Redox Titrations with Potassium Manganate(VII) – Solutions and Titration (Senior 5 Chemistry) | Ganzaa

Solutions and Titration

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Senior 5 · Chemistry

Solutions and Titration

5 lessons

Lesson 3 of 5

Redox Titrations with Potassium Manganate(VII)

25 min

How much iron is in an iron tablet?

Iron tablets are given to treat and prevent anaemia; a common tablet of dried iron(II) sulfate is labelled "200 mg ferrous sulfate, equivalent to 65 mg iron". A quality-control chemist can check that label with a single titration, because iron(II) ions are easily oxidised to iron(III), and potassium manganate(VII) is a strong oxidising agent with a built-in indicator: its own deep purple colour.

Definition

Redox titration: a titration in which the reaction between titrant and analyte is an oxidation–reduction. Moles are related through the electrons transferred.

Self-indicating titration: one in which a reactant's own colour change shows the end point, so no indicator is added. Example: $\mathrm{MnO_4^{-}}$ (purple) → $\mathrm{Mn^{2+}}$ (almost colourless).

Autocatalysis: catalysis of a reaction by one of its own products, e.g. $\mathrm{Mn^{2+}}$ speeding up the reaction of $\mathrm{MnO_4^{-}}$ with ethanedioate ions.

The half-equations

In acid solution, manganate(VII) gains five electrons:

$\mathrm{MnO_4^{-}(aq) + 8H^{+}(aq) + 5e^{-} \rightarrow Mn^{2+}(aq) + 4H_2O(l)}$

Reducing agent analysed	Oxidation half-equation	Overall ratio $\mathrm{MnO_4^{-}}$ : reductant
iron(II)	$\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^{-}}$	1 : 5
ethanedioate (oxalate)	$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}}$	2 : 5
hydrogen peroxide	$\mathrm{H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-}}$	2 : 5
nitrite	$\mathrm{NO_2^{-} + H_2O \rightarrow NO_3^{-} + 2H^{+} + 2e^{-}}$	2 : 5

Balancing the electrons gives, for example:

$\mathrm{MnO_4^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$

$\mathrm{2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$

Conditions, and the reasons for them

Potassium manganate(VII) goes in the burette. Read the top of the meniscus, since the solution is too dark to see the bottom; be consistent for both readings.
Excess dilute sulfuric acid is added to the flask. With too little acid, manganate(VII) is reduced only to manganese(IV) oxide, a brown precipitate, and the 5-electron ratio no longer holds.
Not hydrochloric acid: $\mathrm{MnO_4^{-}}$ oxidises $\mathrm{Cl^{-}}$ to chlorine, so extra titrant is used and the titre is too high. Not nitric acid: it is itself an oxidising agent and can oxidise the analyte. Not ethanoic acid: too weak to supply enough $\mathrm{H^{+}}$ .
Ethanedioate titrations are done warm (about 60 °C). The reaction is slow at first, then speeds up as $\mathrm{Mn^{2+}}$ forms and catalyses it.
No indicator: the end point is the first permanent pale pink (Figure 11.4). Potassium manganate(VII) is not a primary standard, so it is standardised first, against sodium ethanedioate or ammonium iron(II) sulfate.

Figure 11.4: What the flask looks like during a manganate(VII) titration.

Worked example

Problem. 0.1675 g of sodium ethanedioate, $\mathrm{Na_2C_2O_4}$ (M = 134.00), is dissolved in dilute sulfuric acid, warmed to 60 °C and titrated with potassium manganate(VII); the titre is 24.85 cm³. Calculate the concentration of the $\mathrm{KMnO_4}$ .

Solution.

$n(\mathrm{C_2O_4^{2-}}) = \frac{0.1675}{134.00} = 1.250 \times 10^{-3}\,\mathrm{mol} \qquad n(\mathrm{MnO_4^{-}}) = \frac{2}{5} \times 1.250 \times 10^{-3} = 5.000 \times 10^{-4}\,\mathrm{mol}$

$c = \frac{5.000 \times 10^{-4}}{0.02485} = 0.02012\,\mathrm{mol\,dm^{-3}}$

Answer: 0.02012 mol dm⁻³.

Worked example

Problem. One iron tablet is dissolved in dilute sulfuric acid and made up to 100.0 cm³. 25.00 cm³ portions need 11.60 cm³ of 0.005000 mol dm⁻³ $\mathrm{KMnO_4}$ . Calculate the mass of iron in one tablet and compare it with the label value of 65 mg ( $A_{r}$ (Fe) = 55.845).

Solution.

$n(\mathrm{MnO_4^{-}}) = 0.005000 \times 0.01160 = 5.800 \times 10^{-5}\,\mathrm{mol} \qquad n(\mathrm{Fe^{2+}}) = 5 \times 5.800 \times 10^{-5} = 2.900 \times 10^{-4}\,\mathrm{mol} \text{ in } 25.00\,\mathrm{cm^{3}}$

In the whole 100.0 cm³: $4 \times 2.900 \times 10^{-4} = 1.160 \times 10^{-3}\,\mathrm{mol}$ ; mass $= 1.160 \times 10^{-3} \times 55.845 = 0.0648\,\mathrm{g} = 64.8\,\mathrm{mg}$ .

