Science Buddies: "Ask an Expert"

Hi, my names Adam, this is my first post im in year 10 UK (10th grade i believe)

Im predicted an A* in chemistry so this is why my experiment is more complicated.
I changed the charge of the ions, referring to the formula I=nAve, by increasing charge of ions (e), current (I) will increase, which is directly proportional to moles deposited. I weighed the cathode before and after using my 3 electrolytes, which had a charge of +1,+2 and +3. I noticed the trend i would expect, higher charge = higher mass deposited thus higher moles deposited. This may sound stupid, but why is the weight higher, do higher charge ions have a higher mass? are more attracted to the cathode? can you explain the weight difference.

Many Thanks, Adam

It sounds like you are dealing with an electro plating process. One fundamental question you need to answer is "Is the material being depositied in the different experiments chemically the same material?" Different materials have different atomic/molecular weights.

You haven't provided enough details of what chemicals you were actually using for us provide any real clues.

The ionic valences of the plating material both in the solution and in the deposited form will affect how many moles transfer per mole of electrons. If you have metals like iron that have multiple valences, figuring out the proportions of ferrous versus ferric ions that were involved can provide a range of not only molecular weights but moles per electron in the plating process.

Please provide additional details.

Thankyou for replying sorry but much or your last paragraph went over my head

The three electrolytes :
Silver Nitrate AgNO3 +1 ion charge
Copper Chloride CuCl +2 ion charge
Iron Chloride +3 ion charge

I used 100ml of each at 0.1mol.

Could you please explain more about ions to me, the more im reading about them the less im understanding and more confused im getting, thanks.

Adam

Understanding ions and ionization can be confusing. One of the simpler ways to think about it is at the atomic level. If you disolve a compound (molecule) in some solvent (in your case, I presume water or acqueous solution) and the compound ionizes, the electron donor atom gives up some number of electrons (determined by its valance) and becomes a free positive ion and the negative radical or atom (chlorine atom in you case) accepts the extra electrons and becomes a free negative ion. At any given point in time, not all of the molecules are disassociated into ions. As the ions float around in solution, they will be attracted to the opposite charged ions and will form ionic bonds and become a molecule. This back and forth switching between disassociated and associated states will go on at some rate in the absense of some driving force. Once you apply an electric current, the cathode will attract the positive ions and the annode will attract the negative ions. If you start asking questions about the disassociation rates and recombination rates and what affects them, you are asking questions usually covered in an upper level college course on thermodynamics.

Ag (silver) has a single electron in the 5s shell so its valance is +1 (it can donate one electron to become an ion) at low energy states. Unfortunately, silver oxidizes rapidly and has three different oxides so what you are plating can end up being at least four different molecules, pure Ag, plus its three oxides and if there are any negative ions that will plate as crystals on the cathode, the the number of compounds possible is even higher. These compounds can co-exist so determining the molecular weight of a mole of plated molecules will be difficult unless you did the experiment in a solution that was oxygen free and there were no other platable (deposited) ions or platable (deposited) crystals. If you had pure silver plating and measured the result quickly before any oxidation, you should come close to 108 g/mol of electrons used.

Cu (copper) has multiple valances (+1, +2)
I'm assuming you had Copper Chloride CuCl2 (based on the wording and not the formula given). If there was nothing in the solution to donate an electron to convert Cu+2 ions to Cu+1 ions, then you should get one mole of copper atoms plated per two moles of electrons. While copper has multiple oxides, only the surface layer will oxidize and then not vary rapidly so your weight of the plated copper should be close to 63.5 g/mol of copper or 36.7 g/mol of electrons.

Fe (iron) has multiple valances (+2, +3)
With FeCl3 the iron will ionize as Fe+3 so it will take 3 moles of electrons to plate one mole of Fe. Unfortunately, FeCl3 can become FeCl2 in the presence of any competing proton donor. If there is any free oxygen around, you can easily get Fe2O3 and then Fe2O3+FeCl3 -> 3 FeOCl. Figuring out what iron compound or compounds actually got plated in this case will be difficult. Without knowing what the final plated compounds are and the perportion of those compounds and knowing all disolved iron compounds and how many moles of them were produced, you can't balance the family of reactions to convert moles of electrons to predict the result. The best you can do is divide the number of moles of electrons by 3 to determine the best case (pure iron deposition) number of moles of pure iron that might be plated and divide the plated mass by that number and see how close it comes to 55.8 g/mole (or 55.8 g/3 moles of electrons). If it is within experimental measurement error, then you were lucky and the predominant plated material was pure iron.

AdamCFC wrote:I used 100ml of each at 0.1mol.

No! You started with 0.1 mol of some solution. What did you end up with left over? It doubt it was pure water. What happened to the chlorine atoms? The difference between the Ag/Cu/Fe that you started with and what was left in solution is what was used (hopefully all of what you used was plated on your cathode). Hopefully the chlorine ions did not pair up as a poisionous green Cl2 gas and escape, but were left in solution and paired numerically with H+ ions and O2 was released into the atmosphere at the anode and did not form oxides that were plated at the cathode.

I hope you were measuring the current and kept it constant with time and timed the duration carefully so that you can calculate how many moles of electrons you used. If you didn't then you are missing the major driving force in balancing the chemical equations.

A simple explanation for what is going on:
A given quantity of current (i.e., current X time) will be required to electroplate x grams of silver 1+. It will require twice as much quantity of current to electroplate x grams of copper 2+ because it is doubly charged. Similarly, it should require three times as much for iron 3+. However, iron usually cannot be electroplated as the metal in aqueous solution, because an acidic solution is required, and hydrogen gas would instead be evolved from the cathode without depositing any iron. An iron hydroxide may be formed on the cathode under certain conditions. Instead of iron, gold 3+ would be a better choice to illustrate the effect of a triply charged ion, because it can easily be electroplated from acidic chloride solutions.

jackharrar wrote:A given quantity of current (i.e., current X time) will be required to electroplate x grams of silver 1+. It will require twice as much quantity of current to electroplate x grams of copper 2+ because it is doubly charged.

Not quite! Replace "grams" with "moles" and the explaination would be correct. The molecular weight of silver (molecular weight ~108 g/mole) is heavier than copper (63.5 g/mole) so the plated weights for the same amount of electrons transfered would be closer to 3:1 ratio between silver and copper in grams.