[greentea11]

Following cell has been used to determine ka of ammonium ion

Pt (m) | H₂(g) (1 bar) |NH₃ (0.0212M), NH₄Cl (aq) (0.0212M) | AgCl (s), Ag (m)

Identify half reactions and give conventional cell reaction, write down Nernst equation for this process, calculate ka for conditions above

So first half equation is

AgCl (s) + e^-  Ag (s) + Cl^-

But unsure what the other half equation is? using H₂ NH₃ and NH₄Cl

ive got stuck at first hurdle so appreciate any help

[Borek]

You have hydrogen electrode on the left, you just need to find pH.

[Guvanch]

I think second half reaction is the oxidation of hydrogen. From that
you can find H ion concentration. Than it is equal to Ka

[Big-Daddy]

I think you need to know values for the Nernst equation, i.e. EMF and standard electrode potentials. Also K_sp for AgCl (could be within standard electrode potential) ... not to mention possibility for Ag(NH₃)²⁺ to form with very high formation constant. Am I missing something here? Even if we say that Ag(NH₃)²⁺ does not form and so the rest is as said above, I think you still need to use the common ion effect to determine [Cl^-] for use in the Nernst equation (and need EMF and E° for the Nernst equation).

[greentea11]

Sorry left out bits accidentally
the entire question

The following cell has been used to determine the ka of the ammonium ion
Pt (m) | H2(g) (1 bar) |NH3 (0.0212M), NH4Cl (aq) (0.0212M) | AgCl (s), Ag (m)

The measured emf of this cell is 0.8759v at 298k when cell is prepared using the above concentrations

E°cell Ag/AgCl = 0.2223V

1. identify the half reactions and give conventional cell reaction

2. write down nernst equation for this process

3. give an expression for Ka and calculate its conditions under the above conditions
HINT: use Debye Huckel theory to estimate γ Nh4+ and γ Cl-)

[Borek]

What I wrote still holds. NH₃/NH₄⁺ is a buffer solution, which makes calculations even easier.

[Big-Daddy]

What I wrote still holds. NH₃/NH₄⁺ is a buffer solution, which makes calculations even easier.

So we say that [NH₄⁺] = [NH₃] = [Cl^-] = 0.0212 M.

If so then the problem is easy from there, but how do you say so with confidence? AgCl seems to be contributing to Cl^- on top of the original 0.0212 M. Or does the notation mean that the concentrations can be assumed to be somehow fixed at the given values (rather than being initial concentrations necessarily)?

Edit: I got an answer of K_a that is a factor of about 25% off from the expected value given the known value of K_b for NH₃. Is this just lack of activity corrections in my approach (question did ask for them after all)?

[mjc123]

Well, the solubility product of AgCl is ca. 10^-10, so if you have 0.02M Cl^- already, only about 10^-8M of AgCl will dissolve - a negligible change to [Cl^-].

[Big-Daddy]

Well, the solubility product of AgCl is ca. 10^-10, so if you have 0.02M Cl^- already, only about 10^-8M of AgCl will dissolve - a negligible change to [Cl^-].

But I think it isn't the solubility of AgCl we have to worry about but rather the overall cell reaction, which produces Cl^- with E°(cell)=+0.2223 (K = 3.31*10⁷ for n=2 at 298 K)?

Or perhaps this reaction is really slow (seems to be a trend with reactions involving the H⁺/H₂ couple - come to think of it, this is also why it is ok to measure the EMF without using a salt bridge  - but my trouble is, we often see this couple being perfectly fast, e.g. if it is paired with the Na/Na⁺ couple, so this doesn't seem easy to explain) so is not taken to reach (or even significantly approach) its equilibrium, so we only have to look at AgCl dissolution (which reaches equilibrium) for what it adds to the original chloride concentration?

[mjc123]

Oh I get it, you're thinking of the reaction AgCl + e  Ag + Cl^-. Well, as long as the circuit is open there is no (or negligible) reaction, only the potential for reaction (which gives rise to the EMF). The statement of the question implies that you are not performing an electrolysis, to generate material, but measuring the cell potential in order to determine a physical parameter (Ka). For this you would measure the OCV with a high resistance voltmeter drawing minimal current. You don't want significant reaction to occur, as this would (as you say) change the concentrations and the cell potential and complicate your parameter determination.

[Borek]

it isn't the solubility of AgCl we have to worry about but rather the overall cell reaction, which produces Cl^-

That's not the way it works.

You are not reacting anything, you are measuring potential difference. Current that flows depends on the voltmeter resistance, but is in the μA range in the worst case. How much Cl^- can it produce?

Edit: beaten by mjc, we posted exactly the same. Happens.