The hydrogen-oxygen fuel cell is described in the text. For the questions below,
ID: 536980 • Letter: T
Question
The hydrogen-oxygen fuel cell is described in the text. For the questions below, assume the following: Assume that the air is 20% O_2 by volume and that all the O_2 is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior. (a) What volume of H_2(g), stored at 25.0 degree C at a pressure of 167 atm, would be needed to run an electric motor drawing a current of 8.50 A for 3.00 hr? L (b) What volume (in liters) of air at 25.0 degree C and 1.00 atm will have to pass into the cell per minute to run the motor? L/minExplanation / Answer
a) Temperature= 250C
Pressure= 167 atm
Current= 8.50 A
Time= 3 hours
Faraday Constant = 96500 C/mol
The total charge passing through the circuit is
= (8.5 A *3 hours* 3600 seconds) = 9.18*104 C
Converting to number of moles of electrons
9.18*104 C * (1 mol e-/ 96500 C) = 0.951 mol e-
From the anode half reaction, we can find the amount of hydrogen,
0.951 mol e- * (2 mol H2/4 mol e-) = 0.475 mol H2
The volume can be calculated using, PV= nRT
V= nRT/P = (0.475 mol H2 * 0.0821 L.atm/K.mol * 298 K)/ (167 atm) = 0.069 L of H2 would be needed to run
b) The charge passing through the circuit in one minute is
= (8.5 C* 60) = 510 C/min
Converting to number of moles of electrons
510 C/min * (1 mol e-/ 96500 C) = 5.28*10-3 mol e-/min
We can find the amount of oxygen from the cathode half-reaction and the ideal gas equation
5.28*10-3 mol e-/min * (1 mol O2/4 mol e-) = 1.32*10-3 mol O2 /min
The volume can be calculated using, PV= nRT
V= nRT/P = (1.32*10-3 mol O2/min * 0.0821 L.atm/K.mol * 298 K)/ (1 atm) = 0.032 L O2 /min
Air contains 20% O2 by volume (given)
So, (0.032 L O2 /min) * (1 L air/ 0.20 L O2) = 0.16 L of air/min will have to pass into the cell to run the motor.
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