eletrolytic cells


by OmniReader
Tags: cells, eletrolytic
OmniReader
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#1
Oct23-13, 05:12 AM
P: 31
Please tell me how to deduce reliably from galvanic cell representation, the half-cell reactions, for any cell, or vice versa.

Pt(s) | H2(g) | HCl(aq) | Cl2(g) | Pt(s)
Pb(s) | PbCl2(s) | HCl(aq) | H2(g) | Pt(s)
Pb(s) | PbSO4(s) | K2SO4(aq) || KNO3(aq) || KCl(aq) | PbCl2(s) | Pb(s)

have seen the basic guidelines on how to write cells knowing the reaction, but they do not explain why in these representations are so many salts rather than ions. nor is it clear how to approach deducing half-cell reactions from the cells above...
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Borek
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#2
Oct23-13, 06:47 AM
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Salts are listed because solutions are prepared by dissolving them, converting to ions should be a breeze.
OmniReader
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#3
Oct24-13, 09:55 AM
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and for salts who don't dissociate/dissolve, AKA with solid state symbol (solubility rules tells us), e.g. Hg2Cl2, PbSO4, PbCl2, list goes on including from examples used. what reactions for the solid salts, or those like Hg2Cl2 that dissolve but not dissociate?

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#4
Oct24-13, 11:12 AM
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eletrolytic cells


They are dissolved as well, just the concentrations are very low.

Say, you have a silver wire covered with solid AgCl and in contact with KCl solution (Ag(s) | AgCl(s) | KCl(aq)). The real redox system is Ag/Ag+, but concentration of Ag+ is limited by solubility product of AgCl.
OmniReader
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#5
Oct24-13, 05:41 PM
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I thought reaction would be AgCl(s) + e- = Cl-(aq) + Ag(s). So if "solid"(s) salt (or known for low solubility) is there we must use Ksp equation?
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#6
Oct25-13, 02:22 AM
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Quote Quote by OmniReader View Post
I thought reaction would be AgCl(s) + e- = Cl-(aq) + Ag(s).
You won't be able to experimentally differentiate between

AgCl(s) → Ag+(aq) + Cl-(aq) → Ag(s) + Cl-(aq)

and

AgCl(s) → Ag(s) + Cl-(aq)

(e- omitted for brevity)

So if "solid"(s) salt (or known for low solubility) is there we must use Ksp equation?
Yes, Ksp is an important element of the whole picture.

Silver electrode in solution of silver ions is just a concentration cell, with potential given by

[tex]E = E_{0Ag/Ag^+} + \frac {RT}{nF} \ln [Ag^+][/tex]

if the electrode is covered with AgCl(s) concentration of Ag+ near the electrode surface can be calculated from Ksp:

[tex][Ag^+] = \frac {K_{sp}}{[Cl^-]}[/tex]

when plugged into the Nernst equation we get

[tex]\begin{align}E &= E_{0Ag/Ag^+} + \frac {RT}{nF} \ln \frac {K_{sp}}{[Cl^-]}\\

&= E_{0Ag/Ag^+} + \frac {RT}{nF} \ln {K_{sp}} - \frac {RT}{nF} \ln {[Cl^-]}\\

&= E_{0Ag/AgCl} - \frac {RT}{nF} \ln {[Cl^-]}
\end{align}[/tex]

where

[tex]E_{0Ag/AgCl} = E_{0Ag/Ag^+} + \frac {RT}{nF} \ln {K_{sp}}[/tex]

You can easily check that's working by comparing tabulated standard potentials and solubility products.
OmniReader
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#7
Oct26-13, 03:33 AM
P: 31
ok so for my last example is

Pb(s) | PbSO4(s) | K2SO4(aq) || KNO3(aq) || KCl(aq) | PbCl2(s) | Pb(s)

the reactions will be PbSO4 (s) + 2e- = Pb(s) + SO4^2-(aq), PbCl2(s) + 2e- = Pb(s) + 2Cl-(aq), overall. for example Fe (s) | Fe(OH)3(s) | NaOH (aq) reaction will be Fe(OH)3(s) + 3e- = Fe(s) + 3OH-(aq).

purpose of all this is measure equilibrium constants/concentrations for redox reactions, so same reactions should occur without electrodes. how is H2 evolved from protons with electrode but not without?
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#8
Oct26-13, 04:01 AM
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Quote Quote by OmniReader View Post
how is H2 evolved from protons with electrode but not without?
Not sure what the question is, but

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

comes to mind.
OmniReader
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#9
Oct26-13, 05:13 AM
P: 31
now I understand. redox reaction occurs without electrode too, only purpose of electrode is to measure K/conc. can we apply quantitative equilibrium method to solutions w/redoxreactions?
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#10
Oct26-13, 07:19 AM
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Quote Quote by OmniReader View Post
now I understand. redox reaction occurs without electrode too, only purpose of electrode is to measure K/conc
Not the only purpose. By separating half cells you can force the charge to flow through an external circuit and do some work - that's how batteries are made.

can we apply quantitative equilibrium method to solutions w/redoxreactions?
Sure, why not. Although in most cases equilibrium lies far to one side.
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#11
Oct26-13, 05:51 PM
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Quote Quote by Borek View Post
Not the only purpose. By separating half cells you can force the charge to flow through an external circuit and do some work - that's how batteries are made.
I didnot know. this Please provide links w/detailed info& calculations.
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#12
Oct27-13, 02:59 AM
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Quote Quote by OmniReader View Post
Please provide links w/detailed info& calculations.
http://lmgtfy.com/?q=battery+chemistry
OmniReader
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#13
Oct27-13, 09:05 AM
P: 31
please specify calculations: what quantity current will be drawn by cell . Or constant-current transfers what quantity of mass. is irreversibility approximation made? for approximation "only 1reduction and 1 oxidation reaction", how is it chosen when solution has many redox reactions?
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#14
Oct27-13, 10:36 AM
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You do know Faraday's laws of electrolysis? They combine mass and charge.

Actually even if you don't know them, you can easily answer your own question. Just think what the current is, and how many electrons pass through the circuit (hint: electrons can be counted, just like atoms or molecules, mole of electrons expressed in Coulombs is an important proportionality constant).
OmniReader
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#15
Oct27-13, 01:22 PM
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Quote Quote by Borek View Post
Actually even if you don't know them, you can easily answer your own question. Just think what the current is, and how many electrons pass through the circuit (hint: electrons can be counted, just like atoms or molecules, mole of electrons expressed in Coulombs is an important proportionality constant).
constant=F. I googled Faraday's laws.question is: what about situation in which are few possible redox reactions, e.g. aqueous solution. Which is chosen at cathode and anode respectively? same question for EMF in fact ...
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#16
Oct27-13, 01:53 PM
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Sorry, I am not going to write a book for you, and apparently that's what you are looking for.
OmniReader
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#17
Oct27-13, 02:22 PM
P: 31
have read books, none cover the issue. please sugest good title or provide link.

I maybe know why. emf reading is taken for K or conc measurement, not needed to do so in cases when different redox are possible. instead conc is made small for non-main redox and then discarded so "only main" contributes emf. reactions in solution at equilibrium, just use K values. also for SHE or oxygen electrodes common approximation is to make unity of partial pressures. resultantly we find K or conc values by single emf reading, and more complicated solutions, can apply equilibrium approximation. did I understand the books correctly?


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