Question

The International Space Station uses the electrolysis of water to continuously replenish the oxygen in the station’s atmosphere (the hydrogen formed is vented). The station’s electricity is generated by the four arrays of solar modules, which produce a total of 120 kW. Essential heating/cooling, circulation and steering equipment requires 75 kW. If one of the four modules fails completely, this leaves 15 kW for O2 production, which is run at a 2 V potential. How many astronomers can survive on board of the space station if each of them consumes 25 L of O2(g) per hour? Enter your numerical answer (rounded DOWN to the closest whole number, as we cannot have partial astronauts!) in the answer box. Assume 1 atm and 298 K as conditions.

To help you solve this problem, use the following information, suggested steps and hints:

a) Write the balanced half reactions at each electrode and the overall reaction!

b) If the electrolysis is carried out at 2 V potential, how many moles of electrons are produced in 1 s? (Hint: Calculate the Ampere available first using the following relationships: 1 W = 1 J/s; 1 J = 1 V·1 C; 1 C = 1 A·s)

c) Using your results from a) and b), how many moles of O₂ can be produced in 1 h?

d) How many moles of O₂(g) are required each hour for one astronaut, if s/he consumes 25 L of O₂(g) per hour? Assume 1 atm and 298 K.

e) How many astronauts can survive on the O₂ produced in c) if each consumes the number of moles calculated in d)?

Answer 1

a) Half cell reactions:

Cathode (reduction):

Anode (oxidation):

Overall reaction:

b) Power for O2 production = 15kW = 15000 W

O2 is produced at 2V

Power (W) = Voltage (V) * Current (A)

Current for O2 producion = Power / Voltage = (15000 / 2) A = 7500 A

1 Ampere = 1 Coulomb(C) / sec

Coulomb (C) of charge produced in 1 sec = 7500 Ampere * 1 sec = 7500 C

1 C = 6.24x10^18 charged particles ( here, electrons)

Number of electrons produced in 1 sec = 7500 * 6.24x10^18

1 mole = 6.022 × 1023  particles ( here , electrons) { Avogadro Number }

Moles of electrons produced in 1 sec = (7500 * 6.24x10^18) / 6.022 × 1023  moles

Moles of electrons produced in 1 sec = 0.078 moles

c) From the oxidation reaction, for every 2 moles of electrons produced 0.5 moles of O2 is produced.

Moles of O2 produced in 1 sec = Moles of electrons produced in 1 sec / 4

Moles of O2 produced in 1 sec = (0.078 / 4) moles

Moles of O2 produced in 1 sec = 0.019 moles

Moles of O2 produced in 1 hour (3600 sec)= (0.019 * 3600) moles

Moles of O2 produced in 1 hour (3600 sec)= 69.94 moles

d) Volume of O2 consumed by an astronaut in 1 hour = 25 L

Assuming O2 to be an ideal gas,

PV = nRT , { R: universal gas constant, R = 0.0821 atm.L/mol.K}

Plugging in values

1*25 = n * 0,0821 * 298

n = 1.02

Moles of O2 consumed by an astronaut in 1 hour = 1.02 moles

e) Number of astronauts that can survive = O2 produced in 1 hour / O2 consumed by an astronaut in 1 hour

Number of astronauts that can survive = 69.94 / 1.02 = 68.56

68 astronauts can survive with the O2 produced