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Relation between $E^\circ_{cell}$ and $K_{sp}$

Connecting Thermodynamics, Nernst Equation, and Solubility Equilibrium.

By chemca Team • Updated Jan 2026

The Solubility Product ($K_{sp}$) of a sparingly soluble salt can be calculated using electrochemical data. By constructing a suitable galvanic cell where the dissolution process represents the cell reaction, we can derive a direct relationship between the standard cell potential ($E^\circ_{cell}$) and $K_{sp}$.

1. The Thermodynamic Connection

Free Energy Link

We know two fundamental relationships involving the standard Gibbs Free Energy change ($\Delta G^\circ$):

1. From Electrochemistry: $\Delta G^\circ = -nFE^\circ_{cell}$

2. From Equilibrium Thermodynamics: $\Delta G^\circ = -2.303 RT \log K_{eq}$

For a solubility equilibrium, the equilibrium constant $K_{eq}$ is equal to the Solubility Product $K_{sp}$.

Equating the two expressions for $\Delta G^\circ$:

$$ -nFE^\circ_{cell} = -2.303 RT \log K_{sp} $$

2. The Derivation

Standard Equation

Rearranging the equation derived above:

$$ E^\circ_{cell} = \frac{2.303 RT}{nF} \log K_{sp} $$

At Standard Temperature (298 K):

• $R$ (Gas constant) = 8.314 J/K·mol
• $T$ = 298 K
• $F$ (Faraday's constant) = 96487 C/mol

The term $\frac{2.303 RT}{F}$ reduces to approximately 0.0591 V.

$$ E^\circ_{cell} = \frac{0.0591}{n} \log K_{sp} \quad (\text{at } 298 \text{ K}) $$

Or: $\log K_{sp} = \frac{n E^\circ_{cell}}{0.0591}$

3. Example: Silver Chloride

Calculating $K_{sp}$ of $AgCl$

Consider the cell constructed to measure the solubility of AgCl:

Anode (Oxidation): $Ag(s) \rightarrow Ag^+(aq) + e^-$

Cathode (Reduction): $AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq)$

Net Cell Reaction:

$$ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) $$

This is exactly the solubility equilibrium reaction!

Calculation:

Given standard potentials:

• $E^\circ_{Ag^+/Ag} = 0.80 \, V$
• $E^\circ_{Cl^-/AgCl/Ag} = 0.22 \, V$

$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.22 - 0.80 = -0.58 \, V $$

Now, substitute into the formula ($n=1$):

$$ \log K_{sp} = \frac{1 \times (-0.58)}{0.0591} \approx -9.8 $$ $$ K_{sp} = 10^{-9.8} \approx 1.6 \times 10^{-10} $$

Important Note on Signs:
Usually, $E^\circ_{cell}$ for a solubility equilibrium comes out negative because dissolution of sparingly soluble salts is non-spontaneous at standard conditions (1 M concentrations). This results in a very small $K_{sp}$ ($< 1$).

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