Electrolytic cell, copper as an example:
Amperage is important (voltage has little consequence for the most part) and almost becomes unimportant in comparison. amperage determines how much metal moves through the cell, with that and knowing the metal and its molecular weight we can determine how much time to move the copper from the anode to the cathode
The copper anode of fairly high purity is to be dissolved and plated out as a more pure product electrowinning of copper.
The anode is the sight of oxidation (loss of electron) where the copper metal or other metals are forced by your DC power supply to cause the electrochemical reaction that would not occur otherwise, into giving up its electrons, the metal atoms on the loss of electrons dissolve going into solution as Ion's or salt of that metal, along with a salt of the electrolyte used (sulfite ions, from sulfuric in the example using copper sulfate).
The copper anode will have several metals involved, we will call them impurity metals for discussion, some valuable, some not, some that will remain in electrolyte solution as ions some that will fall out of solution as anode slime or mud, as salts or elemental metals.
Of the several metals, it is the amperage (not voltage) or current flow that selects the metals one at a time (in a certain order of oxidation by their (Standard electrode potentials), ( the more reactive the metal the easier it will give up its electrons), zinc, iron, nickel, lead, copper, silver, gold, what does not form ionic solution precipitates as mud anode slimes.
Silver and gold being more reactive than copper will only be affected after copper, and will only be oxidized or released after the copper metal atoms are oxidized into solution and precipitate as mud or anode slime.
More reactive metals Zn, Fe, Ni, and Pb, more easily oxidized than copper atoms will react or goes into solution before copper (into solution as ions if soluble in the medium) (or as mud, such as lead sulfate salts).
After these metals are in solution as ions in an ionic solution, their disposition onto the cathode (the sight of reduction or gain of the electrons) or plating out at the anode, where the metal ions gain electrons to be converted from a salt of the metal to elemental metal atom, this reactions also takes place in a specific order Au, Ag, Cu, Pb, Fe, Zn, this is dependent on current (amps), electrolyte, temperature, voltage, anode metals involved, condition of the anode...
Too high of a current (amps) can cause more or all metals to deposit or plate out (alloying the cathode) or contaminate it.
The more neutral (basic) the easier the more electronegative metals to be deposited.
The same can be said of the electrolyte, the poorer it is in copper (if the copper anode has more impurity metals or metals other than copper), These are more easily oxidized at the anode or more easily to be reduced at the cathode.
The purer the copper anode involved, the purer the electrolyte, the less dense and compact the anode, and the easier the process will go,
Being more important than voltage, The amperage is the driving force of the cell, the amperage is a result of the resistance of the electrolytic in the cell which conditions can change, the electrical connection resistance internal and external, and the voltage of the power supply needed to overcome this resistance and any overvoltage potential to overcome these forces.
The size of the electrode surface area helps us to determine the current or amperage range we need.
Example: CuSO4 electrolyte, 68 degrees, 3 parts CUSO4, 10 parts H2SO4, diluted 10 times water, current then adjusted to plate.
Maximum current limit to 1 amp per 33 centimeters or about 5 square inches, or 30 amps per square foot of cathode area. (The purer anode needs a higher current).
How much time?
Copper, Atomic weight 63.55 grams per mole (mol or mole a measure like a dozen of eggs) (a mole can be a measure of electrons, ions, molecules, sheep going through a gate...).
From the electrode potential of copper (reduction of copper in this case).
Cu2+ + 2e- --> Cu (s) Electrode potential +0.34 volts (reduction or gain of electrons at the cathode in this case.
From this, we see to form one mole of copper (63.55 grams of copper deposit on the cathode). We will need to move 2 moles of electrons or two faradays (2e-).
1 amp current X 1 second of time = 1 coloumb
96485.3383 coulombs = one faraday
1 faraday per mole
one mole elcectons 6.022x1023
each electron carries a negative charge 1.6021x10-19
(one mole elcectons 6.022x1023) X ( each electron carries a negative charge 1.6021x10-19) = onne faraday 96485 coulombs
One faraday oxidizes or reduces one mole of the atomic weight of the substance (copper in our case.
From the equation, we see we need two moles of electrons or two faradays to deposit a mole of copper ( 63.5 grams of copper metal at the cathode.
Cu2+ + 2e- --> Cu (s)
So get to the answer. How much time to deposit an amount of copper?
Example;
Say we are running the cell at 2.5 amps (determined by our cathode size) (voltage of the cell is not important except to overcome cell resistance to give us our desired current (amp) flow through the circuits). Using CuSO4 copper sulfate electrolyte, and running the cell for 50 minutes time how much copper will deposit?
Cu2+ (copper sulfate) + 2e- (2 moles of electrons or two faradays needed to move or 2x96478 coulombs) --> Cu (s) (deposit of a mole of copper metal 63.5grams)
How many coulombs passed?
Coulombs? = 50 minutes X 60seconds/1minute X 2.5 amps/second = 7.50 X 103 coulombs
grams of copper deposited
7.50x103 coulombs x 1mol e-/96478 x 63.5g Cu/2mol e-
=24.7g copper metal deposited onto the cathode, running 2.5 amps for 50 minutes
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