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Topic: Voltage drop  (Read 5720 times)

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Offline HouseMD

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Voltage drop
« on: August 10, 2009, 11:54:43 AM »
I have to answer the following problem, but I have no idea, why the voltage is so slight, because I would expect a voltage of 0,36V:

The problem:
If you combine two half cells:
electrode 1= zinc
electrode 2 = cadmium
electrolyt =  ZnCl2
Why can you measure only 0,2V?

Are there any effects why there is a voltage drop?

Sorry about my english, but I am from France  ;)

HouseMD
« Last Edit: August 10, 2009, 12:17:38 PM by HouseMD »

Offline zxt

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Re: Voltage drop
« Reply #1 on: August 10, 2009, 01:12:48 PM »
Is ZnCl2 the only electrolyte? ZnCl2 is for Zn to form the half electrode reaction:Zn2+/Zn and What's the electrolyte for cadmium? CdCl2 or CdCl or Cd positive ion combined with any other anions? Whether is it under standard condition or not? Without these, I cannot help any more.

Offline HouseMD

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Re
« Reply #2 on: August 10, 2009, 01:23:38 PM »
ZnCl2(1 mol/l) is the only one and yes it is under standard conditions! I think there is something special about this combination, but I don't know what it is, because there is also written in the problem definition: if you use Cu,Ag or Fe instead of cadmium, there is no voltage drop(if the conditions and the electrolyte are the same), so there has to be something special about the combination of zinc with cadmium.

Offline zxt

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Re: Voltage drop
« Reply #3 on: August 10, 2009, 01:41:53 PM »
I don't understand how this cell runs. One half-reaction is Zn2+ + 2e :rarrow:Zn, and what's the other half-reaction? Zinc and Cadmium cannot form a half electrode reaction.

Offline HouseMD

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Re: Voltage drop
« Reply #4 on: August 10, 2009, 02:19:49 PM »
I have enclosed the whole excercise  ::)

Offline zxt

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Re: Voltage drop
« Reply #5 on: August 10, 2009, 02:52:03 PM »
I think it has something to do with Zn(OH)2's Ksp.Zn(OH)2 ::equil::Zn2+ + 2OH-,then Ksp=[Zn2+]*[OH-]2. According to Ksp definition equation, when in acidic environment, [OH-] decreases and [Zn2+] increases, because Ksp is constant under certain temp. So in basic environment, vise versa. Then According to Nerrnst equation, the answer is there.

Offline zxt

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Re: Voltage drop
« Reply #6 on: August 10, 2009, 02:53:03 PM »
Mistake in quoting my words above which has been deleted

Offline zxt

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Re: Re
« Reply #7 on: August 10, 2009, 03:03:41 PM »
ZnCl2(1 mol/l) is the only one and yes it is under standard conditions! I think there is something special about this combination, but I don't know what it is, because there is also written in the problem definition: if you use Cu,Ag or Fe instead of cadmium, there is no voltage drop(if the conditions and the electrolyte are the same), so there has to be something special about the combination of zinc with cadmium.

I can see voltage drop by the other electrodes(excepting Fe). Here "between" refers to all electrodes(including Cd and Zn) but not just Cd and Zn

Offline HouseMD

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Re: Voltage drop
« Reply #8 on: August 10, 2009, 03:41:36 PM »
Yes, you're right, mhmm, why haven't I noticed this earlier? :-[ well, but the problem stays the same, why is there a voltage drop?  ???

Offline zxt

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Re: Voltage drop
« Reply #9 on: August 11, 2009, 05:00:24 AM »
According to Ammeter, the Zn electrode is cathode while M is anode, so  :delta:E=EMn+/M-EZn2+/Zn.
EZn2+/Zn=Eθ+(0.0592/2)lg[Zn2+]. Zn2+ molarity in acidic environment is higher than that in basic one due to hydrolysis intensity of ZnCl2 :Zn2+ + 2H2O ::equil::Zn(OH)2 + 2H+ compared with ionization degree as Zn(OH)42- ::equil::Zn2+ + 4OH-, so looking back to Arrnest equation, EZn2+/Zn in acidic environment is higher than that in basic environment. I think EMn+/M doesn't change though I still don't know how it runs, then  :delta:E will drop in acidic condition excepting when anode is Fe and why it is abnormal for Fe? I don't know. Moreover, Ksp should be considered, while solution's pH is basic, the number of Zn2+ will decrease with pH increaseing.
« Last Edit: August 11, 2009, 05:28:16 AM by zxt »

Offline HouseMD

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Re: Voltage drop
« Reply #10 on: August 11, 2009, 06:35:28 AM »
 ;) As I already said, I guess the answer is rather an effect than an simple explanation, but thanks. Our teacher just said:Hey, I wonder if somebody finds out why there is a voltage drop. So it has to be something very specific. I hope somebody has an idea.

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