3. H2 gas is generated using PEM electrolysis at pH 0 02(g) 1.229 53.2 mW/ph -40

Question

3. H2 gas is generated using PEM electrolysis at pH 0 02(g) 1.229 53.2 mW/ph -40H H20 2H +2e H2(g) .1 14 Acid Base PH Assume rj = 0.00$(V) randule 0.65 + 0.051°gj (V) Mcathole 0.05 + 0.03 logj (V) the electrical current density is measured with units of kA/m (a) Calculate the potential needed to split the water at current density of 1 kA/m2. (10) (b) Calculate the overll ftieney of this Blctrols5) (c) Calculate the electrical energy needed to generate 1 kg H2. (10) (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

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

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