The zirconia sensor and the switch at λ = 1
Turn the engine mixture from lean to rich, record the voltage of a zirconia sensor and find the stoichiometric point; then change the sensor temperature and the yttria content and see what alters the reading.
Aim Measure the sensor voltage as a function of λ, locate the jump, and explain why the voltage depends on temperature but not on the number of oxide vacancies.
Procedure
- At a sensor temperature of 900 K, start at λ = 1.05 and lower it by 0.01 at a time, recording the voltage down to λ = 0.95.
- Between 0.99 and 1.01 switch to steps of 0.001 and find where the voltage crosses 0.45 V.
- Repeat a few key points at 700 K and at 1000 K.
- Finally change the Y₂O₃ content from 3 to 10 mol% and watch what happens to the number of vacancies and to the voltage.
What to notice
- At 900 K the lean side reads only 0.06–0.09 V (λ = 1.05–1.01) and the rich side 0.85–0.91 V (λ = 0.99–0.95); almost the whole jump happens within five parts per thousand of λ = 1.
- The hotter the sensor, the lower the rich-side voltage (at λ = 0.95, 1.00 V at 700 K but 0.86 V at 1000 K): the combustion equilibrium constants fall with temperature, so the equilibrium oxygen pressure rises.
- Changing the Y₂O₃ content changes the fraction of vacancies but leaves the voltage untouched: the Nernst voltage depends only on the oxygen activities on the two sides, as long as the electrolyte conducts oxide ions.
In a real lab
- The sensor and exhaust pipe run at several hundred degrees, so let them cool before handling; an engine run indoors must vent its exhaust outside — carbon monoxide is odourless and deadly.
- Leaded petrol poisons the sensor’s platinum electrodes and the catalytic converter — one reason cars with catalysts must run on unleaded fuel.
- Real controllers take about 0.45 V as the boundary between rich and lean and keep the mixture swinging around the stoichiometric point; an ageing sensor responds more slowly, which the diagnostic system detects.