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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Nomvalt on October 12, 2010, 05:09:53 pm

Title: faraday's first law of electrolysis
Post by: Nomvalt on October 12, 2010, 05:09:53 pm: okay, yesterday we did an experiment involving oxidation and reduction. It was one of those experiments where you make a circuit and with time notice a build up of metal in one of the two electrodes. Anyway, we used copper for the electrodes. here are the results:

Current: 1.0 A
Time: 900 secs
Charge: 900 coloumbs
Initial mass of cathode: 2.58 g
Final mass of cathode: 2.85 g
Mass gain: 0.27 g

1) Calculate the quantity of charge, measured in coulombs, required to deposit 63.5 g (1.0 mol) of copper metal.

pretty standard question, anyway any help would be appreciated.
edit: both electrodes (anode and cathode) were made of copper. this was dunked into copper sulfate solution for 15 mins.
Title: Re: faraday's first law of electrolysis
Post by: sb3700 on October 12, 2010, 05:35:04 pm: First, what do all the terms mean? Most of these are unrelated to the question but you need to know (presumably) the concepts for the rest of the question.

Charge (Q in C): just a way of counting electrons - each Coulomb is some big number of electrons
1 Coulomb = 1 / (1.6e-19) electrons

Current (I in A): charge / second - basically, how many electrons go past each second
I = Q / T

Voltage (V in V): energy / charge - basically how much energy each electron has at that point.
V = E / Q

(Note it is a relative number to 0V, so 10V means each electron at 10V has 10 units more voltage than at 0V. Correctly, each coulomb of electrons at 10 V has 10 J more energy than at 0 V.)

and finally, each mole of electrons has 1 F = 96500 C / mol charge (by definition of a Faraday)
as an equation:
Q = n * F, where n is the number of moles of electrons, F = 96500 C / mol

---

To solve, it work out what do we know?

We know that we have ~~1 mol~~ 2 mol of copper. Each copper reduction reaction uses 1 electron
$Cu^{2+} + 2e^- \to Cu$

So the total number of electrons, n = (1 mol copper) * (2 mol electron / mol copper) = 2 mol electron

so total charge needed = n * F = 2 mol electron * 96500 C / (mol electron)

EDIT: fixed mistake
Title: Re: faraday's first law of electrolysis
Post by: akira88 on October 12, 2010, 05:42:37 pm: Doesn't copper use 2 mol of electrons for every mol of copper metal produced? :|
Cu2+(aq) + 2e ---> Cu (s)
Title: Re: faraday's first law of electrolysis
Post by: toshibaj on October 12, 2010, 06:50:47 pm: Quote from: akira88 on October 12, 2010, 05:42:37 pm
Doesn't copper use 2 mol of electrons for every mol of copper metal produced? :|
Cu2+(aq) + 2e ---> Cu (s)

yep akira's right about that
Title: Re: faraday's first law of electrolysis
Post by: sb3700 on October 12, 2010, 08:13:02 pm: Thanks, I wasn't thinking straight. I'll edit in a correction
Title: Re: faraday's first law of electrolysis
Post by: Nomvalt on October 13, 2010, 07:48:55 am: could you also find the answer by solving for x in:
63.5g/0.27g = x/900 C

?