Stoichiometry and redox titrations
In the school lab you drip purple potassium manganate(VII) from a burette into a flask of iron(II) solution. The colour vanishes on contact, until one single drop turns the flask faintly pink and the colour stays. How many moles of iron were in the flask? That pink drop tells you exactly, and this page shows you how to read it.
Moles from a balanced redox equation
A balanced redox equation is read like any other equation: the coefficients are mole ratios. The only extra job is getting the equation balanced in the first place.
First, balancing by oxidation numbers. In acidic solution, manganate(VII) oxidises iron(II). Mn goes from $+7$ to $+2$ (a change of $-5$) and each Fe goes from $+2$ to $+3$ (a change of $+1$). The changes must cancel, so 5 Fe²⁺ are needed for each MnO₄⁻: $\mathrm{MnO_4^- + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+}}$. Now balance the charge: the left has $-1 + 10 = +9$ and the right has $+2 + 15 = +17$, so add 8 H⁺ to the left. Finally balance the hydrogen and oxygen: 8 H⁺ need 4 H₂O on the right, which also balances the 4 oxygen atoms. Half-equations (the ion-electron method) give the same result. You need them for electrode reactions, and they are shown below.
The same method works for ethanedioate (oxalate) with manganate(VII). In $\mathrm{C_2O_4^{2-}}$ each carbon is $+3$, and in $\mathrm{CO_2}$ it is $+4$, so the two carbons together change by $+2$. Mn changes by $-5$. The lowest common multiple is 10, so 2 MnO₄⁻ react with 5 C₂O₄²⁻: $\mathrm{2MnO_4^- + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2}$. The left has charge $-2 - 10 = -12$ and the right has $+4$, so add 16 H⁺ to the left and 8 H₂O to the right: $\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$.
$$\text{Oxidation: } \mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$$ $$\text{Reduction: } \mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$$ Multiply the oxidation half-equation by 5 so the electrons cancel, then add. This ion-electron method is the one you use for half-equations at electrodes. For balancing a titration equation, oxidation-number changes are the main method.
In the permanganate titration of dissolved iron, the permanganate ion quantitatively oxidises the iron(II) cation in a strongly acidic medium produced with excess…
In the permanganate titration of dissolved iron, the permanganate ion quantitatively oxidises the iron(II) cation in a strongly acidic medium produced with excess sulfuric acid, being reduced to the manganese(II) ion while the iron is oxidised to the iron(III) ion.
- a) Balance, using the ion-electron method, the overall ionic equation for this redox process.
- b) If exactly $12.50\,\mathrm{mL}$ of an aqueous potassium permanganate solution $0.0200\,\mathrm{mol\times L^{-1}}$ is used to titrate a test sample, determine the amount of substance (in moles) of $\mathrm{Fe^{2+}}$ ions present in the sample analysed.
When two species are oxidised: counting electrons
Sometimes one sample contains two things that both lose electrons. Instead of writing one long equation, count electrons. In any redox reaction, the electrons lost by the reducing agent equal the electrons gained by the oxidising agent.
| Half-equation | Electrons per unit |
|---|---|
| $\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$ | gains 5 |
| $\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}$ | loses 1 |
| $\mathrm{Sn^{2+}} \rightarrow \mathrm{Sn^{4+}} + 2\mathrm{e^-}$ | loses 2 |
| $\mathrm{C_2O_4^{2-}} \rightarrow 2\mathrm{CO_2} + 2\mathrm{e^-}$ | loses 2 |
The same idea works when both ions come from one formula unit: a salt $\mathrm{XY}$ gives off the electrons from X and from Y together. For example, iron(II) ethanedioate (oxalate), $\mathrm{FeC_2O_4}$, loses 1 electron from $\mathrm{Fe^{2+}}$ and 2 from $\mathrm{C_2O_4^{2-}}$, so 3 electrons per formula unit. With $\mathrm{MnO_4^-}$ gaining 5, the ratio $\mathrm{MnO_4^-} : \mathrm{FeC_2O_4}$ is $3:5$. For a hydrate such as $\mathrm{FeC_2O_4 \cdot 2H_2O}$, include the water in the molar mass: $55.8 + 24.0 + 64.0 + 36.0 = 179.8\ \text{g/mol}$. Add up the electrons per formula unit, then divide by the electrons gained per oxidising agent to get the mole ratio in the balanced equation.
