evaporation, and iodide is air-oxidised |
| Stopper the flask between steps | limits both iodine loss and atmospheric oxidation |
- Notice that each justification names a specific reaction or property, not a general principle. That is what "in relation to the reactions occurring" means.
Worked ExampleAn iodometric determination of hypochlorite
A student determines the hypochlorite concentration in a bleach sample. 10.0 mL of bleach was diluted to 250.0 mL. A 25.0 mL aliquot was acidified and excess potassium iodide added. The liberated iodine required 21.65 mL of 0.0207 mol L−1 sodium thiosulfate. Calculate the concentration of hypochlorite in the original bleach.
Step 0 — Note the two ratios before starting
- OCl− : I2 is 1 : 1
- I2 : S2O32− is 1 : 2
So overall, OCl− : S2O32− is 1 : 2.
Step 1 — Moles of thiosulfate used
Step 2 — Moles of iodine that were present
From the second equation, I2 : S2O32− is 1 : 2, so there was half as much iodine:
Step 3 — Moles of hypochlorite in the aliquot
From the first equation, OCl− : I2 is 1 : 1:
The iodide was in excess, so it is not limiting and does not appear in the calculation.
Step 4 — Concentration in the diluted bleach
Step 5 — Scale back to the original bleach
The dilution factor is :
Step 6 — Significant figures
The measured quantities are the titre (21.65 mL, 4 sf), the thiosulfate concentration (0.0207 mol L−1, 3 sf) and the aliquot (25.0 mL, 3 sf). The limiting precision is 3 significant figures.
Step 7 — Justify two procedural steps
- Excess iodide was added so that all the hypochlorite was converted to an equivalent amount of iodine. If iodide were limiting, some hypochlorite would remain unreacted and the result would be low.
- The mixture was titrated promptly and kept stoppered, because iodine is volatile and is lost by evaporation, while iodide in acidic solution is slowly oxidised by atmospheric oxygen to give additional iodine. The first error makes the result low and the second makes it high, so a delay introduces error in an unpredictable direction.
Answer: the bleach contains hypochlorite at 0.224 mol L−1.