# Cyclovoltamogram of silver in solutions of its insoluble salts

Hello I was dealing with the following question in my electrochemistry course:

Sketch cyclic voltammograms for the silver oxidation and reduction in different electrolytes. Assume electrochemical reversibility and a concentration of 50 mM of either one of the respective salt in the solutions. The solubility products of the different silver salts in are: $$K_{sp}(AgBr)=5.4\cdot 10^{-13}$$, $$K_{sp}(AgCl)=1.6\cdot 10^{-10}$$, $$K_{sp}(AgI)=8.5\cdot 10^{-17}$$, $$K_{sp}(AgSCN)=1.0\cdot 10^{-12}$$. Sketch for (a) bromide solution, (b) chloride solution, (c) iodide solution, (d) thiocyanate solution. Hint: The concentration of silver ions in the Nernst equation can be substituted with the rearranged equation of the solubility product.

So, to tackle this problem, I thought about two things. First, how to derive the expression for the Nernst equation in this situation, and I assume it is largely how the Ag/AgCl-formal potential is derived:

$$E(Ag/Ag^{+})=E^{0}(Ag)+\frac{RT}{F} \cdot ln(a(Ag^{+}))$$ (1)

But as the silver cation potential is controlled by the $$AgX$$ salt solubility product, $$K_{sp}$$, we get:

$$E(Ag/AgX)=E^{0}(Ag)+\frac{RT}{F} \cdot ln \left (\frac{K_{sp}}{a(X^{-})} \right )$$ (2)

Then we seperate the logarithmic terms:

$$E(Ag/AgX)=E^{0}(Ag)+\frac{RT}{F} \cdot ln (K_{sp}) - \frac{RT}{F} \cdot ln (a(X^{-}))$$ (3)

As the solubility product and $$E^{0}(Ag)$$ are constant, we define those two as $$E^{0}(Ag/AgX)$$ and get:

$$E(Ag/AgX)=E^{0}(Ag/AgX) - \frac{RT}{F} \cdot ln (a(X^{-}))$$ (4)

So, as we have an expression for the $$E^{0}(Ag/AgX)$$, and the reaction is reversible, we know that the peak potentials of the cyclic voltammograms need to be at this value, which can be calculated from the solubility products using equation (3) (the peaks will be seperated by 57 mV from this value, $$\frac{E_{pc}+E_{pa}}{2}$$).

However, I am unsure how to draw the cyclic voltammogram, for you see, the potential actually increases when $$Ag^{+}$$ is formed due to oxidation, because the $$X^{-}$$ crashes out of solution because of the solubility product, in contrast to e.g the $$Fc/Fc^{+}$$ couple. It just increases and never reaches $$E^{0}(Ag/AgX)$$ for the peak in its CV diagram (eq. 4)

In the $$Fc/Fc^{+}$$ redox couple, as $$Fc$$ is oxidized to $$Fc^{+}$$, its potential increases until the ratio (Q) of $$Fc/Fc^{+}$$ is equal to one (take a look at eq. 5) and $$E^{0} (Fc/Fc^{+})$$ is reached for the peak of the CV-curve.

$$E(Fc/Fc^{+})=E^{0}(Fc/Fc^{+}) - \frac{RT}{F} \cdot ln (\frac{a(Fc)}{a(Fc^{+})})$$ So how can I draw the CV here?