You have a pure salt containing only iron(II) oxalate dihydrate, of formula $\mathrm{FeC_2O_4 \times 2\,H_2O}$. In dilute, hot sulfuric acid medium, both constituent…
You have a pure salt containing only iron(II) oxalate dihydrate, of formula $\mathrm{FeC_2O_4 \times 2\,H_2O}$. In dilute, hot sulfuric acid medium, both constituent ions ($\mathrm{Fe^{2+}}$ and $\mathrm{C_2O_4^{2-}}$) are oxidised simultaneously by the permanganate ion:
$\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}$
$\mathrm{C_2O_4^{2-}} \rightarrow 2\,\mathrm{CO_2} + 2\,\mathrm{e^-}$
$\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 5\,\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}$
- a) Deduce the balanced overall ionic equation using the ion-electron method and determine the total number of electrons transferred per formula unit of $\mathrm{FeC_2O_4}$.
- b) If $0.2159\,\mathrm{g}$ of this pure salt is dissolved in excess sulfuric acid medium and titrated at $60\,^\circ\mathrm{C}$, determine the volume (in millilitres) of $\mathrm{KMnO_4}$ $0.0240\,\mathrm{mol\times L^{-1}}$ solution that will be used until the persistent pink end point is reached. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Fe} = 55.85$; $\mathrm{C} = 12.01$; $\mathrm{O} = 16.00$; $\mathrm{H} = 1.01$.
Titrations with potassium manganate(VII)
A titration adds a solution of known concentration from a burette to a measured volume of another solution until the reaction is exactly complete.
Potassium manganate(VII) is not a primary standard: it is slightly impure and decomposes in light. So it is first standardised against a primary standard, such as sodium oxalate. It is also its own indicator. The end point is the first faint pink that lasts for about 30 seconds.
The titration of oxalic acid with potassium permanganate requires controlled prior heating of the conical flask to about $60\,^\circ\mathrm{C}$ to speed up the slow…
The titration of oxalic acid with potassium permanganate requires controlled prior heating of the conical flask to about $60\,^\circ\mathrm{C}$ to speed up the slow initial kinetics before the $\mathrm{Mn^{2+}}$ ion produced acts as an autocatalyst. The balanced reaction is:
$2\,\mathrm{MnO_4^-} + 5\,\mathrm{H_2C_2O_4} + 6\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$
$20.00\,\mathrm{mL}$ of a standard solution of $\mathrm{H_2C_2O_4}$ $0.0500\,\mathrm{mol\times L^{-1}}$ is titrated, using $16.00\,\mathrm{mL}$ of $\mathrm{KMnO_4}$ solution. Calculate the molar concentration of the potassium permanganate solution used.
You have an aliquot of $25.0\,\mathrm{mL}$ of an aqueous solution containing $\mathrm{Fe^{2+}}$, treated with excess sulfuric acid. It is titrated by permanganate…
You have an aliquot of $25.0\,\mathrm{mL}$ of an aqueous solution containing $\mathrm{Fe^{2+}}$, treated with excess sulfuric acid. It is titrated by permanganate titration with a standard solution of $\mathrm{KMnO_4}$ $0.0200\,\mathrm{mol\times L^{-1}}$. In the experimental diagram below you can see the initial and final meniscus readings of the solution in the burette, corresponding to $3.10\,\mathrm{mL}$ and $18.10\,\mathrm{mL}$ respectively:
Consider the balanced ionic reaction:
$\mathrm{MnO_4^-} + 5\,\mathrm{Fe^{2+}} + 8\,\mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 5\,\mathrm{Fe^{3+}} + 4\,\mathrm{H_2O}$
- a) Determine the molar concentration of $\mathrm{Fe^{2+}}$ ions in the solution analysed.
- b) Justify the experimental criterion that defines the end point in this permanganate titration and state why hydrochloric acid should not be used instead of sulfuric acid to acidify the medium.
