Lesson 3 · 30 min

Solid electrolytes: the oxygen sensor in a car exhaust

Yttria-doped zirconia conducts O²⁻ through its oxide vacancies, and the difference in oxygen pressure across it gives a Nernst voltage. Compute the oxygen pressure of exhaust from combustion equilibria and see why it jumps by more than ten orders of magnitude at λ = 1.

In this lesson

Explain why yttria-doped zirconia conducts oxide ions, and compute the fraction of vacant oxide sites from its composition
Write the Nernst equation of an oxygen concentration cell and compute its voltage from the two oxygen pressures
Explain from combustion equilibria why the oxygen pressure of exhaust jumps by more than ten orders of magnitude at λ = 1, and why the three-way catalyst depends on it

A ceramic sensor the thickness of a finger is screwed into the exhaust pipe of every petrol car. Moment by moment it tells the engine computer whether the mixture is rich (too much fuel) or lean (too much air), and the computer trims the fuel injection accordingly, holding combustion right at the stoichiometric point — otherwise the catalytic converter cannot do its job.

The heart of the sensor is a zirconia ceramic: at a few hundred degrees Celsius oxide ions can move through it, making it a solid electrolyte. One face meets the exhaust, the other the air, and the difference in oxygen pressure between them produces a voltage — a concentration cell made of solid.

Predict first

Replace some of the Zr⁴⁺ in ZrO₂ by Y³⁺. What must happen for the crystal to stay neutral?

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

Some O²⁻ sites are left empty Extra free electrons appear in the crystal Some Zr⁴⁺ becomes Zr⁵⁺ Nothing needs to happen

Ions that move through a solid

Pure ZrO₂ is monoclinic at room temperature; with about 8 mol% Y₂O₃ the cubic fluorite structure of high temperatures is “stabilised” down to room temperature (less gives the tetragonal form) — yttria-stabilised zirconia (YSZ). In fluorite the cations form a face-centred cubic array and the oxide ions fill all the tetrahedral holes (as in CaF₂). The vacancies left by the Y³⁺ are scattered at random over the oxide sublattice.

To hop into a neighbouring vacancy an ion must cross an energy barrier, and the hopping rate grows with temperature according to Arrhenius. That is why an oxygen sensor has to be several hundred degrees hot before it works; modern sensors carry their own heater, so they start working soon after a cold start.

(1-y)\,\mathrm{ZrO_2} + y\,\mathrm{Y_2O_3} \;\rightarrow\; \mathrm{Zr}_{1-x}\mathrm{Y}_x\mathrm{O}_{2-x/2},\qquad x = \frac{2y}{1+y}

Each formula unit has x/2 oxide vacancies; there are two oxide sites, so a fraction x/4 of them is empty.

$y$: mole fraction of Y₂O₃
$x$: fraction of cation sites holding Y³⁺

Worked example

Vacancies in 8YSZ

Oxygen sensors often use zirconia with 8 mol% Y₂O₃ (8YSZ). Find x and the fraction of oxide sites that are empty.

A concentration cell for oxygen

Coat both faces with porous platinum electrodes. On the air side, oxygen molecules take up electrons and enter the electrolyte as oxide ions; on the exhaust side, oxide ions give up electrons and leave as oxygen molecules:

\mathrm{O_2(g)} + 4\,e^- \rightleftharpoons 2\,\mathrm{O^{2-}}

The reaction is the same on both sides and only the oxygen pressure differs, so this is a concentration cell with a standard emf of zero. Four electrons are transferred per O₂, so the open-circuit voltage is given by the Nernst equation below. Practically no current is drawn and there is no net flow of ions — the voltage reports the equilibrium the two sides reach.

E = \frac{RT}{4F}\,\ln\frac{p_{\mathrm{O_2}}^{\text{air}}}{p_{\mathrm{O_2}}^{\text{exhaust}}}

The reference side is air with p(O₂) = 0.2095 bar. At 900 K, RT/4F = 19.4 mV: every tenfold difference in oxygen pressure changes the voltage by 44.6 mV.

