Animated Solution for Chemistry - Electrochemistry: Given the data at $25^{\circ }\text{C}$, $Ag+I^{-}⟶AgI+e^{-}$; $E^{\circ }=0.152\text{ V}$ $Ag⟶A{g}^{+}+e^{-}$; $E^{\circ }=-0.8\text{ V}$ What is the value of $\log K_{sp}$ for $AgI$ ? $\left(\frac{2.303RT}{F}=0.059\text{ V}\right)$

Imagine a beaker containing solid silver iodide ($AgI$) at the bottom. A very tiny fraction of this solid dissolves into the water, establishing a dynamic equilibrium with its constituent ions:

This equilibrium is mathematically described by the solubility product constant, $K_{sp}$. But how do we calculate this purely from electrochemical data? The secret lies in the standard Gibbs free energy change ($\Delta G^{\circ }$). Thermodynamics tells us that $\Delta G^{\circ }=-2.303RT\log K_{sp}$, while electrochemistry states that $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$.

By equating these two fundamental laws, we derive a master equation that beautifully links the macroscopic voltage to the microscopic equilibrium:

Given that $\frac{2.303RT}{F}=0.059\text{ V}$ at $25^{\circ }\text{C}$, the equation simplifies to:

To use our master equation, we first need to find the standard cell potential ($E_{cell}^{\circ }$) for the exact $K_{sp}$ reaction. Let's look at the data provided:

1) $Ag+I^{-}⟶AgI+e^{-}{E}_{ox1}^{\circ }=0.152\text{ V}$ 2) $Ag⟶A{g}^{+}+e^{-}{E}_{ox2}^{\circ }=-0.8\text{ V}$

Notice that both are written as oxidation reactions (electrons are on the product side). However, our target $K_{sp}$ reaction requires solid $AgI$ on the reactant side. To achieve this, we must reverse the first reaction.

When we reverse an oxidation reaction, it becomes a reduction reaction, and the sign of its standard potential flips:

Reduction: $AgI+e^{-}⟶Ag+I^{-}{E}_{red1}^{\circ }=-0.152\text{ V}$ Oxidation: $Ag⟶A{g}^{+}+e^{-}{E}_{ox2}^{\circ }=-0.8\text{ V}$

Now, we add the two modified half-reactions together. The solid silver ($Ag$) and the electrons ($e^{-}$) cancel out perfectly on both sides, yielding our desired target reaction:

The standard cell potential for this overall reaction is simply the sum of the reduction and oxidation potentials:

With $E_{cell}^{\circ }$ in hand, and knowing that exactly one electron ($n=1$) was transferred during the process, we substitute these values back into our master equation:

Solving for $\log K_{sp}$:

What does this mean physically? A $\log K_{sp}$ of $-16.135$ means the actual $K_{sp}$ is $10^{-16.135}$. This incredibly tiny number confirms that silver iodide is highly insoluble in water. The negative cell potential ($-0.952\text{ V}$) also tells us the same story: the forward reaction (dissolving) is highly non-spontaneous!