An electrolyzer consists of 100 cells, each with 1 m^2 of effective area. When c

Question

An electrolyzer consists of 100 cells, each with 1 m^2 of effective area. When current density is 1000 A/m^2, the voltage required by each cell is 1310 V. The electrolyzer is installed in a cubical room (3 m edge) whose walls, door and ceiling conduct heat at a rate of 50 W/m^2 per kelvin of temperature difference. No other source or sink of heat is in the room. Outside temperature is 30 degree C. When a current of 1000 A is forced through the cells, How many kg of H_2 are produced per day? How many kg of O_2 are produced per day? How many kg of H_2O are consumed per day? What is the equilibrium temperature of the room? What current causes the room to reach the lowest possible temperature?

Explanation / Answer

We know 1 A = 1 coloumb / sec

1000 A = 1000 coloumb / sec

also 1.6 * 10-19 electrons = 1 Coloumb

hence

1000A = 1000/1.6 * 10-19 = 6.24 * 1021 electrons/sec

We require 2 electrons to get 1 molecule of H2

hence H2 produced per day will be = 1/2 * 6.24 * 1021 /6.02 * 1026 kmoles/second

= 5.18 * 10-6 kmoles/second

converting second to days

= 5.18 * 10-6 * 86400 = 0.448 kmoles/day

it will be - 0.9 kg/day

so for 100 cells we get 90 kg/day

b) molecular weight of oxygen = 16

molecular weight of hydrogen = 2

ratio = 8:1

so weight of oxygen produced will be 720 kg/day

c)

water consumed will be = 0.9 + 7.2 = 8.1 kg/day

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.