Question 13.1: An ideal H2 – O2 fuel cell (refer to Figure 13.1) is operati......
An ideal H_{2} – O_{2} fuel cell (refer to Figure 13.1) is operating at 800 K and the product water is (i) in vapor state or (ii) in liquid state. Compute the maximum reversible cell voltage and its maximum efficiency.
1. The molar specific heats at constant pressure are
c_{p,\mathrm{ H_{2}O}}\,=\,38.7\mathrm{\,\mathrm{J/(mol\,K}}),c_{p,\mathrm{H_{2}}}\,=\,29.6\ \mathrm{J/(mol\,K}),c_{p,\mathrm{O_{2}}}\,=\,33.7\, \mathrm{J/(m\mathbf{o}l\,K)}2. The change in isobaric specific heat c_{p} is
\Delta c_{p}\,=\,c_{p,\mathrm{H_{2}O} }\,-\,c_{p,\mathrm{H_{2}}}\,-\,0.5\,c_{p,O_{2} }\Delta c_{p}\,=\,c_{p,\mathrm{H_{2}O} }\,-\,c_{p,\mathrm{H_{2}}}\,-\,0.5\,c_{p,O_{2} }= 38.7 – 2\mathbf{9}.6-0.5\times33.7
= -7.75 J/(mol K)
3. The enthalpy and the Gibbs free energy changes at standard state (T_{0} = 298 K) are
a. For water in vapor state
b. For water in liquid state
\Delta H_{0}\,=\,-285.8\,\mathrm{k}\mathrm{J}/ \mathrm{mol} \quad\Delta G_{0}\,=\,-237.1\,\mathrm{kJ}/\mathrm{mol}4. The change in the Gibbs free energy is
\Delta G\,=\,\Delta H_{0}-T/T_{0}(\Delta H_{0}\,-\,\Delta G_{0})-\Delta c_{p}T\left[\ln\,\left(T/T_{0}\right)+\left(T_{0}/T-1\right)\right]a. For water in vapor state
\Delta G=-241.8-(800/298)\times(-241.8+228.6)-(-0.00775)\times800\times\left[\ln\,\,(800/298)+(298/800-1)\right]=-204.1\,{ }{\mathrm{kJ/mol}}
b. For water in liquid state
\Delta G\,=\,-285.8-(800/298)\times(-285.8+237.1)-(-0.00775)\times\mathrm{800}\times\left[\ln\,\,(800/298)+(298/800-1)\right]=-152.8\,{ }{\mathrm{kJ/mol}}
5. With n_{O_{2}} = 0.5 for fuel H_{2}, the maximum reversible cell voltage is
V_{\mathrm{rev}}\,=\,-\Delta G/(4\,n_{O_{2}}F)\,=\,-\Delta G/(2\times96.487)a. For water in vapor state, V_{rev} = 1.058 V, and (b) for water in liquid state,
V_{rev} = 0.792 V.
6. For the change in enthalpy \Delta H = \Delta H_{0} + \Delta c_{p}\,(T – T_{0}) and the maximum cell efficiency \eta_{rev} = \Delta G/\Delta H, we obtain
a. For water in vapor state
\Delta H=-241.8+(-0.00775)\times(800-298)=-245.7\mathrm{~kJ} /\mathrm{mol~and}{\bf h}_{\mathrm{rev}}\,=\,-204.\mathbf{1}/(-245.7)=\,0.83
b. For water in liquid state
\Delta H=-285.8+(-0.00775)\times(800-298)=-289.7\ \mathrm{kJ}/\mathrm{mol\, and}{\bf h}_{\mathrm{rev}}\,=\,-152.8/(-289.7)\,=\,0.528