Question

(a) Write the reaction for the overall oxidation of propane (C3H8) with oxygen to produce CO2 and water.
(b) Write a second expression for the use of air as the oxidant. Assuming air use, and assuming C3H8 is the only electrochemically active species, determine the exit species concentrations in the anode and cathode compartments for the following cases:
(c) 85% fuel utilization and 25% oxidant utilization in a fuel cell that conducts oxygen ions (O= ), and (d) 90% fuel utilization and 50% oxidant utilization in a fuel cell that conducts hydrogen ions (H+ ). (e) Assuming a reversible potential of 1.3V at cell operating conditions of 1 atmosphere and 298 K in both cases, determine the operating voltage limit for each of these two cases using the Nernst equation.

Answer 1

Answer:

Oxidation is the loss of electrons, that is, addition of electronegetive elements, example is addition of oxygen. Also, removal of electropositive elements, example is removal of hydrogen.

Explanation: a) In the presence of excess oxygen, propane burns in air, which gives the following chemical equation:

C3H8 + 5O2⇒ 3CO2 + 4H2O +Heat

b) When insufficient oxygen or too much oxygen is present for complete combustion, the following equation is given:

2C3H8 + 9O2 ⇒ 4CO2 + 2CO + 8H2O + Heat

c) At the anode( negative terminal): O∧2- ⇒ O + e

Oxygen accepts electron.

d) At cathode ( positive terminal): H∧+ + e∧- ⇒ H

Hydrogen donates electron

d) Nernst equation for reversal potential is given as follows:

E= RT/zF  In{ion outside cell}/{ion inside cell}= 2.303 RT/zF In{ion outside cell}/{ion inside cell}