Question

Anode: CH2OH + H2O + 6H+ 3 2 Cathode:-02 +6H+ +6e + 340 Overall reaction: CH20H + 302 – 24,0 + CO2 G H kl/mol kJ/mol Methanol(Overall raction) -685.3 -638.5 a) How many MJ of electricity would be produced in an ideal fuel cell at 25°C and 1 atm? (calculate the Energy per kg methanol) b) How would the amount of electricity generated change if air were used instead of pure oxygen? Assume atmospheric pressure and that that the partial pressure of oxygen in air is 0.2 atm. Assume all other reactants remain in liquid phase. c) How much work is produced in the form of heat? (Since this is an ideal fuel cell, it can absorb heat from the environment as opposed to releasing heat) Answer in kJ/mol methanol. d) What is the reversible voltage of an ideal methanol fuel cell? (assume molecular oxygen is used)

Answer 1

A) Electrical energy = 20545.54 kJ/kg

B) ∆G = -652.3 kJ/mol

C) q = 46.8 kJ/mol

D) E​​​​​​ocell = 1.184 V

Overall cell reaction & CH₃OH + 302 2H₂O + CO2 Logo = -658+3 kJ/mol Hr = -638.5kJ/mol La Gibbs free energy of cell convert into electricity. Thus electrical energy produce by one mule of mathanol = 658.3 kJ In 1 kg of mathanol, the number of mole mass in am 'molar mass molar mass g/mol l = 39.049/mal) 1000 g - 31.21 mol 32.04 g/mol energy produce py 1kg of methanol = 650. 3K J X 31.21 per TE = 20545.54 KJ kg-1 I b gt pure O2 replaced by oxigen = OG= 0Go - RT lng OG=-658.3 kJ/mol - 8.314 x 298K In 10273/2 DG = -658.3K 5/mut 6 kJ/mol 06 = - 652.3kJ/mol

o Heat produce = OG=OH-TOS DG-OH = -TOS or Tos = OH-OG = -63815 + 685.3 TDS = 46.8 kJ/mol TOS = q = 46.0 KJ/mol do Gibbs free energy and voltage of cell DG = NFE - 685.381035 = -6x9 6500 XEO to = 1.1840