Consider the following two half-cell reactions along with the standard reduction potential given : CO2 + 6H+ + 6e- CH3OH + H2O E red = 0.02 V 12O2 + 2H+ + 2e- H2O E red = 1.23 V A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of CH3OH is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, V = m3. (nearest integer) Given : F = 96500 C mol-1
Answer
560
Solution
In the fuel cell CH3OH is oxidised (anode) and O2 is reduced (cathode):
CH3OH + 32O2 CO2 + 2H2O, n = 6
E cell = 1.23 - 0.02 = 1.21 V
G = -nFE cell = -6 96500 1.21 = -700590 J
Useful work at 80% efficiency = 0.80 700590 = 560472 J
Work on the gas: W = Pext| V|
⇒ | V| = 5604721000 = 560.47 m3 560 m3 (decrease in volume)
Answer: 560