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Redox titrations (SACE Stage 2 Chemistry)

Syllabus dot point

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

9 min read

Quick answer

A redox titration finds the concentration of an oxidising or reducing analyte by reacting it with a standard solution of the opposite type. You combine the two half-equations to get the overall mole ratio, calculate moles of the standard from n=cVn = cV, apply the ratio, then divide by the sample volume. Permanganate titrations are self-indicating: the first permanent pink shows the endpoint.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. Self-indicating endpoints
  4. Building the overall equation
  5. Iodine/thiosulfate (iodometric) titrations
  6. Why it matters for monitoring
  7. Exam-style questions

What this dot point is asking

SACE expects you to construct the balanced overall equation from half-equations, read off the correct (often non 1:11:1) ratio, and complete the stoichiometric calculation, including multi-step iodine/thiosulfate analyses.

Lead worked calculation

Permanganate titration of oxalate

A 0.200 g0.200\ \text{g} sample of impure sodium oxalate was dissolved, acidified, and titrated with 0.0150 mol L−10.0150\ \text{mol L}^{-1} KMnO4\text{KMnO}_4; the average titre was 28.20 mL28.20\ \text{mL}. Find the percentage purity. (M(Na2C2O4)=134.00 g mol−1M(\text{Na}_2\text{C}_2\text{O}_4) = 134.00\ \text{g mol}^{-1}.)

The key challenge in a redox titration calculation is finding the correct mole ratio before doing any arithmetic. We derive the ratio from the half-equations, then work through the stoichiometry.

Step 1: Combine the half-equations to find the mole ratio

Write both half-equations with the electrons transferred:

MnO4−+8H++5e−→Mn2++4H2O\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

C2O42−→2CO2+2e−\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-

Electrons must balance: multiply the reduction half by 22 and the oxidation half by 55, giving the overall equation:

2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O.2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O.}

The mole ratio is MnO4−:C2O42−=2:5\text{MnO}_4^- : \text{C}_2\text{O}_4^{2-} = 2:5.

Step 2: Calculate moles of permanganate from the titre

Use n=cVn = cV, converting the titre to litres:

n(MnO4−)=0.0150×0.02820=4.230×10−4 mol.n(\text{MnO}_4^-) = 0.0150 \times 0.02820 = 4.230 \times 10^{-4}\ \text{mol.}

Step 3: Apply the mole ratio to find moles of oxalate

For every 2 moles of MnO4−\text{MnO}_4^- there are 5 moles of C2O42−\text{C}_2\text{O}_4^{2-}, so multiply by 52\dfrac{5}{2}:

n(C2O42−)=52×4.230×10−4=1.058×10−3 mol.n(\text{C}_2\text{O}_4^{2-}) = \frac{5}{2} \times 4.230 \times 10^{-4} = 1.058 \times 10^{-3}\ \text{mol.}

Step 4: Convert moles to mass and find percentage purity

Multiply by the molar mass of sodium oxalate, then divide by the sample mass and convert to a percentage:

m=nM=1.058×10−3×134.00=0.1417 g.m = nM = 1.058 \times 10^{-3} \times 134.00 = 0.1417\ \text{g.}

% purity=0.14170.200×100=70.9%.\%\text{ purity} = \frac{0.1417}{0.200} \times 100 = 70.9\%.

Final answer: The sample is 70.9%70.9\% pure sodium oxalate.

Self-indicating endpoints

Potassium permanganate is its own indicator. MnO4−\text{MnO}_4^- is intensely purple, while the product Mn2+\text{Mn}^{2+} is almost colourless. While reductant remains, each drop is decolourised; at the endpoint the first tiny excess of MnO4−\text{MnO}_4^- gives the solution a faint permanent pink. No separate indicator is needed. Iodine titrations are nearly self-indicating (the brown I2\text{I}_2 fades), but a starch indicator is added near the endpoint to give a sharp blue-black to colourless change.

Acidifying a permanganate titration

Permanganate titrations are carried out in excess dilute sulfuric acid, not hydrochloric or nitric acid. Hydrochloric acid would be oxidised to chlorine by MnO4−\text{MnO}_4^-, consuming permanganate and giving a falsely high titre, while nitric acid is itself an oxidiser. Sulfuric acid supplies the H+\text{H}^+ needed by the half-equation without interfering.

Building the overall equation

The crux of redox titration calculations is the ratio, which is rarely 1:11:1.

  1. Write each half-equation with its electrons.
  2. Multiply each so the electrons lost equal the electrons gained.
  3. Add and cancel electrons, H+\text{H}^+ and H2O\text{H}_2\text{O} where possible.
  4. Read the mole ratio of analyte to titrant directly from the coefficients.

