Hi Ferdinand,

Just to clarify, you got a reaction to occur at 1 V even though theoretically it shouldn't occur until 1.22 V and you want to why, correct?

Firstly, from the values you have above the Eo would be 2.00 V. From a table of stadard potentials I have Fe2+ + 2e- --> Fe having an Eo of -0.41 V, giving an overall potential of -1.25 V for the cell you're creating/

If that's the case the reaction will only occur at 1.25 V under standard conditions (1 M solutions of electrolyte being the major factor). Are you starting with Fe2+ in solution? If not, the reaction will occur at lower potentials as per the Nernst Equation:

E=Eo-RT/nF*log(cred/cox)

If that isn't the case can you provide more information about how you've set up the cell (electrodes, electrolytes, etc.)?

Cheers,

Laurence

Thank you sir

first, sorry I forget that 0.34 is for copper reduction, and sorry for my english

I just want to know that the Fe dissolution (as anode)  still occur even though I set the potential at 1V (theoretically i should set the potential at 1.22 ~ 1.25 V). 

I don't use electrolyte, I used 500 mL electroplating wastewater.

maybe you need this information about the waste characteristic.

pH

Hi Ferdinand,

No need to apologise, your English is better than my ability in any other languages.

Based on your solution composition, it's fairly clear that is the reason for the lower required voltage. The fact that there is not already a significant amount of Fe2+ in solution will drive the oxidation reaction forward. If the solutions in your cell were 1 M Fe2+ and 1 M Al3+ respectively, then you would have the reaction occurring at the calculated Eo value. Also if you don't have other elements in the cathodic half cell solution you may not be reducing aluminium. 

The Nernst equation as given above would allow you to calculate the actual expected voltage required with your solutions. From the equation above the terms are:

E = Actual potential

Eo = Standard potential

R = Universal gas constant (8.314 J/mol.K)

T = Temperature in K

n = number of electrons in the reaction

F = Faraday's constant (96485 C/mol)

cred = concentration of the reducing agent (the species being oxidised)

cox = concentration of oxidising agent (the species being reduced)

There are plenty of examples online of how to use this.

Hope this helps.

Cheers,

Laurence

Thank you sir..

"Based on your solution composition, it's fairly clear that is the reason for the lower required voltage."

this mean that the required voltage could be lower due to the composition of my waste?

"The fact that there is not already a significant amount of Fe2+ in solution will drive the oxidation reaction forward."

this mean, under 1Volt the oxidation of Fe (dissolution) in my case  is occur? 1V is enough to make oxidation happened?

"If the solutions in your cell were 1 M Fe2+ and 1 M Al3+ respectively, then you would have the reaction occurring at the calculated Eo value."

this mean that 1.25V is required under standard condition (1M of corresponding metal). If the solution isn't under standard condition the required potential could be higher or lower. am i right sir?

"Also if you don't have other elements in the cathodic half cell solution you may not be reducing aluminium. "

I don't really understand about this sir. if I have no Al in my solution, the reduction of it couldn't occur?

"The Nernst equation as given above would allow you to calculate the actual expected voltage required with your solutions."

yes sir, I will learn about it more.

Thank you so much sir. If what I respond to you was correct then I'm quite understand.

Hi Ferdinand,

1) Yes, the voltage will change based on the solution you use (in your case the waste water).

2) That's correct, a lower voltage will work where no/little Fe2+ is in the anode solution.

3) Exactly, different solutions will produce a different potential.

4) You can't reduce metallic aluminium you need Al3+ ions to reduce to metallic Al, so just because the cathodic electrode is aluminium, doesn't mean that the reaction will be aluminium, it means something else is reducing at your cathode.

5) Excellent, there are plenty of good resources.

Cheers,

Laurence

Hi Sir, Thank You again

4) I misunderstood about that sentence sir.

yes sir....It can't reduce Al if It doesn't have Al on it.   I'm sorry sir, my bad. 

Thank You so much sir, this is such a relife.

Best Regards,

Ferdinand

Dear Ferdinand Dos Santos,

in most electrochemical measurements[1], it is necessary[2] to keep one of the electrodes at a constant (reference) potential. A reference electrode (RE) has a known[3] electrode potential and is stable. The high stability of RE is achieved by employing the redox system, which must contain (buffered or saturated) concentrations in each of the participating solutions of the reaction. It's used as a half cell in the electrochemical cell to allow for the determination[1-4] of the other (your Al and Fe) half's cell potential. So, try, please, to use, at least temporarily, an available[2,3] reference (third) electrode in order to measure the potential(s) of your (Al and Fe, under possible corrosion[5]) electrodes vs. the RE.

1.   Reference Electrodes   https://www.gamry.com/application-notes/electrodes-cells/reference-electrodes

2.   Reference Electrode HydroFlex   https://www.referenceelectrode.info/

3.   Standard electrode potentials   https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)   ;    Standard Electrode Potentials  http://www.corrosion-doctors.org/Corrosion-Thermodynamics/Standard-potential.htm#decreasing

4.   https://www.researchgate.net/post/What_is_the_actual_role_of_reference_electrode_and_counter_electrode_in_3-electrode_system_in_electrodeposition2

5.   Corrosion cells and reactions   http://www.chem1.com/acad/webtext/elchem/ec7.html