3. H2 gas is generated using PEM electrolysis at pH -0. 1.229 02(g) 0.401 H20 LL

Question

3. H2 gas is generated using PEM electrolysis at pH -0. 1.229 02(g) 0.401 H20 LLU H2(g) 0.829 .1 Base 14 Acid pH Assume rj = 0.00$(V) node = 0.65 + 0.05|Ogj (V) ncathode -0.05 + 0.03logj (V) the electrical current density is measured with units of kA/m2 (a) Calculate the potential needed to split the water at current density of 1 kA/m2. (10) (b) Calculate the overall efficiency of this electrolysis. (5) e Calculate the lericl energ neded to gene) (d) Based on your calculation, estimate the energy density of H2. (5)

Explanation / Answer

The question has been asked from water electrolysis. For understanding the fundamental of water electrolysis, the pH Vs. potential (E) pourbiax diagram is shown.

Question (a)

If I understand correctly, the question (a) is for over potential (cell) calculation in overall cell.

Since, the Tafel equation for both anodic: =0.65+0.05logj and cathodic: =0.05+0.03logj are provided.

Hence, the overall cell needs overpotential () of cell = cathodic + anodic

cell = 0.65+0.05logj+0.05+0.03logj

            =0.70+0.08logj

The given current is 1kA/m2,

=0.70+0.08log1000

=0.70+0.08log1000

=0.70+0.080*3

= 0.94 V

Question (b)

The overall cell voltage efficiency is (Eanode-Ecathode)/Ecell

Potential at anode side, the given current is 1kA/m2

anode=0.65+0.05logj

anode=0.65+0.05log1000

anode=0.8 V

cathode =0.05+0.03logj

cathode =0.05+0.03log1000

cathode = 0.14

The voltage efficiency = (0.94-0.14)/0.94 *100

                                    = 85.1% Voltage efficiency for hydrogen evolution

Question (c)

Electrical energy needed for 1 kg of hydrogen generation

1 kg of hydrogen is nothing but 500 moles of hydrogen, it is equivalent to 2*500M*96485C/M

= We need 96485kC charge to generate 1kg of Hydrogen, If we run the above electrolyser for 96485 s, we can get 1 kg of hydrogen.

Question (d)

Energy density of this produced hydrogen,

When we pass 0.94 over potential to generate 1 kA current in water electrolyser

The net potential of this electrolyser is 1.23 + 0.94=2.17

The current is 1000 A

Hence the power is P = V*I= 2.17*1000

                                           =2.17kJ/s

1000kA is equivalent to 0.01 M/s of hydrogen evolution

So, 217 kJ/M is the hydrogen energy density from this water electrolysis

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