Chemistry help

[helloworld123]

if u find the mass formed on the cathode of a particular element and divide it by molar mass will it give u the same as Q=I*t and then working the number of mole by n(e-)= Q/F?

will n=m/M give u the same answer as Q=I*t, n(e-)=Q/F, compare electrons to element and do stoich will this give the same mol?

[nisha]

n(mass at cathode)= n(element) only if the ratios are 1:1
You can find n(e-) using the equations from the electrochemical series. n(e-) does not always equal n(element cation) as the ratios are different.
Eg:
Ca2+ +2e- -> Ca (s)
n(Ca2+)= 0.05 mol
Therefore n(e-)=2*0.05=0.1 mol

Does that, sort of make sense?

[helloworld123]

we were given a chart with Charge and time and also the amount that had accumulated on the cathode in grams but i did not do n=m/M i did Q=It and n(e-)=Q/F and compared the ratios with my half equations on the table provided so was wondering if i would get the same amount of mol if i did it this way?

thanks for ur reply:)

[nisha]

Yes, finding the moles through n=m/Mr, is the same as finding the moles through Q=It, then n(e-)=Q/F, then n(cation)= multiple in the electrochemical series *n(e-).

[helloworld123]

by this "then n(cation)= multiple in the electrochemical series *n(e-)." do u mean say if i find the n(e-) = 0.002 then for example say the half eqn is cu2+ + 2ei ----> cu then n(cu) = n (e-)/2 ??

[nisha]

by this "then n(cation)= multiple in the electrochemical series *n(e-)." do u mean say if i find the n(e-) = 0.002 then for example say the half eqn is cu2+ + 2ei ----> cu then n(cu) = n (e-)/2 ??

Yes, i mean exactly that:)

[helloworld123]

thankyou so much! q. Draw a diagram for a copper plating cell, labelling the cathode, anode, polarity of electrodes, direction of electron flow in wires, direction of movement of the ions in the electrolyte and the metal used for the anode. im not 100% what this would look like?

also there is a question i cant really remember the full bit so it will be difficult:P 
"the charge of an electron is 1.062*110^-19C. calculate the value of avagadros constant using the value of faraday you determined in q9"
dont have the sheet with me but any ideas whatsoever on how this may look to work out plz help!

thanks

[Lasercookie]

"the charge of an electron is 1.062*110^-19C. calculate the value of avagadros constant using the value of faraday you determined in q9"

So you would have calculated a value for Faraday's constant. Faraday's constant is the charge on one mole of electrons right?
We also know the charge on a single electron. So if we want to know how many electrons are in a mole, then we can divide the value of charge on one mole by the charge on the single electron.

e is the charge on the electron.

[helloworld123]

fo in the question before hand i would use the formula F=Q/n(e-) and for this next part  calculate the value of avagadros constant using the value of faraday you determined in q9" i would use  the formula for NA which u have provided?

[charmanderp]

I always find it better with Avogadro's number to realise the theory behind it. Saves you have to memorise a formula. Also remember that the charge on a single electron is in the data booklet.

[helloworld123]

what does 'the charge of one mole of electrons, in coulomb' mean?

& Q. Allowing for experimental error, what is the relationship between the charge on the metal ion present in the electrolyte in a cell and the charge, in coulombs, needed to deposit one mole of metal?

[charmanderp]

It's exactly that; the electrical charge on 6.022*10^23 particles worth of electrons.