April 21, 2019, 04:48:22 AM
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### Topic: Calculation Equilibrium Potential  (Read 388 times)

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#### SeanB123

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##### Calculation Equilibrium Potential
« on: February 06, 2019, 06:13:39 AM »
I am working on calculating the equilibrium potential, E_eq, of four reactions in an electrolytic cell, two occurring at a steel cathode and two occurring at a titanium anode:

Cathode:
Water Electrolysis
2H2O + 2e- -> H2 + 2OH-
Oxygen Reduction
2H2O + O2 + 4e- -> 4OH-

Anode:
Water Electrolysis
2H2O -> O2 + 4H+ + 4e-
Hydroxide Oxidation
4OH- -> 2H2O + O2 + 4e-

The cell has a potential difference of -15V.

I understand that E_cell = E^0 - (RT/nF)*lnQ

Solving this equation however is the problem.

Any advice and knowledge that can be shed on this would be greatly appreciated.

Thanks

#### chenbeier

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##### Re: Calculation Equilibrium Potential
« Reply #1 on: February 06, 2019, 06:29:56 AM »
The given voltage of -15 V cannot be by calculation, its to high. Or do you have some cells in series?

Values from table book

2H2O + 2e- -> H2 + 2OH-  +0,828V

2H2O + O2 + 4e- -> 4OH- -0,401 V

2H2O -> O2 + 4H+ + 4e- +1,229 V

4OH- -> 2H2O + O2 + 4e- +0,401 V

« Last Edit: February 06, 2019, 06:56:33 AM by chenbeier »

#### SeanB123

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##### Re: Calculation Equilibrium Potential
« Reply #2 on: February 06, 2019, 06:38:22 AM »
Thanks fir the reply. Would I be right in saying that the values from the book you've quoted are half-cell potentials rather than equilibrium potentials?

#### chenbeier

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##### Re: Calculation Equilibrium Potential
« Reply #3 on: February 06, 2019, 06:55:16 AM »
The equation is a half cell reaction to see on the electrons e-. So the values are the half cell potentials.

#### SeanB123

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##### Re: Calculation Equilibrium Potential
« Reply #4 on: February 06, 2019, 07:17:41 AM »
Ok so this is where I become confused. Im working on a problem which is similar to cathodic protection. The reactions in the problem I'm following are iron oxidation,
oxygen reduction, and hydrogen evolution. These are occurring at the cathode.

Fe → Fe2+ + 2e-
O2 + 2H2O + 4e- → 4OH-
2H2O + 2e- → H2 + 2OH-

Using the tables the half-cell potentials for these reactions are+0.41, +0.401 and  −0.8277 respectively.
However the equilibrium potentials used are -0.68V, +0.189V and -0.76V

This is what's confusing me. How do they arrive at these values?

#### chenbeier

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##### Re: Calculation Equilibrium Potential
« Reply #5 on: February 06, 2019, 07:46:30 AM »
At a cathode  will be reduction, so no iron oxidation. Only the two other equations can work.
Where you get the equilibrium potential numbers, and what is definition of it.

Normaly you get a resulting voltage between a cathode and a anode. The half potentials normaly measured against a hydrogen gas elecrode.

The word equilibrium potential is only found by talking about membranes?