Electrolysis When calculating amount of charge from an amount of

[Whalstib]

Electrolysis

When calculating amount of charge from an amount of product in an electrolysis the typical equations proceeds for instance:
.404g Cu * (1 mole Cu/63.6g) * (2 moles electrons/1 moles Cu) * (96485/1 mole e) = 1.226 E3 C

BUT when solving for amount of product for instance we go like this:

5.11C *(1 mole e/96485) * ( 1 mole I2/ 2 mole e) * (254 g I2/1 mol I2) = 6.73 E-3 g I2

My question is in regards to the Faraday Number which is highlighted.

Why is the fraction associated with the Faraday Number inverted? Since there are 5 variable to solve for how does one know which one to use?

I have 2 reference books Ebbing’s and Zumdahls textbooks and neither explains this.

Any clues and how I can tell which to use to say solve for time give the other variables?

Thanks,

W

 

[Yanick]

Faraday's number is the magnitude of the charge of 1 mole of electrons. IE find the charge on 1 electron and multiply by avogadro's number to find the charge on 1 mol of electrons. Another point to know about the number is the fact that an electron's charge is negative whereas the Faraday constant is typically written without the negative which is really only describing the magnitude of charge of 1 mol of electrons.

The reason it is inverted is because it is being used as a conversion factor, just as we may use g/mol or mol/g when converting from mol to grams or vice versa. Its called, I believe, dimensional analysis and it is a math 'trick' to save some time when doing calculations that involve lots of conversions of units and such. Instead of using a page of calculations with proportions and such, you can do the necessary conversions in 1 long line of calculations which will ultimately yield the appropriate number and units. The trick with setting up these calculations is to put all the conversion factors so the units that you don't want end up canceling out. IE when going from grams of substance you multiply that number by mol/gram to yield number of moles;

Xg * 1mol/Yg = X/Y mol

Its a very useful skill to attain if you plan on doing any kind of work in science which requires calculations.

Its the same for the Faraday constant. If you know the total charge (coulomb's) you multiply by 1 mol e/96485 C so that the coulomb's cancel and you are left with mol of electrons. And if you know mol's of electrons you mulitply by the reciprocal so that the mols of electrons cancel and you know the total charge.

Getting a hang of these types of calculations will mean the end of memorizing the algorithms for solving such questions. I almost purposefully don't memorize the procedure but try and understand where it all came from so that if I ever have to do it and don't have any references handy I can pretty much set everything up based on the concepts. You should have a clear picture in your head what each step of that procedure that you listed accomplishes. IE why are you multiplying mols of Cu by 2mol e/1mol Cu? What did that number just get you? What would you need to do to that number to progress toward your ultimate goal? Its more of a pain to really learn the stuff than just memorize it, but if you want to do anything even remotely scientific you need to start thinking about these things.