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SAChemistrySyllabus dot point

How can redox titrations measure the concentration of oxidising or reducing species in a sample?

Use redox titrations, including permanganate titrations, to determine the concentration of an analyte from balanced half-equations and stoichiometry.

How redox titrations such as permanganate titrations determine analyte concentration using self-indicating endpoints, balanced half-equations and mole-ratio stoichiometry.

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  1. What this dot point is asking
  2. How a redox titration works
  3. Permanganate as a self-indicating titrant
  4. Worked stoichiometry
  5. Reliability of the result
  6. Other redox titrants

What this dot point is asking

You must be able to set up and interpret a redox titration, write and combine half-equations, and carry the stoichiometry through to a concentration. Permanganate titrations are the classic SACE example.

How a redox titration works

A redox titration is just like an acid-base titration except the reaction is electron transfer rather than proton transfer. A burette delivers a solution of known concentration (the titrant or standard) into a measured volume of the analyte until the reaction is exactly complete (the equivalence point).

Permanganate as a self-indicating titrant

Potassium permanganate, KMnO4\text{KMnO}_4, is the most common SACE titrant because it is self-indicating. In acidic solution the permanganate ion is reduced:

MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)

MnO4\text{MnO}_4^- is intensely purple; Mn2+\text{Mn}^{2+} is almost colourless. While analyte remains, each added drop is decolourised. At the endpoint the first slight excess of MnO4\text{MnO}_4^- turns the solution a faint, permanent pink - no separate indicator is needed.

Worked stoichiometry

A common analyte is iron(II), which is oxidised:

Fe2+(aq)Fe3+(aq)+e\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-

To combine the half-equations, balance the electrons. Multiply the iron half-equation by 5 so it matches the 5 electrons gained by permanganate, then add:

MnO4(aq)+8H+(aq)+5Fe2+(aq)Mn2+(aq)+5Fe3+(aq)+4H2O(l)\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)

The mole ratio is therefore 1MnO4:5Fe2+1\,\text{MnO}_4^- : 5\,\text{Fe}^{2+}.

Reliability of the result

As with any titration, use the average of concordant titres (within 0.10 mL0.10\ \text{mL} of each other) and discard the rough trial. Rinse the burette with the titrant and the pipette with the analyte so residual water does not dilute either solution.

Other redox titrants

  • Iodine/thiosulfate titrations measure oxidising agents indirectly; starch is added near the endpoint to sharpen the blue-black to colourless change.
  • Dichromate (Cr2O72\text{Cr}_2\text{O}_7^{2-}) is a strong oxidiser used with a separate redox indicator because its colour change is less distinct.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 SACE Stage 22 marksA titration determined Fe3+ in a 20.00 mL wastewater sample. Excess I- was added (2Fe3+ + 2I- -> I2 + 2Fe2+) and the I2 titrated with 0.9930 mol L-1 S2O3 2- (I2 + 2S2O3 2- -> 2I- + S4O6 2-). The average titre was 3.51 mL. Calculate the number of moles of S2O3 2- in the average titre.
Show worked answer →

Use n = cV with the volume in litres.

n(S2O3 2-) = c x V = 0.9930 x (3.51 / 1000)
n(S2O3 2-) = 0.9930 x 0.00351 = 3.485 x 10^-3 mol

Rounded to an appropriate number of significant figures, n(S2O3 2-) = 3.49 x 10^-3 mol. One mark for correct substitution (converting mL to L), one mark for the answer. (The average titre of 3.51 mL excludes the outlier 3.75 mL trial.)

2024 SACE Stage 23 marksUsing 2Fe3+ + 2I- -> I2 + 2Fe2+ and I2 + 2S2O3 2- -> 2I- + S4O6 2-, and n(S2O3 2-) = 3.49 x 10^-3 mol in a titre, determine the concentration of Fe3+, in mol L-1, in the 20.00 mL wastewater sample. Express the answer to the appropriate number of significant figures.
Show worked answer →

Work back through the two mole ratios.

  1. From I2 + 2S2O3 2-, n(I2) = n(S2O3 2-) / 2 = 3.49 x 10^-3 / 2 = 1.74 x 10^-3 mol.

  2. From 2Fe3+ + 2I- -> I2, n(Fe3+) = 2 x n(I2) = 2 x 1.74 x 10^-3 = 3.49 x 10^-3 mol.

  3. c(Fe3+) = n / V = 3.49 x 10^-3 / 0.02000 = 0.174 mol L-1 (3 significant figures). One mark per ratio step, one for the final concentration.

2023 SACE Stage 23 marksA back-titration found ethanol in a distillate. Excess Cr2O7 2- oxidised the ethanol, and the leftover Cr2O7 2- was titrated with 0.344 mol L-1 Fe2+ (Cr2O7 2- + 6Fe2+ + 14H+ -> 2Cr3+ + 6Fe3+ + 7H2O). Calculate the number of moles of Fe2+ that reacted if the titre was 20.73 mL. Express your answer to the appropriate number of significant figures.
Show worked answer →

Apply n = cV with volume in litres.

n(Fe2+) = c x V = 0.344 x (20.73 / 1000)
n(Fe2+) = 0.344 x 0.02073 = 7.131 x 10^-3 mol

To 3 significant figures (matching the data), n(Fe2+) = 7.13 x 10^-3 mol. One mark for converting volume to litres, one for substitution, one for the correctly rounded answer.