For the galvanic cell
$$\ce{Ag | AgCl(s), KCl (\pu{0.2 M}) || KBr (\pu{0.001 M}), AgBr(s) | Ag}$$
find the EMF generated given $K_\mathrm{sp}(\ce{AgCl}) = \pu{2.8e-10},$ $K_\mathrm{sp}(\ce{AgBr}) = \pu{3.3e-13}.$
This is the question from JEE exam (1992).
How to start solving the problem since the $E^°$ of individual half reactions is not given? How to write the $E^°$ for the cell without it? Or is it not needed?
How to start solving the problem since the $E^\circ$ of individual half reactions is not given?
This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).
AgCl(s) electrode $$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$ $$\ce{Ag+(aq) + e- -> Ag(s)} $$
AgBr(s) electrode $$\ce{Ag(s) -> Ag+(aq) + e-}$$
$$\ce{Ag+(aq) + Br-(aq) <=> AgBr(s)}$$
The standard reduction potentials will cancel out, i.e. $E^\circ (\mathrm{cell}) = 0$.
Further thoughts
[Comment by EdV] The Ag|AgCl electrodes I have are commercial, but making them in the lab is just a matter of oxidizing Ag wire in a chloride solution, so the electrode is Ag wire with an adherent coating of AgCl... I have never seen one of these made by just sticking an Ag wire in the Ag halide, but I guess it would work.
I added more to the answer prompted by that thoughtful comment.
[...my own comment] I pictured the silver electrode submerged in the solution, with the solid halide on the bottom. I am puzzled now too. Does it make a difference if the electrode touches the solid, the liquid, or both?
What are the actual reduction potentials?
$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−}\ \ \ \ E^\circ_\mathrm{red} = \pu{+0.22233 V}\tag{1}$$ $$\ce{AgBr(s) + e− <=> Ag(s) + Br−}\ \ \ \ E^\circ_\mathrm{red} = \pu{ +0.07133 V}\tag{2}$$ $$\ce{Ag+ + e− <=> Ag(s)}\ \ \ \ E^\circ_\mathrm{red} = \pu{ +0.7996 V}\tag{3}$$
Are they related?
If you subtract (3) from (1), you get the dissolution reaction of AgCl, if you subtract (3) from (2), you get the dissolution reaction of AgBr. Thus, standard reduction potentials for (1) and (3) should be different by
$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgCl})$$
and standard reduction potentials of (2) and (3) should be different by
$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgBr})$$
Finally, standard reduction potentials (1) and (2) should be different by
$$ -\frac{RT}{zF} \ln \frac{K_\mathrm{sp}(\ce{AgBr})}{K_\mathrm{sp}(\ce{AgCl})}$$
Numerical answer using half reaction (1) and (2)
$$\ce{AgCl(s) + Br-(aq) <=> AgBr(s) + Cl-(aq)}$$
$$Q = \frac{[\ce{Cl-}]}{[\ce{Br-}]} = 200$$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$= \pu{(0.22233 V− 0.07133 V) - 0.13612 V = 0.0149 V}$$
Numerical answer using half reaction (3) twice
$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$
"c" stands for chloride side, and "b" stands for bromide side. For consistency, I am using the following values for the solubility products (derived from difference of standard reduction potentials of half reactions (1), (2) and (3)).
$$K_\mathrm{sp}(\ce{AgCl}) = \pu{1.74e−10}$$ $$K_\mathrm{sp}(\ce{AgBr}) = \pu{4.89e−13}$$
$$[\ce{Ag+}]_c = K_\mathrm{sp}(\ce{AgCl}) / [\ce{Cl-}] = \pu{8,27e−10} $$ $$[\ce{Ag+}]_b = K_\mathrm{sp}(\ce{AgBr}) / [\ce{Br-}] = \pu{4.89e−10} $$
$$ Q = \frac{[\ce{Ag+}]_c}{[\ce{Ag+}]_b} = 0.560 $$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$\pu{= 0 - (-0.0149 V) = 0.0149 V}$$
Even though this question 1) has an answer with multiple upvotes (and I was the first upvote), 2) the OP has accepted the answer and 3) I have great respect for @Karsten Theis, having co-taught a quant class with him back in 2008 and knowing, first hand, that he is an excellent scientist and teacher, nonetheless, I have several problems with this trick exam question.
First, the $K_{sp}(AgCl)$ is about $1.8 x 10^{-10}$. This is the trivial problem, nothing more than a typo. More importantly, the $K_{sp}$ solution does not tell the whole story. So suppose T = 298.15K, i.e., standard temperature, n = 1 equivalent/mole, and all activity coefficients are assumed to be unity, so molar concentrations can be used in place of (tossing the molarity units) the unitless activities. Then, under standard state conditions, we have the following voltaic cell:
Thus AgCl(s) will be reduced to Ag(s) plus $Cl^-$ ions, on the right hand side (RHS) of fig. 1, and Ag(s) will be oxidized to AgBr(s), on the left hand side (LHS) of fig. 1.
But what happens if [$Cl^-$] = 0.2 M and [$Br^-$] = 0.001 M? Then both reduction potentials increase (become more positive), but the new Ag|AgCl reduction potential is still the most positive, so that is where reduction is spontaneous wrt (with respect to) the other half cell reaction reduction potential. This is shown is fig. 2 below:
In this figure, n = 1 equivalent/mole, T = 298.15K, R = 8.314472 J/(mole K), F = 96485.3383 C/equivalent, so RT/nF = 25.6926 mV and (ln10)•RT/nF = 59.1594 mV. Plugging in the numbers, as per the $E_{cell}$ equation in fig. 2, yields +0.0149V. If [$Br^-$] = 0.0001 M, then the cell potential would be -0.044V. This means the Ag|AgCl electrode would be the anode and the Ag|AgBr electrode would be the cathode. Note that the OP’s question actually showed the Ag|AgCl electrode as the anode, since this is the standard convention. But it wasn’t! So this was a rather nasty trick question, in my professional judgment.