Iodine–thiosulfate titrations
Add colourless bleach to colourless potassium iodide solution and the mixture turns dark brown at once: iodine has formed. How can we use that brown iodine to measure the bleach, when we cannot titrate the bleach directly?
Sodium thiosulfate is not a primary standard either. It is standardised using a primary standard such as potassium iodate, which releases a known amount of iodine from excess iodide: $$\mathrm{IO_3^- + 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O}$$ One mole of iodate gives 3 moles of $\mathrm{I_2}$, and each mole of $\mathrm{I_2}$ needs 2 moles of thiosulfate. So $n(\mathrm{S_2O_3^{2-}}) = 6 \times n(\mathrm{IO_3^-})$.
The sodium hypochlorite ($\mathrm{NaClO}$) content of a commercial bleaching product is analysed by indirect iodometric titration. A $5.00\,\mathrm{mL}$ sample of bleach…
The sodium hypochlorite ($\mathrm{NaClO}$) content of a commercial bleaching product is analysed by indirect iodometric titration. A $5.00\,\mathrm{mL}$ sample of bleach is diluted to $250.0\,\mathrm{mL}$ with distilled water. An aliquot of $20.00\,\mathrm{mL}$ of this diluted solution is acidified with acetic acid and an excess of potassium iodide is added, producing the quantitative reaction:
$\mathrm{ClO^-} + 2\,\mathrm{I^-} + 2\,\mathrm{H^+} \rightarrow \mathrm{I_2} + \mathrm{Cl^-} + \mathrm{H_2O}$
The iodine produced is titrated with sodium thiosulfate $0.0400\,\mathrm{mol\times L^{-1}}$, requiring $18.50\,\mathrm{mL}$ until the iodine-starch complex is decolourised:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
- a) Calculate the moles of sodium hypochlorite in the aliquot titrated.
- b) Determine the concentration of sodium hypochlorite in the original commercial solution in grams per litre. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Na} = 23.0$; $\mathrm{Cl} = 35.5$; $\mathrm{O} = 16.0$.
In the synthesis of an inorganic product, a solution is obtained containing iodate ($\mathrm{IO_3^-}$) and iodide ($\mathrm{I^-}$) ions. When strongly acidified with…
In the synthesis of an inorganic product, a solution is obtained containing iodate ($\mathrm{IO_3^-}$) and iodide ($\mathrm{I^-}$) ions. When strongly acidified with sulfuric acid, comproportionation of both ions occurs, forming free molecular iodine ($\mathrm{I_2}$):
$\mathrm{IO_3^-} + 5\,\mathrm{I^-} + 6\,\mathrm{H^+} \rightarrow 3\,\mathrm{I_2} + 3\,\mathrm{H_2O}$
The starting point is a solution initially containing $1,200\,\mathrm{mmol}$ of $\mathrm{KIO_3}$ and an excess of $\mathrm{KI}$. After quantitatively collecting the iodine recovered from the process, it is titrated with sodium thiosulfate according to:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
Given that a solution of $\mathrm{Na_2S_2O_3}$ $0.1500\,\mathrm{mol\times L^{-1}}$ was used and that the net volume consumed was $43.20\,\mathrm{mL}$:
- a) Calculate the theoretical amount of $\mathrm{I_2}$ that should have formed, assuming a yield of $100\,\%$ in the comproportionation reaction.
- b) Determine the actual moles of $\mathrm{I_2}$ obtained from the titration and deduce the overall percentage yield of iodine production and recovery.
Concentration and percentage purity of a commercial sample
Hydrogen peroxide acts as a reducing agent towards potassium permanganate in sulfuric acid medium. To determine the concentration of a commercial hydrogen peroxide…
Hydrogen peroxide acts as a reducing agent towards potassium permanganate in sulfuric acid medium. To determine the concentration of a commercial hydrogen peroxide solution whose density is $1,015\,\mathrm{g\times mL^{-1}}$, $2.00\,\mathrm{mL}$ of the sample is taken and diluted in a volumetric flask to $100.0\,\mathrm{mL}$. An aliquot of $20.00\,\mathrm{mL}$ of this solution is then titrated with $\mathrm{KMnO_4}$ $0.0150\,\mathrm{mol\times L^{-1}}$, using $19.60\,\mathrm{mL}$ of titrant.