Common ratios to know: MnO4−:Fe2+=1:5\text{MnO}_4^- : \text{Fe}^{2+} = 1:5; MnO4−:C2O42−=2:5\text{MnO}_4^- : \text{C}_2\text{O}_4^{2-} = 2:5; Cr2O72−:Fe2+=1:6\text{Cr}_2\text{O}_7^{2-} : \text{Fe}^{2+} = 1:6; I2:S2O32−=1:2\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2.

Iodine/thiosulfate (iodometric) titrations

Many oxidisers are measured indirectly. The oxidiser first liberates iodine from excess iodide (oxidiser+I−→I2\text{oxidiser} + \text{I}^- \rightarrow \text{I}_2), and the liberated I2\text{I}_2 is then titrated with standard thiosulfate. Because two steps are chained, you must apply two ratios in sequence, as in the dissolved-oxygen example above. This indirect approach lets a single standard (thiosulfate) measure many different oxidising analytes.

Mixing up the direction of the ratio

Students frequently invert the mole ratio, multiplying when they should divide. Always anchor on the substance whose moles you know (the titrant) and read the ratio as "analyte per titrant". For MnO4−:Fe2+=1:5\text{MnO}_4^- : \text{Fe}^{2+} = 1:5, there are five Fe2+\text{Fe}^{2+} per permanganate, so multiply the permanganate moles by 55. A second frequent slip is using hydrochloric acid to acidify a permanganate titration, which oxidises the chloride and ruins the result.

Why it matters for monitoring

Redox titrations quantify species that acid-base methods cannot, such as dissolved oxygen, chemical oxygen demand, iron content in water and ores, and the concentration of disinfectants. The Winkler dissolved-oxygen analysis and permanganate-based oxygen-demand tests are standard tools in environmental water monitoring.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks
A 25.00 mL25.00\ \text{mL} sample of a solution containing iron(II) ions was acidified and titrated with 0.0200 mol L−10.0200\ \text{mol L}^{-1} potassium permanganate. The average titre was 21.30 mL21.30\ \text{mL}. The reaction is MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}. Calculate the concentration of Fe2+\text{Fe}^{2+} in the sample.
Show worked answer →

Step 1: n(MnO4−)=cV=0.0200×0.02130=4.260×10−4 moln(\text{MnO}_4^-) = cV = 0.0200 \times 0.02130 = 4.260 \times 10^{-4}\ \text{mol}. (1 mark)

Step 2: the equation gives a 1:51:5 ratio of MnO4−\text{MnO}_4^- to Fe2+\text{Fe}^{2+}, so n(Fe2+)=5×4.260×10−4=2.130×10−3 moln(\text{Fe}^{2+}) = 5 \times 4.260 \times 10^{-4} = 2.130 \times 10^{-3}\ \text{mol}. (2 marks)

Step 3: c(Fe2+)=nV=2.130×10−30.02500=0.0852 mol L−1c(\text{Fe}^{2+}) = \dfrac{n}{V} = \dfrac{2.130 \times 10^{-3}}{0.02500} = 0.0852\ \text{mol L}^{-1}. (2 marks)

SACE 2020-style5 marks
The dissolved oxygen in a water sample was measured by a Winkler-type method in which the liberated iodine was titrated with sodium thiosulfate: I2+2S2O32−→2I−+S4O62−\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}. A 50.0 mL50.0\ \text{mL} sample required 8.40 mL8.40\ \text{mL} of 0.0100 mol L−10.0100\ \text{mol L}^{-1} thiosulfate. Given that O2+4I−+4H+→2I2+2H2O\text{O}_2 + 4\text{I}^- + 4\text{H}^+ \rightarrow 2\text{I}_2 + 2\text{H}_2\text{O}, calculate the concentration of dissolved oxygen in mol L−1\text{mol L}^{-1}.
Show worked answer →

Step 1: n(S2O32−)=0.0100×0.00840=8.40×10−5 moln(\text{S}_2\text{O}_3^{2-}) = 0.0100 \times 0.00840 = 8.40 \times 10^{-5}\ \text{mol}. (1 mark)

Step 2: I2:S2O32−=1:2\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2, so n(I2)=12(8.40×10−5)=4.20×10−5 moln(\text{I}_2) = \tfrac{1}{2}(8.40 \times 10^{-5}) = 4.20 \times 10^{-5}\ \text{mol}. (1 mark)

Step 3: from O2→2I2\text{O}_2 \rightarrow 2\text{I}_2, the ratio is 1:21:2, so n(O2)=12(4.20×10−5)=2.10×10−5 moln(\text{O}_2) = \tfrac{1}{2}(4.20 \times 10^{-5}) = 2.10 \times 10^{-5}\ \text{mol}. (1 mark)

Step 4: c(O2)=2.10×10−50.0500=4.20×10−4 mol L−1c(\text{O}_2) = \dfrac{2.10 \times 10^{-5}}{0.0500} = 4.20 \times 10^{-4}\ \text{mol L}^{-1}. (2 marks)

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Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained