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Redox titration calculations

Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II

Redox titration calculations follow the same route as acid-base ones, but the ratios come from half-equations and are rarely 1 : 1. In October 2024, about a third of candidates got the titration calculation right; most errors were in the ratio or the scaling.

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Ratios to know

Reaction	Ratio
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+	MnO4- : Fe2+ = 1 : 5
2MnO4- + 16H+ + 5C2O4²- -> 2Mn2+ + 8H2O + 10CO2	MnO4- : C2O4²- = 2 : 5
I2 + 2S2O3²- -> 2I- + S4O6²-	I2 : S2O3²- = 1 : 2
2Cu2+ + 4I- -> 2CuI + I2	Cu2+ : S2O3²- = 1 : 1 overall
IO3- + 5I- + 6H+ -> 3I2 + 3H2O	IO3- : S2O3²- = 1 : 6 overall
FeC2O4: both Fe2+ and C2O4²- are oxidised	MnO4- : FeC2O4 = 3 : 5

The route

1. n(titrant) = c x V.

2. Ratio to the substance in the flask (for iodometric titrations, go through I2).

3. Scale up to the whole solution if an aliquot was taken.

4. Then mass (n x M), percentage by mass, concentration or x in a formula.

Keep full figures until the end; give the answer to 3 s.f. (or a whole number for x).

Example: iron tablets

Tablets weighing 1.50 g in total are dissolved in dilute sulfuric acid and made up to 100.0 cm³. 25.0 cm³ portions need 12.40 cm³ of 0.0100 mol dm^-3 KMnO4.

n(MnO4-) = 1.24 x 10^-4 mol; n(Fe2+) = 6.20 x 10^-4 mol per portion; x 4 = 2.48 x 10^-3 mol; mass = 0.138 g; % iron = 9.23%.

Worked example

2.02 g of FeC2O4·xH2O is made up to 250.0 cm³. 25.0 cm³ portions need 34.25 cm³ of 0.0195 mol dm^-3 KMnO4. MnO4- : FeC2O4 = 3 : 5. Find x.

n(MnO4-) = 0.03425 x 0.0195 = 6.6788e-04 mol
n(FeC2O4) per portion = 5/3 x that; total = 0.01113 mol
M = 2.02 / 0.01113 = 181.5; FeC2O4 = 143.8
x = (181.5 - 143.8) / 18.0 = 2.09

Answer: x = 2

Key idea: Use the half-equation ratio (1 : 5 for MnO4- : Fe2+; 1 : 2 for I2 : S2O3²-; 1 : 1 for Cu2+ : S2O3²-). Scale for the aliquot. Do not round until the end.

Check yourself

25.0 cm³ of an iron(II) solution needs 21.60 cm³ of 0.0200 mol dm^-3 KMnO4. Calculate [Fe2+].
Show answer
0.0864 mol dm^-3. n(MnO4-) = 4.32 x 10^-4 mol; n(Fe2+) = 5 x that = 2.16 x 10^-3 mol; c = 2.16 x 10^-3 / 0.0250 = 0.0864 mol dm^-3.
Iron tablets with a total mass of 1.50 g are dissolved and made up to 100.0 cm³. 25.0 cm³ portions need 12.40 cm³ of 0.0100 mol dm^-3 KMnO4. Calculate the percentage by mass of iron (Fe = 55.8).
Show answer
9.23 %. n(Fe2+) per portion = 5 x 1.24 x 10^-4 = 6.20 x 10^-4; total 2.48 x 10^-3 mol; mass = 0.1384 g; 9.23%.
1.80 g of FeC2O4·xH2O is made up to 250.0 cm³. 25.0 cm³ portions need 30.00 cm³ of 0.0200 mol dm^-3 KMnO4 (MnO4- : FeC2O4 = 3 : 5). Find x (FeC2O4 = 143.8, H2O = 18.0).
Show answer
2. n(FeC2O4) total = 0.0300 x 0.0200 x 5/3 x 10 = 0.0100 mol; M = 180.0; x = 2.01, so 2.
0.250 g of brass is dissolved, neutralised and treated with excess KI. The iodine released needs 25.60 cm³ of 0.100 mol dm^-3 sodium thiosulfate. Calculate the percentage of copper in the brass (Cu = 63.5).
Show answer
65.0 %. n(S2O3²-) = 2.56 x 10^-3 mol = n(Cu2+); mass = 0.1626 g; 65.0%.
25.0 cm³ of potassium iodate(V) solution is added to excess acidified KI. The iodine needs 24.00 cm³ of 0.100 mol dm^-3 thiosulfate. Calculate [IO3-].
Show answer
0.0160 mol dm^-3. n(S2O3²-) = 2.40 x 10^-3; n(IO3-) = that / 6 = 4.00 x 10^-4 mol; c = 0.0160 mol dm^-3.
25.0 cm³ of ethanedioic acid solution needs 20.00 cm³ of 0.0200 mol dm^-3 KMnO4. Calculate the concentration of the ethanedioic acid.
Show answer
0.0400 mol dm^-3. n(MnO4-) = 4.00 x 10^-4; n(C2O4²-) = 5/2 x that = 1.00 x 10^-3 mol; c = 0.0400 mol dm^-3.
In the determination of copper by iodine-thiosulfate titration, what is the overall mole ratio Cu2+ : S2O3²-?
- 1 : 2
- 1 : 1
- 2 : 1
- 1 : 5
Show answer
1 : 1. 2Cu2+ gives 1 I2, which reacts with 2 S2O3²-.

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Last updated 8 November 2026. Spotted a mistake? Tell us.