Question

The overall reaction and equilibrium constant value fora hydrogen-oxygen fuel cell at 298 K is
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(l\right) K=1.28\times {10}^{83}$
a. Calculate E^o and$\Delta G^o$at 298 K for the fuel-cellreaction.
b. Predict the signs of$\Delta H^o\&\Delta G^o$for the fuel-cellreaction.
c. As temperature increases, does the maximumamount of work obtained from the fuel-cell reactionincrease, decrease, or remain the same? Explain.

Short answer

(a) Thevalues of$E^o,\Delta G^o$for the given reaction are 1.23 V, -474kJ respectively.

(b) Electricity is produced from the energy released. Thus, $\Delta H^o$is negative. The gas molecule number decreases to zero from three when there is a conversion from reactants to products. Therefore, there is a decrease in standard entropy and $\Delta S$is negative.

(c) There is a loss in energy and the work reduces. There will be a heat loss when the temperature increases as it is an exothermic process.

Step-by-step solution

Calculation of cell potential

The Gibbs free energy can be denoted as:

$\Delta G^o=\sum \Delta {G^o}_{f\left(products\right)}-\sum \Delta {G^o}_{f\left(reac\tan ts\right)}$

The overall reaction can be split up into two half-reactions.

$$\begin{gathered} H_2+2O{H}^{-}\to 2H{}_{2}O+2e^{-} \\ {H}_2+O_2+4e^{-}\to 4O{H}^{-} \end{gathered}$$

The Gibbs free energy can be calculated as:

$$\begin{gathered} \Delta G^o=-RT \ln K \\ \Delta G^o=-(8.314\times {10}^{-3} kJ/mol/K)\left(298\right)\times 2.303 \log (1.28\times {10}^{83}) \\ \Delta G^o=-474 kJ \end{gathered}$$

The cell potential can be denoted as:

$$\begin{gathered} {E^o}_{cell}=\frac{\Delta G^o}{-nF} \\ {E^o}_{cell}=\frac{-474 kJ}{-4mol(96.5 kJ/mol/K)} \\ {E^o}_{cell}=1.23 V \end{gathered}$$

Signs for the fuel cell reaction

The conversion of chemical energy in the form of potential energy to electrical energy by fuel cell is the fuel cell reaction. The cell will use oxygen and hydrogen gas as fuel in proton exchange membrane. The products are heat, water, and electricity.

Electricity will be produced from the energy released. Thus, $\Delta H^o$is negative. The gas molecule number decreases to zero from three when there is a conversion from reactants to products. Therefore, there is a decrease in standard entropy and $\Delta S$is negative.

There will be heat loss as it is an exothermic process

There is a loss in energy and work will reduce. There will be a heat loss when temperature increases as it is an exothermic process.