Transition metals have variable oxidation states, making them highly effective in redox chemistry. In this lesson, we cover their roles as catalysts in industrial and biological systems and study how redox titrations are used to determine iron concentration quantitatively.
A heterogeneous catalyst is a catalyst that is in a different physical state (or phase) than the reactants. The reaction occurs on the surface of the catalyst.
A homogeneous catalyst is a catalyst that is in the same physical state (or phase) as the reactants. It works by forming an intermediate species that changes oxidation state before regenerating.
Transition Metals as Catalysts
Transition metals make excellent catalysts because they can easily donate and accept electrons due to their variable oxidation states, which provides alternative reaction pathways with lower activation energies.
1. Heterogeneous Catalysis
In heterogeneous catalysis, reactants are adsorbed onto active sites on the solid catalyst surface. The bonds in the reactant molecules are weakened, bringing the reactants close together in the correct orientation. After the reaction, the product molecules desorb from the surface.
Key industrial examples include:
- Iron (\(\text{Fe}\)) in the Haber Process: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \xrightarrow{\text{Fe(s)}} 2\text{NH}_3(\text{g})\)
- Vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) in the Contact Process: \(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \xrightarrow{\text{V}_2\text{O}_5\text{(s)}} 2\text{SO}_3(\text{g})\)
2. Homogeneous Catalysis
Homogeneous catalysts act by changing oxidation state. A classic exam example is the reaction between peroxodisulfate (\(\text{S}_2\text{O}_8^{2-}\)) and iodide (\(\text{I}^-\)) ions, catalyzed by iron(II) or iron(III) ions:
Without a catalyst, this reaction is extremely slow because it requires two negatively charged ions to collide. The electrostatic repulsion between the anions creates a very high activation energy.
Iron ions catalyze the reaction by providing a two-step mechanism that avoids anion-anion collisions:
- Step 1: Iron(II) reduces peroxodisulfate to sulfate, becoming oxidised to iron(III): \[ \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq}) \]
- Step 2: The newly formed iron(III) then oxidises iodide ions to iodine, regenerating the iron(II) catalyst: \[ 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \]
Because each step involves a collision between oppositely charged species (an anion and a cation), the activation energy for both steps is much lower, significantly increasing the rate of reaction.
Potassium Manganate(VII) Redox Titrations
Potassium manganate(VII), \(\text{KMnO}_4\), is a powerful oxidising agent used in redox titrations to find the concentration of reducing agents, most commonly iron(II) ions (\(\text{Fe}^{2+}\)).
Redox Half-Equations
In acidic conditions, the purple manganate(VII) ion (\(\text{MnO}_4^-\)) is reduced to the virtually colourless manganese(II) ion (\(\text{Mn}^{2+}\)):
Iron(II) is oxidised to iron(III):
Multiplying the iron half-equation by 5 and combining them yields the overall redox equation:
🔑 Key Principle: Manganate titration stoichiometry
The reacting ratio of \(\text{MnO}_4^-\) to \(\text{Fe}^{2+}\) is always 1:5. This means: \[ \text{Moles of } \text{Fe}^{2+} = 5 \times \text{Moles of } \text{MnO}_4^- \] This stoichiometry is the core of all titration calculations for this topic.
Titration Methodology
- Fill the burette with a standard solution of potassium manganate(VII) (\(\text{KMnO}_4\)).
- Pipette a known volume (typically \(25.0\text{ cm}^3\)) of the \(\text{Fe}^{2+}\) solution into a conical flask.
- Acidify the conical flask by adding an excess of dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)).
- Titrate by adding the purple \(\text{KMnO}_4\) solution from the burette. The purple colour fades instantly upon entering the flask as \(\text{Mn}^{2+}\) forms.
- End Point: The end point is reached when a single drop of \(\text{KMnO}_4\) remains unreacted, turning the solution in the flask permanently pale pink. No indicator is needed because \(\text{KMnO}_4\) is self-indicating.
You must know why only dilute sulfuric acid (\(\text{H}_2\text{O}_4\)) is used to acidify the titration. The exam frequently asks why other acids are unsuitable:
- Hydrochloric acid (\(\text{HCl}\)): Unsuitable because the chloride ions (\(\text{Cl}^-\)) will be oxidised to chlorine gas (\(\text{Cl}_2\)) by the powerful \(\text{MnO}_4^-\) ions. This would use up extra manganate, resulting in an artificially high titre.
- Nitric acid (\(\text{HNO}_3\)): Unsuitable because it is a strong oxidising agent itself. It would oxidise the \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) before the titration begins, resulting in an artificially low titre.
- Ethanoic acid (\(\text{CH}_3\text{COOH}\)): Unsuitable because it is a weak acid and does not dissociate fully, meaning it cannot provide the high concentration of \(\text{H}^+\) ions required for the reduction of manganate.
Potassium Dichromate(VI) Redox Titrations
In some circumstances, potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), is used as an oxidising agent instead of manganate. The reduction half-equation of the orange dichromate ion to green chromium(III) is:
Combining this with the iron(II) half-equation gives a reacting mole ratio of 1:6 for dichromate to iron(II):
Step 1: Calculate the moles of manganate(VII) ions used
\[ \text{Moles of } \text{MnO}_4^- = \text{concentration} \times \text{volume (dm}^3) \]
\[ \text{Moles of } \text{MnO}_4^- = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\text{ mol} \]
Step 2: Use the reaction ratio to find moles of iron(II)
The stoichiometric ratio between \(\text{MnO}_4^-\) and \(\text{Fe}^{2+}\) is 1:5.
\[ \text{Moles of } \text{Fe}^{2+} = 5 \times (4.50 \times 10^{-4}) = 2.25 \times 10^{-3}\text{ mol} \]
Step 3: Calculate the concentration of the iron(II) ions
The sample volume was \(25.0\text{ cm}^3\), which is \(0.0250\text{ dm}^3\).
\[ \text{Concentration of } \text{Fe}^{2+} = \frac{\text{moles}}{\text{volume}} = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900\text{ mol dm}^{-3} \]
Step 1: Calculate the moles of manganate(VII) in the titration
\[ \text{Moles of } \text{MnO}_4^- = 0.00500 \times \frac{18.40}{1000} = 9.20 \times 10^{-5}\text{ mol} \]
Step 2: Calculate the moles of iron(II) in the titrated \(25.0\text{ cm}^3\) aliquot
\[ \text{Moles of } \text{Fe}^{2+} \text{ in } 25.0\text{ cm}^3 = 5 \times (9.20 \times 10^{-5}) = 4.60 \times 10^{-4}\text{ mol} \]
Step 3: Calculate the moles of iron(II) in the original \(250\text{ cm}^3\) solution
The volumetric flask contained ten times the volume of the aliquot.
\[ \text{Moles of } \text{Fe}^{2+} \text{ in } 250\text{ cm}^3 = 4.60 \times 10^{-4} \times 10 = 4.60 \times 10^{-3}\text{ mol} \]
Step 4: Calculate the mass of iron in the tablet
\[ \text{Mass of iron} = \text{moles} \times A_r = (4.60 \times 10^{-3}) \times 55.8 = 0.2567\text{ g} \]
Step 5: Calculate the percentage by mass
\[ \text{Percentage mass} = \frac{0.2567\text{ g}}{0.350\text{ g}} \times 100 = 73.3\% \]
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