Redox titrations are quantitative analytical procedures used to determine the concentration of an unknown reducing or oxidising agent by reacting it with a standard solution. The most common redox titration encountered at A-Level involves the use of potassium manganate(VII), \( \text{KMnO}_4 \), as a powerful oxidising agent under acidic conditions.
🔑 Key Principle
Manganate(VII) titrations are self-indicating. The manganate(VII) ion, \( \text{MnO}_4^- \), is a intense deep purple colour, whereas the reduced manganese(II) ion, \( \text{Mn}^{2+} \), is virtually colourless. The end point is reached when a single extra drop of \( \text{MnO}_4^- \) added remains unreacted, turning the solution in the conical flask permanently pale pink.
The Chemistry of Manganate(VII) Titrations
In acidic solution, the purple manganate(VII) ion is reduced to the colourless manganese(II) ion. The half-equation for this reduction is:
\[ \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \to \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \]A titration in which the reaction between the analyte and the titrant is an oxidation-reduction reaction, involving a transfer of electrons.
A titration where one of the reacting species changes colour at the equivalence point, removing the need for an external chemical indicator.
Titration with Iron(II) Ions
When determining the concentration of iron(II) ions, \( \text{Fe}^{2+} \) is oxidised to iron(III) ions, \( \text{Fe}^{3+} \):
\[ \text{Fe}^{2+}(\text{aq}) \to \text{Fe}^{3+}(\text{aq}) + e^- \]Multiplying the iron half-equation by 5 and combining it with the manganate(VII) half-equation yields the overall ionic equation:
\[ \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \to \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \]This shows a clear 1:5 mole ratio between \( \text{MnO}_4^- \) and \( \text{Fe}^{2+} \).
The Critical Choice of Acid
To ensure that the reduction of \( \text{MnO}_4^- \) goes to completion (with manganese forming \( \text{Mn}^{2+} \)), a high concentration of \( \text{H}^+ \) ions is required. However, the choice of acid is extremely restricted because the acid must not participate in any redox reaction of its own.
⚠️ Acid Suitability Summary
- Dilute Sulfuric Acid (\( \text{H}_2\text{SO}_4 \)): The only suitable acid. The sulfate ion, \( \text{SO}_4^{2-} \), is stable and cannot be further oxidised by manganate(VII), nor is it a strong enough oxidising agent to react with the analyte.
- Hydrochloric Acid (\( \text{HCl} \)): Unsuitable. Manganate(VII) is a strong enough oxidising agent to oxidise the chloride ions (\( \text{Cl}^- \)) to toxic chlorine gas (\( \text{Cl}_2 \)): \[ 2\text{Cl}^-(\text{aq}) \to \text{Cl}_2(\text{g}) + 2e^- \] This would consume extra manganate(VII) solution, leading to an artificially high titre.
- Nitric Acid (\( \text{HNO}_3 \)): Unsuitable. The nitrate ion (\( \text{NO}_3^- \)) is itself a strong oxidising agent and would oxidise the \( \text{Fe}^{2+} \) ions to \( \text{Fe}^{3+} \) before the titration even begins, leading to an artificially low titre.
- Ethanoic Acid (\( \text{CH}_3\text{COOH} \)): Unsuitable. It is a weak acid and does not dissociate fully to provide a high enough concentration of \( \text{H}^+ \) ions. This leads to incomplete reduction of manganate(VII), forming a brown precipitate of manganese dioxide (\( \text{MnO}_2 \)), which obscures the end point: \[ \text{MnO}_4^-(\text{aq}) + 4\text{H}^+(\text{aq}) + 3e^- \to \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \]
Questions asking why a specific acid cannot be used are extremely common. When explaining why \( \text{HCl} \) is not used, you must explicitly state that the chloride ions are oxidised by the manganate(VII) ions, and explain that this would lead to an over-estimation of the volume of manganate needed (a larger titre).
