Why is Electrochemistry often the method of choice for determining Ksp values? [closed]








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I don't understand the relationship they have with one another. Please, someone explain.




Perhaps a simple example will illustrate how a solubility product constant, $\ce{K\_{sp}}$, may be determined via measurement of an appropriate voltaic cell potential. Consider the estimation of the $\ce{K\_{sp}}$ value for the following equilibrium:


$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq) \quad K = K\_{sp}} \tag{1}$$


This equilibrium lies far to the left because AgCl has very poor solubility in water and, as a direct consequence, the concentrations of silver ions and chloride ions are very low. Therefore, the $\ce{K\_{sp}}$ value is small, i.e., much less than unity, and a direct measurement of the $\ce{K\_{sp}}$ value is difficult due to the low solubility of AgCl.


However, we have the following two half-reactions, with their standard reduction potentials (from standard tables):


$$\ce{Ag+ +  e− <=> Ag(s)} \quad E^\circ\_\mathrm{red} = \pu{+0.7996 V} \tag{2}$$


$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−} \quad E^\circ\_\mathrm{red} = \pu{+0.22233 V} \tag{3}$$


Half-reaction (2) has the more positive standard reduction potential, hence more negative $\ce{\Delta G}$, so it will be the spontaneous reduction half-reaction at the cathode in a voltaic cell that can be constructed. Reversing half-reaction (3) yields the oxidation half-reaction at the anode in the constructed voltaic cell:


$$\ce{Ag(s) + Cl− <=> AgCl(s) + e−} \tag{4}$$


So the voltaic cell notation is 


$$\ce{Ag(s)|Cl^-(aq;\pu{1 M})||Ag^+(aq;\pu{1 M})|Ag(s)}$$


and the net cell reaction is obtained by adding half-reactions (2) and (4), resulting in


$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s) \quad K = 1/K\_{sp}} \tag{5}$$


This equilibrium lies far to the right because it is simply the reverse of half-reaction (1). Therefore, the $\ce{K}$ value is large. Indeed, it is simply the reciprocal of half-reaction (1)'s $\ce{K\_{sp}}$ value. This is consistent with the fact that mixing aqueous solutions containing silver ions and chloride ions immediately results in spontaneous formation of AgCl solid that precipitates out of solution.


The **Nernst equation**, where $\ce{Q}$ is the reaction quotient, is


$$ {E\_\mathrm{cell} = E^\circ\_\mathrm{cell} - \frac{R T}{n F} \ln(Q) } \tag{6}$$


The natural logarithm term is basically a correction term that takes into account the fact that voltaic cells are usually not at standard state, i.e., *they are not dead yet*. This is accomplished using $\ce{Q}$. At standard state, $\ce{Q} = 1$, and at equilibrium, $\ce{Q} = K$, where $\ce{K}$ is the equilibrium constant. In the present case, $\ce{K}$ pertains to equation (5).


Now, at equilibrium (and ignoring units), $\ce{\Delta G\_{cell} = 0, E\_{cell} = 0 and Q = K}$, so 


$$E^\circ\_\mathrm{cell} = \frac{R T}{n F} \ln(K) = {E^\circ\_\mathrm{cathode} - E^\circ\_\mathrm{anode} = \pu{+0.7996 V} -(\pu{+0.22233 V}) = \pu{+0.5773 V}} \tag{7}$$ 


Note that the standard cell potential is *defined* as


$$E^\circ\_\mathrm{cell} = E^\circ\_\mathrm{cathode} - E^\circ\_\mathrm{anode} \tag{8}$$


With $\pu{T = 298.15 K}$ and $\pu{n = 1}$, $\pu{RT/nF \approx 0.02569 V}$, so equation (7) can be solved for $\ce{K}$:


$$\mathrm{K = e^{nfE^\circ \_{cell} /RT} \approx e^{+0.5773V/+0.02569V} = 5.74 \times 10^{9} \tag{9}}$$


Hence $$\pu{K\_{sp} = 1/K = 1.74 \times 10^{-10}} \tag{10}$$
which is close to typically quoted values in the literature.


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The standard half cell potential, $\mathrm{E^0\_{red}}$, is for a 1.000 molar solution for the ion being reduced. For other concentrations the [Nernst Equation](https://en.wikipedia.org/wiki/Nernst_equation) gives the half-cell potential, $\mathrm{E\_{red}}$.
$$\mathrm{E\_{red}} = \mathrm{E^0\_{red}} - \dfrac{\mathrm{RT}}{\mathrm{zF}}\ln{\dfrac{a\_{\mathrm{Red}}}{a\_\mathrm{Ox}}}$$


Note that the Nernst equation should use [activities](https://en.wikipedia.org/wiki/Thermodynamic_activity), $a$, but for low quantities of a solute the concentrations are often be used. 


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