# eletrolytic cells

Tags: cells, eletrolytic
 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...
 Admin P: 21,701 Salts are listed because solutions are prepared by dissolving them, converting to ions should be a breeze.
 P: 31 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?
P: 21,701

## 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.
 P: 31 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?
P: 21,701
 Quote by OmniReader 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

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

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

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

when plugged into the Nernst equation we get

\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}

where

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

You can easily check that's working by comparing tabulated standard potentials and solubility products.
 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?
P: 21,701
 Quote by OmniReader 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.
 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?
P: 21,701
 Quote by OmniReader 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.
P: 31
 Quote by Borek 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.