Titration with Ethanedioate Ions
Manganate(VII) can also be used to determine the concentration of ethanedioic acid, \( \text{H}_2\text{C}_2\text{O}_4 \), or ethanedioate ions, \( \text{C}_2\text{O}_4^{2-} \). During this reaction, the ethanedioate ions are oxidised to carbon dioxide gas:
\[ \text{C}_2\text{O}_4^{2-}(\text{aq}) \to 2\text{CO}_2(\text{g}) + 2e^- \]Combining the two half-equations (multiplying the reduction by 2 and the oxidation by 5) gives:
\[ 2\text{MnO}_4^-(\text{aq}) + 16\text{H}^+(\text{aq}) + 5\text{C}_2\text{O}_4^{2-}(\text{aq}) \to 2\text{Mn}^{2+}(\text{aq}) + 10\text{CO}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l}) \]This reaction displays a 2:5 mole ratio between \( \text{MnO}_4^- \) and \( \text{C}_2\text{O}_4^{2-} \).
The reaction between \( \text{MnO}_4^- \) and \( \text{C}_2\text{O}_4^{2-} \) is very slow at room temperature because both reactants are negatively charged ions, resulting in a high activation energy due to electrostatic repulsion. To overcome this, the reaction mixture must be warmed to about 60 °C at the start. Once the reaction begins, the manganese(II) ions (\( \text{Mn}^{2+} \)) produced act as a homogeneous catalyst. This is a classic example of autocatalysis.
Redox Titration Calculations
Calculations follow the standard stoichiometry structure: calculate the moles of the known species, use the balanced chemical equation to find the moles of the unknown species, and then convert that to the required quantity (concentration, mass, or percentage purity).
Calculate the percentage by mass of \( \text{FeSO}_4 \) (\( M_r = 152.0 \)) in the iron tablet.
Step 1: Write down the reaction stoichiometry. From the combined equation: \[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \] The mole ratio of \( \text{MnO}_4^- : \text{Fe}^{2+} \) is 1:5.
Step 2: Calculate the moles of manganate(VII) used in the titration. \[ n(\text{MnO}_4^-) = c \times V = 0.0200 \text{ mol dm}^{-3} \times \left(\frac{18.50}{1000}\right) \text{ dm}^3 \] \[ n(\text{MnO}_4^-) = 3.70 \times 10^{-4} \text{ mol} \]
Step 3: Calculate the moles of \( \text{Fe}^{2+} \) in the 25.0 cm\(^3\) sample. Using the 1:5 mole ratio: \[ n(\text{Fe}^{2+}) \text{ in } 25.0 \text{ cm}^3 = 5 \times n(\text{MnO}_4^-) = 5 \times (3.70 \times 10^{-4}) = 1.85 \times 10^{-3} \text{ mol} \]
Step 4: Scale up to find the moles of \( \text{Fe}^{2+} \) in the original 250.0 cm\(^3\) solution. \[ n(\text{Fe}^{2+}) \text{ in } 250.0 \text{ cm}^3 = 1.85 \times 10^{-3} \text{ mol} \times \left(\frac{250.0}{25.0}\right) = 0.0185 \text{ mol} \]
Step 5: Calculate the mass of \( \text{FeSO}_4 \) in the tablet. Each mole of \( \text{Fe}^{2+} \) corresponds to one mole of \( \text{FeSO}_4 \): \[ m(\text{FeSO}_4) = n \times M_r = 0.0185 \text{ mol} \times 152.0 \text{ g mol}^{-1} = 2.812 \text{ g} \] Wait, this mass exceeds the 1.42 g of the tablet. Let's re-verify the values. Ah, let's recalculate: \( m(\text{FeSO}_4) = 0.0185 \times 152.0 = 2.812 \text{ g} \). Let's adjust the tablet mass in the problem description to a logical value, say 4.50 g, to avoid a percentage purity greater than 100%. Let's change the tablet mass to 4.50 g. \[ \text{Percentage purity} = \left(\frac{2.812 \text{ g}}{4.50 \text{ g}}\right) \times 100\% = 62.5\% \]
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