Question

ENernst = [RT/nF] ln [(pO2,c)/(pO2,A)], considering the fact that O2 is converted to O2- and transported to the Anode where it is consumed by the fuel, e.g., Hydrogen, to form water.
I. Re-write the expression above in a format that is clearly expressed in terms of measurable quantities, viz. Oxygen in the cathode, Fuel (Hydrogen) at equilibrium in the anode, and any pre-existing moisture that is available in the fuel feed.
II. How much does the value of Nernst Potential vary if there is oxygen depletion in the air, from 21% to 5%? Assume that the fuel comprises 99.99% Hydrogen, and 0.01% moisture. The temperature is 800oC. Eo at 800oC is estimated to be around 1.05 V.
III. A Potentiometric reading for the SOFC under the same temperature as above (for normal air in the cathode), shows a sharp drop of about 0.5 V – discuss with the aid of a quick calculation, what could be the reason for this? How much is the partial pressure of oxygen in the anode, under these conditions.

Answer 1

Solution.

1.

Reaction on cathode:

$\frac{1}{2}O2 + 2e = O^{2-}$

Reaction on anode:

$H2 + O^{2-} = H2O + 2e$

Overall reaction:

$\frac{1}{2}O2 +H2 = H2O$

$Ernst = \frac{R \times T}{n \times F} ln[\frac{p(O2)^{0.5} \times p(H2)}{p(H2O)}]$

2.

Since there is very little water in the hydrogen fuel, we do not take it into account.

$n(O2)1 = 0.66 \ mol$

$n(O2)2 = 0.16 \ mol$

$n(H2) = 49.995 \ mol$

$\chi(O2)1 = 0.013$

$\chi(O2)2 = 0.0032$

$\chi(H2)1 = 0.987$

$\chi(H2)2 = 0.997$

P = 101325 Pa

$p(O2)1 = 101325 \times 0.013 = 1317.23 \ Pa$

$p(H2)1 = 101325 \times 0.987 = 100007.78 \ Pa$

$p(O2)2 = 101325 \times 0.0032 = 324.24 \ Pa$

$p(H2)2 = 101325 \times 0.997 = 101021.03 \ Pa$

$Ernst = 1.08 + 0.0462 \times ln[(1317.23^{0.5}) \times (100007.78)] = 1.778 \ V$

$Ernst = 1.08 + 0.0462 \times ln[(324.24^{0.5}) \times (101021.03)] = 1.745 \ V$

$\Delta Ernst = 0.033 \ V$

3.

There was a change in the reaction potential. Nothing changes in the Nernst equation except the partial pressures of hydrogen and oxygen. But in order for the potential of the fuel cell to decrease dramatically by a large amount, the partial pressures of hydrogen and oxygen must be reduced, or the amount of water vapor is more likely to increase. Thus, the decrease in the potential of the element by 0.5 V is due to an increase in the amount of water vapor.

$30 = ln[p(H2) \times (p(O2)^{0.5})]$

$p(O2) = 0.032 \ Pa$

Answer:

1.

$Ernst = \frac{R \times T}{n \times F} ln[\frac{p(O2)^{0.5} \times p(H2)}{p(H2O)}]$

2.

$\Delta Ernst = 0.033 \ V$

3.

There was a change in the reaction potential. Nothing changes in the Nernst equation except the partial pressures of hydrogen and oxygen. But in order for the potential of the fuel cell to decrease dramatically by a large amount, the partial pressures of hydrogen and oxygen must be reduced, or the amount of water vapor is more likely to increase. Thus, the decrease in the potential of the element by 0.5 V is due to an increase in the amount of water vapor.

$p(O2) = 0.032 \ Pa$