I have encountered cell emf having a reactant and product in the subscript. But this is the first timeI have witnessed 3 elements/compounds/ ions. How do i interpret this and what would be the solution? If possible do give the reaction equation.enter image description here

EDIT:- This is what I tried:-

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I made out the first part of question but stuck at comparing whose EMF will be higher.


Half reactions

The species in the subscript describe the half reaction going on at the electrode. It contains all the species except for the electrons. You can figure out from the oxidation states whether species are on the reactant or product side. For example,

$$E_\ce{Ag/Ag+}:\ \ \ce{Ag+(aq) + e- -> Ag(s)}$$

Your examples are a bit more complicated. The first has species $\ce{Ag, AgCl, Cl-}$. The half reaction for that is:

$$\ce{AgCl(s) + e- -> Ag(s) + Cl-(aq)}$$

You should be able to figure out the one for bromide. The one with iodide is strange (not balanced). Iodide would react with AgCl:

$$\ce{AgCl + I-(aq) -> AgI + Cl-(aq)}$$

So then we could just write:

$$\ce{AgI(s) + e- -> Ag(s) + I-(aq)}$$

But in that case, it is not clear what the standard set of concentrations would be. Maybe a typo?

Cell potentials

The (silver/silver cation) standard potential is for silver cations at 1 M concentration (you would have to take a soluble silver salt like silver nitrate). The (silver/silver halide) standard potential is for the halide ion at 1 M concentration, resulting in a much lower concentration of silver cation. All you have to do is calculate that concentration, and get the (silver/silver cation) potential at that concentration from the standard potential.

For example for chloride, the silver cation activity of a saturated solution (with chloride activity of 1 in solution - I am using activities so the units work out) is:

$$a_\ce{Ag+} = \frac{K_\ce{AgCl}}{a_\ce{Cl-}} = K_\ce{AgCl} = \pu{1e-10}$$

For the half reaction $\ce{Ag+(aq) + e- -> Ag(s)}$, the potential is:

$$E^\circ_\ce{Ag/AgCl/Cl-} = E^\circ_\ce{Ag/Ag+} - \frac{RT}{zF} \ln{Q}$$

With Q = 1/$a_\ce{Ag+} = 1/K_\ce{AgCl}$, the potential comes out as:

$$ E^\circ_\ce{Ag/AgCl/Cl-} = E^\circ_\ce{Ag/Ag+} + \frac{RT}{zF} \ln{K_\ce{AgCl}} = E^\circ_\ce{Ag/Ag+} - \pu{0.59 V}$$

So the reduction potential is lower (by 0.59 V) because the concentration of silver cation in the presence of 1 M chloride is lower (by a factor of 10 billion). It is even lower for 1 M bromide, and the lowest for 1 M iodide because of the respective solubility products.

[OP] ... $E_3$ ...

In the OP's answer, the dissociation reaction is denoted by $E_3$, but it is not a redox reaction, so it should not have a reduction potential. You could solve this problem by calculating the free energy for each step (with a common second half reaction as a reference), and then going back to cell potentials. However, I think the path to a solution given here is easier to follow.

It is odd that under standard conditions, not all concentration are 1 M. Because the aqueous silver ions are not part of the half reaction for the saturated silver halides, however, this is just fine.

  • $\begingroup$ Thanks i now got the equation. But still I am stuck at comparing emfs as asked in question. I have added the try. $\endgroup$ – Aditya suresh Apr 14 '20 at 17:33

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