Question

(a) Write cathode and anode half-reactions, and a net reaction (balanced) for a low-temperature direct methanol fuel cell (assuming complete oxidation of fuel).

could you please explain the problem with answers and explanations please ?

Answer 1

a) The Anode and cathode half cell reactions and the overall reaction are given below

Equation CH3OH + H2O +6H+ +6e + CO2 Anode oxidation 3 202 +6 + +6e + 3 H2O Cathode reduction Overall reaction CH,OH+02 +2 H20+ CO2 redox reaction

$\Delta$H_r = $\Delta$H_f (CO₂) + 2*$\Delta$H_f (H₂O) + $\Delta$H_f (CH₃OH)

= -394 - 572 + 238

= -728 kJ/mole

= -728 x 10³ J/mole

$\Delta$S_r = 2*S(H₂O) + S(CO2) - S(CH3OH) - 1.5*S(O2)

= 2*70 + 214 - 127 - 1.5*205

= -80.5 JK^-1mole^-1

$\Delta$G_r at 25^oC = $\Delta$H_r - T$\Delta$S_r

= -728 x 10³ - 298 x -80.5

= -728 x 10³ + 23989

=-704011 J/mole

= -704.011 kJ/mole

$\Delta$G_r at 185^oC = $\Delta$H_r - T$\Delta$S_r

= -728 x 10³ - 458 x -80.5

= -728 x 10³ + 36869

= -691131 J/mole

= -691.131 kJ/mole

$\Delta$G = -nFE

at 298 K,

E = -704011/(-6*96485.3) (since 6 electrons are transferred in the reaction, n=6, F = 96485.3 (given))

E at 298K = 1.216 V

at 458K,

E = -691131/(-6*96485.3)

E at 458K = 1.193 V

Q2)

Given, Over potential of anode, $\eta$_anode = 0.270 V and

Over potential of cathode, $\eta$_cathode = 0.230 V

Cell potential, E' = E - $\eta$_anode - $\eta$_cathode

At 298 K

E' at 298 K = 1.216 - 0.27 - 0.23

E' at 298 K = 0.716 V

Thermodynamic efficiency at 298 K = E'x100/E = 0.716x100/1.216

Thermodynamic efficiency at 298 K = 58.88%