Material Science Question about Conductivity.

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SUMMARY

The discussion focuses on calculating the conductivity of ZrO2, an ionic conductor used in automobile engine control systems, at a temperature of 800°C. The mobility values provided are 0.01 m²/V·s for Zr4+ ions and 0.04 m²/V·s for O2- ions, with concentrations of 4.5 x 1017 mobile Zr4+ ions/m³ and 2.0 x 1018 mobile O2- ions/m³. The conductivity calculation using the formula σ=ni*|e|*μe + ni*|e|*μh results in a value of approximately 0.01352 S/m. The discussion highlights the importance of considering the charge of the ions in the conductivity calculation.

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ZrO2 is an ionic conductor used in automobile engine control systems. At 800oC the mobility of Zr4+ is 0.01m2/V.sec, and of O2- it is 0.04 m2/V.sec., there are 4.5 x 1017 mobile Zr4+ ions/m3, and 2.0 x 1018 mobile O2- ions/m3. What is the conductivity of ZrO2 at this temperature, in S.m-1?

Homework Equations


σ=ni*|e|*μe + ni*|e|*μh

The Attempt at a Solution


(.01)(4.5E17)(1.6E-19)+(.04)(2E18)(1.6E-19)=0.01352 (ions*C)/(mVs)~>(ions)/(Ωm)

Am I supposed to cancel out ions somehow. I am confused about how this is supposed to work however. I think I am missing a few fundamental concepts here.
 
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It seems to me that you forgot to take into account the charge: four elementary charges on Zr4+ and two on O2-.
 

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