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

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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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  • $\begingroup$ I think I am going to need simpler words. I don't think I understand, sorry. $\endgroup$ – Ofelia Miranda Mar 19 at 0:25
  • $\begingroup$ If either answer is acceptable or helpful to you, please consider giving it the green checkmark. It encourages people to put some thought and time into crafting answers that are correct, understandable and helpful to future questioners. Thanks for considering this! $\endgroup$ – Ed V May 25 at 23:34
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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 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, $a$, but for low quantities of a solute the concentrations are often be used.

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