Answer: 64.8 mg, within 0.4 % of the label. The method assumes all the iron is still iron(II); any iron(III) formed by air oxidation is not counted.

Warning

Safety. Solid potassium manganate(VII) is a strong oxidiser: keep it away from organic materials, which it can ignite. It stains skin and clothes brown. Dilute sulfuric acid irritates skin and eyes; wear goggles.

First aid (CDC/NIOSH Pocket Guide, sulfuric acid): eyes, irrigate with water immediately; skin, flush with water immediately; swallowed, get medical attention immediately.

Tip

Remember the number 5 for manganate(VII): Mn falls from +7 to +2, a change of 5 electrons per $\mathrm{MnO_4^{-}}$ . Divide 5 by the electrons released per mole of the other reactant and you have the ratio: $\mathrm{Fe^{2+}}$ gives 1, so 1 : 5; $\mathrm{C_2O_4^{2-}}$ and $\mathrm{H_2O_2}$ give 2, so 2 : 5.

Rugby School Chemistry: Redox Titration between MnO₄⁻ and Fe²⁺. Watch the purple colour vanish as each portion is added, and the first permanent pale pink at the end point; note that the top of the meniscus is read.

✦

Practice

Write the half-equation for $\mathrm{MnO_4^{-}}$ in acid, and the overall equation for its reaction with hydrogen peroxide.
Give two reasons why dilute sulfuric acid, and not hydrochloric acid, is used to acidify manganate(VII) titrations.
25.00 cm³ of iron(II) sulfate solution needs 22.40 cm³ of 0.02000 mol dm⁻³ $\mathrm{KMnO_4}$ . Calculate $[\mathrm{Fe^{2+}}]$ .
A solution of hydrogen peroxide is diluted tenfold. 25.00 cm³ of the diluted solution needs 17.65 cm³ of 0.02000 mol dm⁻³ $\mathrm{KMnO_4}$ . Calculate the concentration of the original $\mathrm{H_2O_2}$ in mol dm⁻³.
Explain why the first portions of manganate(VII) added to warm ethanedioate solution decolourise slowly, and later portions decolourise almost instantly.
Calculate the concentration of excess $\mathrm{MnO_4^{-}}$ when one drop (0.05 cm³) of 0.0200 mol dm⁻³ $\mathrm{KMnO_4}$ is added to 50 cm³ of solution after the end point. Explain why this makes an indicator unnecessary.
A student titrates an iron(II) solution acidified with hydrochloric acid and obtains a higher titre than a classmate who used sulfuric acid. Evaluate which result is more reliable.

Show answersHide answers

$\mathrm{MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O}$ ; $\mathrm{2MnO_4^{-} + 5H_2O_2 + 6H^{+} \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$ .
Sulfuric acid supplies plenty of $\mathrm{H^{+}}$ and its $\mathrm{SO_4^{2-}}$ ion is not oxidised by manganate(VII). Chloride would be oxidised to chlorine, using extra titrant.
$n(\mathrm{MnO_4^{-}}) = 4.480 \times 10^{-4}$ ; $n(\mathrm{Fe^{2+}}) = 2.240 \times 10^{-3}$ ; $c = 0.08960\,\mathrm{mol\,dm^{-3}}$ .
$n(\mathrm{MnO_4^{-}}) = 3.530 \times 10^{-4}$ ; $n(\mathrm{H_2O_2}) = \frac{5}{2} \times 3.530 \times 10^{-4} = 8.825 \times 10^{-4}$ in 25.00 cm³, i.e. 0.03530 mol dm⁻³; original = 0.353 mol dm⁻³.
The reaction is slow until some $\mathrm{Mn^{2+}}$ has formed; $\mathrm{Mn^{2+}}$ catalyses the reaction (autocatalysis), so later portions react quickly.
$n = 0.0200 \times 0.05 \times 10^{-3} = 1.0 \times 10^{-6}\,\mathrm{mol}$ in 0.050 dm³: $2.0 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ . Solutions this dilute are already visibly pink, so the first excess drop shows itself.
The sulfuric acid result is more reliable. In hydrochloric acid some manganate(VII) oxidises chloride ions to chlorine, so the titre includes manganate that did not react with iron(II), and the iron content would be overestimated.

Sources and credits

Iron tablet label value: British National Formulary entry for ferrous sulfate (200 mg dried ferrous sulfate ≈ 65 mg iron).
First aid: NIOSH Pocket Guide to Chemical Hazards: Sulfuric acid, US Centers for Disease Control and Prevention.
Image: Potassium permanganate solutions 2.JPG by Leiem, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Redox Titration between MnO₄⁻ and Fe²⁺, Rugby School Chemistry, embedded from YouTube.
The text, Figure 11.4, the table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1How much iron is in an iron tablet?

2The half-equations

3Conditions, and the reasons for them

How much iron is in an iron tablet?

The half-equations

Conditions, and the reasons for them