Lesson 4 · 30 min

Catalysts and materials

Iron makes ammonia, vanadium makes sulfuric acid, and platinum, palladium and rhodium clean exhaust gas. Why are transition metals such good catalysts? Then see how ferromagnetism and alloying make metals more useful still.

In this lesson

Explain how transition metals act as homogeneous catalysts: variable oxidation states let them take electrons and hand them back
Use electrode potentials to judge whether a couple can catalyse a slow redox reaction
Describe the adsorption mechanism of heterogeneous catalysis, with the Haber process, the contact process and car catalytic converters as examples
Explain paramagnetism and ferromagnetism by unpaired electrons, and distinguish substitutional from interstitial alloys

Car exhaust carries poisonous CO and smog-forming NO. The reaction between them, $\ce{2CO} + \ce{2NO} \mathrel{\ce{->}} \ce{2CO2} + \ce{N2}$ , releases an enormous amount of energy and has an unimaginably large equilibrium constant — yet in the exhaust pipe it simply does not happen. Fit a catalytic converter containing a few grams of platinum, palladium and rhodium, and more than 90 % of the harmful gases are converted in well under a second. Why are transition metals such good catalysts?

Predict first

$\ce{S2O8^2-} + \ce{2I-} \mathrel{\ce{->}} \ce{2SO4^2-} + \ce{I2}$ is slow, but a little $\ce{Fe^2+}$ speeds it up markedly. Which property of the iron ions matters most?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

The Fe³⁺/Fe²⁺ potential lies between the two couples of the reaction, so iron can be oxidised and then reduced again Fe²⁺ is small enough to squeeze between the two anions Fe²⁺ solutions are coloured and absorb light for energy Iron is magnetic

Two ways to catalyse

Homogeneous catalysis: the catalyst is in the same phase as the reactants (usually in solution). A transition-metal ion can switch back and forth between neighbouring oxidation states, so it can split one electron transfer that is hard to achieve directly into two easy ones: take electrons from one reactant, then hand them to the other. The condition is that the catalyst’s couple lies between the two couples of the reaction, so both steps run downhill.

Heterogeneous catalysis: the catalyst is a solid and the reaction happens on its surface. Gas molecules adsorb, forming bonds to surface atoms that weaken their own — $\ce{N2}$ on iron and $\ce{H2}$ on nickel or platinum even split into atoms. The partly filled d orbitals of transition metals can both accept and donate back electron density, which makes them outstanding adsorbers. Adsorption must be just right (the Sabatier principle): too weak and the molecules are not activated; too strong and the products refuse to leave, poisoning the surface.

Haber process: iron (promoted with $\ce{K2O}$ and $\ce{Al2O3}$ ) catalyses ammonia synthesis.
Contact process: $\ce{V2O5}$ oxidises $\ce{SO2}$ , dropping to +4, and is reoxidised to +5 by $\ce{O2}$ — variable oxidation states again.
Catalytic converters: platinum and palladium oxidise CO and unburnt hydrocarbons; rhodium reduces NO to $\ce{N2}$ .
Hydrogenation and polymerisation: nickel catalyses the hydrogenation of vegetable oils; titanium and zirconium compounds polymerise ethene to polyethylene.

E^\circ_{\text{oxidant}} > E^\circ_{\text{catalyst}} > E^\circ_{\text{reductant}}

The thermodynamic condition for a homogeneous redox catalyst: its potential must lie between the reaction’s two couples, so that both steps have negative ΔG = −nFE°.

$E^\circ_{\text{oxidant}}$: potential of the couple of the reactant being reduced
$E^\circ_{\text{catalyst}}$: potential of the catalyst couple (such as Fe³⁺/Fe²⁺)
$E^\circ_{\text{reductant}}$: potential of the couple of the reactant being oxidised

Choosing a homogeneous catalyst from standard potentials

How variable oxidation states catalyse a reaction

$\ce{2H2O2 -> 2H2O + O2}$ — It needs a go-between that can first accept and then hand back electrons.