$E$: open-circuit voltage
$F$: Faraday constant
$p_{\mathrm{O_2}}$: partial pressure of oxygen

Try the calculation

At a sensor temperature of 900 K the exhaust side has p(O₂) = 1.0 × 10⁻²⁰ bar and the reference is air (0.2095 bar). What is the sensor voltage (in V)?

Rich, lean and the jump at λ = 1

The air ratio λ is the air actually supplied divided by the air needed for complete combustion. Burning iso-octane, C₈H₁₈, completely takes 12.5 mol of O₂ — 15.1 kg of air per kilogram of fuel.

Lean (λ > 1): oxygen is left over, the exhaust holds from a few tenths of a per cent to a couple of per cent of free O₂, p(O₂) ≈ 10⁻³–10⁻² bar, and the voltage is only about 0.05–0.1 V. Rich (λ < 1): there is too little oxygen, and CO and H₂ remain. On platinum, CO + ½O₂ ⇌ CO₂ and H₂ + ½O₂ ⇌ H₂O reach equilibrium, with constants of about 7 × 10¹¹ and 3 × 10¹¹ at 900 K; a few tenths of a per cent of CO and H₂ press the equilibrium oxygen pressure down to about 10⁻²⁰ bar, and the voltage is about 0.85 V. Between the two, within less than 1 % of λ = 1, the oxygen pressure jumps by more than ten orders of magnitude.

Equilibrium exhaust from NIST-JANAF data · voltage from the Nernst equation

The oxygen sensor in a car exhaust

Air ratio λ = 0.990 (air-to-fuel 14.98) Sensor temperature 900 K (627 °C) Y₂O₃ in the zirconia: 8 mol %

Cross-section of the sensor

Sensor voltage

CO, H₂ and O₂ in the equilibrium exhaust

CO
H₂
O₂

Equilibrium exhaust: CO₂ 12.45 %, H₂O 13.85 %, CO 0.1101 %, H₂ 0.2824 %, O₂ 1.4e-15 %; p(O₂) = 2.45e-20 bar.

E = (RT/4F) ln[0.2095/p(O₂)] = 0.845 V — rich: extra fuel, and almost no free oxygen left.

8 mol % Y₂O₃ leaves 3.70 % of the oxide sites empty (Zr₁₋ₓYₓO₂₋ₓ/₂, x = 0.148): O²⁻ moves by hopping into these vacancies.

Drag λ slowly from 0.99 to 1.01: the voltage falls from 0.85 V to 0.09 V, almost all of it within a few parts per thousand of λ = 1. The engine computer uses this steep switch to keep the mixture swinging back and forth between rich and lean. Now change the temperature: the lean side barely moves, the rich side shifts, because the equilibrium constants change with temperature.

Concept check

Why does the engine computer hold the mixture near λ = 1 rather than on the more economical lean side?

The three-way catalyst can oxidise CO and hydrocarbons and reduce NOₓ at the same time only near λ = 1 A lean mixture damages the sensor Combustion is coolest at λ = 1 A lean mixture makes more CO

Key takeaways

Y³⁺ replacing Zr⁴⁺ leaves oxide vacancies (Zr₁₋ₓYₓO₂₋ₓ/₂): in 8YSZ about 3.7 % of the oxide sites are empty, and O²⁻ moves by hopping into them, which needs a few hundred degrees.
The oxygen concentration cell gives E = (RT/4F) ln[p(O₂, air)/p(O₂, exhaust)] — 44.6 mV per tenfold ratio at 900 K.
For exhaust equilibrated on platinum the oxygen pressure jumps by more than ten orders of magnitude at λ = 1, and the voltage drops from about 0.85 V to about 0.1 V: this switch keeps the engine in the working window of the three-way catalyst.

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Data and references

M. W. Chase, Jr., NIST-JANAF Thermochemical Tables, 4th ed., J. Phys. Chem. Ref. Data Monograph 9 (1998), via the Shomate fits of the NIST Chemistry WebBook.
J. Riegel, H. Neumann & H.-M. Wiedenmann, “Exhaust gas sensors for automotive emission control”, Solid State Ionics 152–153, 783–800 (2002).
E. Ivers-Tiffée, K. H. Härdtl, W. Menesklou & J. Riegel, “Principles of solid state oxygen sensors for lean combustion gas control”, Electrochim. Acta 47, 807–814 (2001).