Enthalpy of electrochemical redox half reactions with O2, H2 and H2O

Question

I am working with numerical analysis of solid oxide fuel cells. In such systems we have an oxidation reaction taking place in the anode, and a reduction reaction taking place in the cathode.

Oxidation: $\ce{H2 + O^2- \rightleftharpoons H2O + 2e-}$

Reduction: $\ce{\frac{1}{2}\,O2 + 2e- \rightleftharpoons O^2-}$

I am considering solid oxide fuel cells, so the $\ce{O^2−}$ ions are transported through the electrolyte. The materials are: yittria-stabilized zirconia (YSZ) for the electrolyte; nickel zirconia (Ni-YSZ) for the anode; and strontium-doped lanthanum manganite (LSM) for the cathode. Everything operates at $800 ^\circ $C. $\ce{O2}$ enters the system on the cathode side, it is transported through the porous cathode to be reduced at the cathode-electrolyte interface; $\ce{H2}$ enters the system on the anode side, it is transported through the porous anode to be oxidized at the anode-electrolyte interface; $\ce{H2O}$ exits the system on the anode-side.

I believe that the first reaction is exothermic and the second one is endothermic. But I need the actual energy values in $\pu{J/mol}$ to include as thermal loads in my numerical analysis.

I was trying to compute these using thermodynamics properties, for example, for $T = 1100 \pu{K}$, the enthalpy of formation of $\ce{H2O}$ is $-248.46 \pu{kJ/mol}$ [2]. I believe that this is the heat generated by the exothermic chemical reaction: $\ce{H2 + 1/2 O2 \rightleftharpoons H2O}$. Which would be the net generated heat of both reactions (oxidation of $\ce{H2}$, and reduction of $\ce{O2}$). However, I need the separate values for each reaction, since they happen in different locations of the fuel cell (so the thermal load is not simply the sum of the values, two independent thermal loads must be defined).

Answer 1

The reaction mentioned in the question seems to be convoluted. Rewriting the same as:
At anode: $\ce{1/2H_2(g) <=> H^+(aq) + e^-}$
At cathode: $\ce{1/2O_2(g) + 2H^+(aq) + 2e^- <=> H_2O(l)}$
The enthalpies for individual half-cells can be calculated from Hess' law.
For anode: $$\begin{align} \Delta_{\text{r}}H(\text{anode}) &= \Delta_{\text{f}}H(\ce{H^+(aq)})- \frac{1}{2}\Delta_{\text{f}}H(\ce{H_2(g)}) \\ \Delta_{\text{diss}}H &= \Delta_{\text{f}}H(\ce{H^+(aq)})-\Delta_{\text{f}}H(\ce{H^+(g)}) \\ \Delta_{\text{r}}H(\text{anode}) &= \Delta_{\text{f}}H(\ce{H^+(g)})-\frac{1}{2}\Delta_{\text{f}}H(\ce{H_2(g)})+\Delta_{\text{diss}}H \end{align}$$
For cathode: $$\begin{align} \Delta_{\text{r}}H(\text{cathode}) &= \Delta_{\text{f}}H(\ce{H_2O(l)})-2\ \Delta_{\text{f}}H(\ce{H^+(aq)})-\frac{1}{2}\Delta_{\text{f}}H(\ce{O_2(g)}) \\ &= \Delta_{\text{f}}H(\ce{H_2O(l)})-2\ \Delta_{\text{f}}H(\ce{H^+(g)})-\frac{1}{2}\Delta_{\text{f}}H(\ce{O_2(g)})-2\ \Delta_{\text{diss}}H \end{align}$$
The net reaction can be verified by summing up anode and cathode half-cells: $$ \Delta_{\text{r}}H=2\ \Delta_{\text{r}}H(\text{anode})+\Delta_{\text{r}}H(\text{cathode})=\Delta_{\text{f}}H(\ce{H_2O(l)})-\Delta_{\text{f}}H(\ce{H_2(g)})-\frac{1}{2}\Delta_{\text{f}}H(\ce{O_2(g)}) $$
The dissolution enthalpy is necessary as ions (here $\ce{H^+}$) is in aqueous medium.