[Rutherford]

A solution of 0.01M SnCl₂ and 0.1M KCl was suspected to contain PbCl₂. The lead content was determined by electrodeposition using a Pt cathode and a Ag /AgCl anode. The electrolysis was stopped when the Sn just started to deposit on the cathode (the shiny metallic lustre of Sn makes it visually possible). Calculate the final Pb²⁺concentration in the solution.

I got the idea that the EMF of Ag/Ag⁺//Sn²⁺(0.02M)/Sn should be the same as the EMF of Ag/Ag⁺//Pb²⁺(x)/Pb and it gives the correct answer, but I have no good explanation for this prediction. How to explain the validness of this prediction?

[Borek]

What reacts first? And why?

[Rutherford]

First happens the reaction with the biggest EMF, but that depends on concentration. Sn²⁺ has a bigger concentration than Pb²⁺, while their red-ox potentials are -0.14 and -0.13 V respectively. How to know whether the concentration of Sn²⁺ is bigger enough, so it reacts first or not?

[Borek]

How does the potential depend on the concentration?

[Rutherford]

I know that, it's expressed by the Nernst equation.

[Borek]

So you know everything you need.

[Rutherford]

I need some explanation, why my assumption is correct. By some logic Pb²⁺ should react, first and after the first sign of Sn deposition appears, the electrolysis should be stopped (something like obtaining the end point of a volumetric titration). If Pb²⁺ reacts first then the EMF of its reaction with silver should be bigger than the EMF reaction of Sn²⁺ with silver (which is -1V). But how much bigger? And if I assume that it is equal I get a close answer.

[Big-Daddy]

You said it yourself - "First happens the reaction with the biggest EMF, but that depends on concentration". So let's rephrase - we will approximate that "the reaction with the biggest EMF is the one driven forward by the electrolysis". To be precise further, the reduction with the most positive reduction EMF and the oxidation with the most positive oxidation EMF i.e. most negative reduction EMF will be that for the reaction driven forward by electrolysis.

So, my surmise is that the lead ion concentration was once of such value that the EMF for the reaction involving the lead couple was more positive than that of the tin couple, but once the lead ion concentration has dropped sufficiently it will no longer be the more positive EMF reaction - then, the tin couple will be the more positive EMF reaction. The point at which this instantaneous change is occurs is noted to be as soon as the tin couple reaction becomes the one with more positive EMF, i.e. the two couples will have the same EMF at that point.

[Rutherford]

So, my surmise is that the lead ion concentration was once of such value that the EMF for the reaction involving the lead couple was more positive than that of the tin couple

So what's your suggestion of its calculation? When the EMF become equal, some amount of Pb²⁺ was already electrolysed.

[Borek]

At what potential does the Sn²⁺ start to deposit? What is concentration of Pb²⁺ at this moment?

[Rutherford]

It is -0.19V for the Sn²⁺ reduction. For this potential I get a concentration of Pb²⁺ to be 9.3·10^-3M. So how to determine the starting concentration?

[Borek]

So how to determine the starting concentration?

What for? You were asked about the final concentration, and you just gave it.

[Rutherford]

I see now that the starting concentration can't be calculated. Thank you for the help.

[Big-Daddy]

Only a range of possible values (any that allow the EMF for the lead reaction to be more positive than the initial EMF for the tin reaction) can be found. For anymore, you need to have information about charge passed and current efficiency (and then back-track using the final concentration).