Reaction: $2\,\mathrm{MnO_4^-} + 5\,\mathrm{H_2O_2} + 6\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 5\,\mathrm{O_2} + 8\,\mathrm{H_2O}$
- a) Calculate the molar concentration of $\mathrm{H_2O_2}$ in the commercial hydrogen peroxide solution.
- b) Determine the percentage by mass (% w/w) of $\mathrm{H_2O_2}$ in this commercial sample. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{H} = 1.0$; $\mathrm{O} = 16.0$.
Blank tests and back titrations
Sometimes a direct titration does not work: the sample is a slowly reacting solid, or there is no clear end point. Then a back titration is used. A known excess of reagent is added to the sample. The reagent left over is then titrated, and the analyte is found by difference: $$n_{\text{analyte reacted}} = n_{\text{reagent added}} - n_{\text{reagent left over}}$$
A blank test repeats the whole procedure with every reagent except the analyte. Impurities in the water or reagents use up a little titrant. That volume is subtracted from the titre before you calculate.
In quantitative chemical analysis, a "blank test" is often carried out to correct for the volume of reagent consumed by impurities in the solvents or by the matrix…
In quantitative chemical analysis, a "blank test" is often carried out to correct for the volume of reagent consumed by impurities in the solvents or by the matrix itself.
In an iodometric titration with sodium thiosulfate according to the reaction:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
a test solution is titrated using $24.35\,\mathrm{mL}$ of $\mathrm{Na_2S_2O_3}$ $0.0100\,\mathrm{mol\times L^{-1}}$. A blank test with all the reagents except the analyte requires $0.15\,\mathrm{mL}$ of the same titrant solution.
- a) Justify the need to subtract the blank volume from the total volume measured.
- b) Calculate the mass of iodine ($\mathrm{I_2}$) present in the test sample. Data: $M(\mathrm{I}) = 126.9\,\mathrm{g\times mol^{-1}}$.
The purity of a mineral containing pyrolusite ($\mathrm{MnO_2}$) is analysed by a redox back titration. $0.2450\,\mathrm{g}$ of powdered mineral is weighed and treated…
The purity of a mineral containing pyrolusite ($\mathrm{MnO_2}$) is analysed by a redox back titration. $0.2450\,\mathrm{g}$ of powdered mineral is weighed and treated while hot with $50.00\,\mathrm{mL}$ of a standard solution of sodium oxalate ($\mathrm{Na_2C_2O_4}$) $0.0500\,\mathrm{mol\times L^{-1}}$ in excess sulfuric acid medium, producing complete dissolution of the manganese dioxide according to the reaction:
$\mathrm{MnO_2} + \mathrm{C_2O_4^{2-}} + 4\,\mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 2\,\mathrm{CO_2} + 2\,\mathrm{H_2O}$
The excess oxalate that did not react is titrated while hot at $60\,^\circ\mathrm{C}$ with $\mathrm{KMnO_4}$ $0.0200\,\mathrm{mol\times L^{-1}}$, using $17.85\,\mathrm{mL}$. A blank test of the auxiliary reagents, with the same acidic medium but without mineral or added oxalate, requires $0.35\,\mathrm{mL}$ of $\mathrm{KMnO_4}$.
Titration reaction: $2\,\mathrm{MnO_4^-} + 5\,\mathrm{C_2O_4^{2-}} + 16\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$
- a) Calculate the net moles of oxalate that actually reacted with the manganese dioxide of the mineral.
- b) Determine the percentage by mass of $\mathrm{MnO_2}$ in the mineral analysed. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Mn} = 54.94$; $\mathrm{O} = 16.00$. Assume that the gangue is inert and that the blank consumption corresponds exclusively to reducing impurities in the auxiliary reagents.
Back to the start: the volume from the burette gives the moles of titrant ($n = cV$). The coefficients turn these into moles of analyte. Dilution factors take you back to the original sample, and the molar mass and density give you $\text{g/L}$, $\%\ \text{w/v}$ or $\%\ \text{w/w}$.