Do you know what a faraday is?

[estairella]

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Just asking.. because I didn't. Well I do now, but I've never seen it come up before (it was a discrete question); does anyone know if it's something one needs to know for the MCAT?

(I initially thought it was the same as a farad... made the question very very confusing).

 

[DrBowtie]

I didn't know that. I don't think you need it for the MCAT.

http://en.wikipedia.org/wiki/Faraday

 

[carn311]

Looks to be obsolete.
Helps to be aware of it I guess...

 

[carn311]

I didn't know that. I don't think you need it for the MCAT.

http://en.wikipedia.org/wiki/Faraday

You beat me to it Brett.

It seems my wikipedia skills are eroding

 

[DrCurious]

It is mentioned in the TPR books, and it was a discrete at teh end of one of the AAMC's. Just rememeber it is C per mol of e

 

[odddodo]

I actually didn't know about it until I tutored a high school student in general chemistry earlier this summer and saw it in his textbook...

 

[jpostbac]

yeah, i believe you had to know faraday's constant to answer a discrete at the end of AAMC 7R PS... yet another reason i hated 7R -

 

[Sol Rosenberg]

Pretty sure that they would give it to you in a passage. It should only pop up in the deltaG = -nFE equation, where F is the Faraday. I wouldn't worry about it too much.

 

[solitude]

Weird, we did actually cover this in my physics class. I have no idea why.

 

[superwillis]

Make sure not to confuse Farady's constant with the Farad.
The F in the equation G=-nFE is Faraday's constant (96500 C/mole), which is different from the Farad. The Farad is the SI unit of capacitance, usually in terms of microfarads.

EDIT: Oh, oops, got confused myself.

 

[87138]

We used the Faraday quite a bit in Gen Chem II while talking about electrochem.

 

[Sol Rosenberg]

Make sure not to confuse Farady's constant with the Farad.
The F in the equation G=-nFE is Faraday's constant (96500 C/mole), which is different from the Farad. The Farad is the SI unit of capacitance, usually in terms of microfarads.

Hello McFly, that's what we're talking about here.

While you likely don't need to memorize the Faraday/Faraday's constant, you obviously DO need to know about capacitance and Farads.

 

[grapeflavorsoda]

Just asking.. because I didn't. Well I do now, but I've never seen it come up before (it was a discrete question); does anyone know if it's something one needs to know for the MCAT?

(I initially thought it was the same as a farad... made the question very very confusing).

faraday's law is about the relationship between magnetic flux and the induced electric field(which is non-conservative).

there was a question about that on the mcat i took.

just know that changing magnetic flux through a closed loop can cause electric current in that loop, which is a special form of faraday's law.

i dont think direction of induced current would be important, but you could study lenz's law(which states that induced current will form such that magnetic field formed by it will oppose the change in magnetic flux) if you feel neurotic about it. 😛

 

[odddodo]

faraday's law is about the relationship between magnetic flux and the induced electric field(which is non-conservative).

there was a question about that on the mcat i took.

just know that changing magnetic flux through a closed loop can cause electric current in that loop, which is a special form of faraday's law.

i dont think direction of induced current would be important, but you could study lenz's law(which states that induced current will form such that magnetic field formed by it will oppose the change in magnetic flux) if you feel neurotic about it. 😛

Not Faraday's law but the Faraday. I'm guessing that all we need to know is that one Faraday is the charge of one mole of electrons. Nothing more.

 

[CA-MedStudent]

can someone please explain the question on Faraday that was on the PS of AAMC7 it was like the second to last question. I figured out that all they did was take the 0.1 and divide by 3 because of the charge of the ion, but why? whats the reason behind that? thanks.

 

[87138]

can someone please explain the question on Faraday that was on the PS of AAMC7 it was like the second to last question. I figured out that all they did was take the 0.1 and divide by 3 because of the charge of the ion, but why? whats the reason behind that? thanks.

As another poster pointed out, 96,500 (or 96,485) Coulombs/mole is Faraday's constant. This means that for every 96,500 Coulombs, 1 mole of ELECTRONS is plated out. By telling you that the amount is "0.1 Faraday" or however they phrase it, they're making it easier than telling you "9,650 Coulombs."

So, 0.1 Faraday, by definition, means you plated out 0.1 moles of electrons. If the ion you were using had a +1 charge, then you could also say you plated out 0.1 moles of this ion's metal (i.e. Ag+ to Ag). Since it takes 3 electrons to fully reduce Al +3 (that was the element they used, correct? I'm doing this from memory), that means that for each mole of aluminum you want to plate out, you need three moles of electrons. Thus, you take the moles of electrons divided by 3 and you will arrive at the correct number.


I hope that makes sense and isn't too wordy/confusing.

 

[Dr. AK]

As another poster pointed out, 96,500 (or 96,485) Coulombs/mole is Faraday's constant. This means that for every 96,500 Coulombs, 1 mole of ELECTRONS is plated out. By telling you that the amount is "0.1 Faraday" or however they phrase it, they're making it easier than telling you "9,650 Coulombs."

So, 0.1 Faraday, by definition, means you plated out 0.1 moles of electrons. If the ion you were using had a +1 charge, then you could also say you plated out 0.1 moles of this ion's metal (i.e. Ag+ to Ag). Since it takes 3 electrons to fully reduce Al +3 (that was the element they used, correct? I'm doing this from memory), that means that for each mole of aluminum you want to plate out, you need three moles of electrons. Thus, you take the moles of electrons divided by 3 and you will arrive at the correct number.


I hope that makes sense and isn't too wordy/confusing.

Good explanation.

Two more questions…

If they gave you 9650 C would you be able to solve the problem with that value in Coulombs or would you have to convert to F (i.e. 9650 is 10 % of 96500 therefore 0.1 F) before calculating the amount of electrons and the later the amount of moles reduced.

Also, in this problem when you divide 0.1 F by the number of electrons needed per one Al (that is 3), you are essentially dividing by charge and therefore end up with moles…correct?

 

[87138]

Good explanation.

Two more questions…

If they gave you 9650 C would you be able to solve the problem with that value in Coulombs or would you have to convert to F (i.e. 9650 is 10 % of 96500 therefore 0.1 F) before calculating the amount of electrons and the later the amount of moles reduced.

Yes, though it gets a little more "messy." The only reason I said it was better that they gave you the 0.1 figure was that it doesn't require you to know Faraday's constant. If they said specifically "9,650", then, assuming you didn't instantly realize that was 10% of Faraday's constant, you could divide that number by the charge of an electron:

9,650 C/m
--------- = 6.03E22
1.6E-19 C


At that point you would then realize that that number was also 10% of one mole (or 0.1 mole).

Basically, there are those several ways to go about it, but the ultimate realization is that you have 0.1 mole of electrons being used. Regardless of your approach, this necessary information. They were nice enough to bypass all that other stuff and just give you the "0.1" figure from the beginning.