Electrolysis is a means of generating hydrogen and oxygen gases from water by me

Question

Electrolysis is a means of generating hydrogen and oxygen gases from water by means of an electrochemical cell. Answer the following questions about a proposed design for a steady-state electrolysis unit.

a. What is the minimum required steady-state electrical work (in kW) to turn 1 mole/s of liquid water into separate streams of pure hydrogen and oxygen gases (each at 1 bar pressure) by hydrolysis at 25 C?

b. It is not convenient to transport gases at 1 bar. Therefore the gases from part a must be compressed to 50 bar. This is to be done by a series of compressors and heat exchangers so that the overall process for the separate hydrogen and oxygen streams can each be approximated as a single isothermal compression. What is the minimum required shaft work (report the total amount in kW) to take the two output streams from part a and compress them? To make the calculation go faster, you may assume ideal gases.

c. Instead of operating the electrolysis unit at 1 bar, it is instead operated at 50 bar, meaning that liquid water feed and product gas streams are all at 50 bar. This is done in an effort to avoid the need for expensive gas compressors downstream from the electrolysis unit. Repeat the calculation from part a at this higher pressure.

What is the minimum required steady-state electrical work (in kW)? To save time, you may assume ideal gases and that the Poynting factor for liquid water is one. How does this answer compare to the total work from parts a and b?

d. What is the minimum electrochemical cell voltage required in part c?

Explanation / Answer

The reaction is H2O---> H2+1/2O2

Change in number of moles = 1.5

minimum work =PdV ( where dV= volume change during the reaction)

1 mole of any gas at 273.15 K occupies 22.4*10-3m3

at 298.15K the volume is 22.4*10-3*298.15/273.15

Work done =101.3*103 Pa *1.5moles/s* 22.4*10-3*298.15/273.15=3713 Joules= 3.713 Kj/s= 3.713KW

b) The work done during isothermal compression = nRT*ln(P2/P1)= 1.5*8.314*298.15*ln(50/1)=48.7 Joules/Sec=0.0487 Kw

c) when the pressure is 50 bar , work done for separation = 50*101.3*1.5*103*22.4*10-3*298.15/273.15=18.565Kw

d) gibbs free energy change is the minium work done and this = -nFE= 18.565*1000 J/sec

-4*96500*E= -18.565*1000

E= -18.565*1000/4*96500=-0.048V

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