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OCR Physics B Practical Investigation HELP ME PLEAAAASE!

  • Thread starter shinichii
  • Start date
5
0
Okay, so I'm panicking cause I was supposed to give my draft in two weeks ago... and I haven't.... Cause I got stuck... And began to procrastinate... Have started it again now... AND I'M STILL STUCKKKKK!!!

So if someone could just give me the basic answers to these questions, I'll know I'm on the right track, or if I'm screwing it all up.

1) How would the resistance of a solution vary with the distance between the electrodes and WHYYYY?!?! (like linear, exponential etc...)

2) How would resistance vary with depth of the electrodes and why???

Obviously this is electrochemical cell stuff... Erm, just so you know its CuSO4 with graphite electrodes... And any relevant equations would be very much appreciated.

If someone can help me I will be eternally grateful to you!!!

xxx
 
Last edited:
5
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Please help, I havent had any replies on student room or yahoo answers... I need this in by midnight!!!
 
5
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okay so this is what i've got so far:

1) Resistance increases linearly as distance apart increases. As the distance increases, there is further for the ions to travel from the anode to the cathode, meaning the resistance is greater. But I cant work out WHY it's linear?!? Is there an equation linking distance and resistance???

2)The graph I have looks like a perfect exponential, with resistance decreasing as the depth increases. I have got this so far:
Surface area of electrode increases linearly with depth
As The surface area increases, there are more ions travelling from the anode to the cathode (this is also linear?)
Then obviously as there are more ions, there is less resistance... but i have no idea why it becomes an exponential??

I hope you now know I dont want you to do this for me, and I HAVE already spent 3 hours trawling the internet trying to find out

thanks
 
5
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cmonnnn just a hint??? :cry:
 

gneill

Mentor
20,497
2,616
Take a look at the formula for the resistance of a wire given its length, cross sectional area, and inherent resistivity of the conductor. I think you can draw some parallels.
 
5
0
Okayyyyy,, so I've got it to

R = ρL/(πdL + πr^2)

Is that right? is that the direction I should be taking? cause it doesnt look it :confused:
 

gneill

Mentor
20,497
2,616
You don't need a circular cross section, just the cross sectional area will do.
$$R = \rho \frac{L}{A}$$
and in your case ##A = W \cdot d##, where W is the plate width and d is the depth.

This will naturally be an approximation given that the cell (tank) probably has a cross sectional area larger than the plate areas, so there will be some "fringe effects" for the current path (there will be some current outside the straight-line volume connecting the plate surfaces).
 

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