Candidate catalyst (couple)

Step 1: H₂O₂ oxidises Br⁻ to Br₂
E°cell = 0.710 V, ΔG° = −137 kJ (per 2 mol e⁻)

Step 2: Br₂ oxidises H₂O₂ and turns back into Br⁻
E°cell = 0.371 V, ΔG° = −72 kJ (per 2 mol e⁻)

Overall E°cell = 1.081 V

Both steps are downhill: Br₂/Br⁻ can act as the catalyst — oxidised, then reduced, and returned unchanged.

For the decomposition of hydrogen peroxide, try each candidate couple: Fe³⁺/Fe²⁺ and Br₂/Br⁻ lie between +0.695 V and +1.776 V, so both steps run downhill; Co³⁺/Co²⁺ (+1.92 V) is too high and Cu²⁺/Cu⁺ too low. Then return to the peroxodisulfate reaction and find the other couples that would work.

Worked example

How badly does the converter reaction want to happen?

From $\Delta_\mathrm{f}G^\circ$ (kJ/mol): CO −137.2, NO +87.6, $\ce{CO2}$ −394.4, find $\Delta_\mathrm{r}G^\circ$ and $K$ for $\ce{2CO} + \ce{2NO} \mathrel{\ce{->}} \ce{2CO2} + \ce{N2}$ at 25 °C.

Magnetism

Substances with unpaired electrons are drawn into a magnetic field (paramagnetism). Counting spin only, the magnetic moment is

\mu = \sqrt{n(n+2)}\ \mu_\mathrm{B}

where n is the number of unpaired electrons and $\mu_\mathrm{B}$ the Bohr magneton. High-spin $\ce{Mn^2+}$ and $\ce{Fe^3+}$ (d⁵) have five unpaired electrons and moments of about 5.9 $\mu_\mathrm{B}$ ; d¹⁰ $\ce{Zn^2+}$ has none and is diamagnetic. Ligands can change the number of unpaired electrons (high and low spin; see crystal-field theory).

In iron, cobalt and nickel (and gadolinium below 20 °C) the spins of neighbouring atoms line up spontaneously into domains: this is ferromagnetism. Above the Curie temperature thermal motion scrambles the order: 770 °C for iron, 1115 °C for cobalt and 354 °C for nickel.

Alloys

Pure metals are often too soft. In substitutional alloys, atoms of similar size (radii within about 15 %) take each other’s places in the lattice: brass (copper and zinc), bronze (copper and tin), nichrome (heating elements). In interstitial alloys, small atoms squeeze into holes in the metal lattice: steel has carbon atoms lodged in iron, where they block the slip of atomic layers and harden the metal (see close packing and holes). Stainless steel combines both: chromium forms the passive film and nickel stabilises the face-centred cubic structure — the familiar “18/8” grade contains about 18 % chromium and 8 % nickel.

Concept check

Counting spin only, which of these ions has the largest magnetic moment?

High-spin

\ce{Mn^2+}

(d⁵) High-spin

\ce{Fe^2+}

(d⁶)

\ce{Cu^2+}

(d⁹)

\ce{Zn^2+}

(d¹⁰)

Try the calculation

What is the spin-only magnetic moment of high-spin $\ce{Fe^3+}$ (d⁵), in Bohr magnetons?

Key takeaways

Variable oxidation states let a transition-metal ion split a hard electron transfer into two easy ones — provided its potential lies between the two couples of the reaction.
Heterogeneous catalysts weaken the reactants’ bonds by adsorption, which must be neither too weak nor too strong (Sabatier). Fe, V₂O₅ and Pt/Pd/Rh serve the Haber process, the contact process and exhaust clean-up.
A catalyst changes the rate, not the equilibrium constant or ΔG°.
Unpaired electrons give paramagnetism (μ = √(n(n+2)) μB); Fe, Co and Ni are ferromagnetic below their Curie temperatures; alloys are substitutional or interstitial.

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Data and references

P. Vanýsek, “Electrochemical series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010).
J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989); CRC Handbook, “Standard thermodynamic properties of chemical substances”.
I. Chorkendorff, J. W. Niemantsverdriet, Concepts of Modern Catalysis and Kinetics, 3rd ed., Wiley-VCH (2017).
C. E. Housecroft, A. G. Sharpe, Inorganic Chemistry, 5th ed. (2018), ch. 20 (magnetism) and ch. 25 (catalysis).
CRC Handbook of Chemistry and Physics, “Curie and Néel temperatures of magnetic materials”.