Concept Explanation: Relating Solubility Product (Ksp) to Standard Electrode Potentials

This problem requires finding the value of pX/3 (where pX = -log X and X is Ksp for Hg₂Cl₂) using given standard electrode potentials. The key is to recognize that the difference in standard electrode potentials relates to the equilibrium constant for the dissolution reaction, which is Ksp.

Step-by-Step Solution:

Step 1: Write the dissolution reaction and its relation to Ksp

The solubility product for Hg₂Cl₂(s) corresponds to the reaction:

${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)\rightleftharpoons {\mathrm{Hg}}_2^{2+}\left(aq\right)+2{\mathrm{Cl}}^{-}\left(aq\right)$

Therefore, $K_{sp}=\left[{\mathrm{Hg}}_2^{2+}\right][{\mathrm{Cl}}^{-}{]}^2=X$

Step 2: Identify the relevant half-cell reactions

We are given two half-reactions:

1. ${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)+2e^{-}\to 2\mathrm{Hg}\left(l\right)+2{\mathrm{Cl}}^{-}$;   $E^{\circ }=0.27\text{ V}$

2. ${\mathrm{Hg}}_2^{2+}+2e^{-}\to 2\mathrm{Hg}\left(l\right)$;   $E^{\circ }=0.81\text{ V}$

Step 3: Construct the cell reaction that represents the dissolution

The dissolution reaction can be obtained by reversing the second half-reaction and adding it to the first:

First half-reaction (reduction): ${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)+2e^{-}\to 2\mathrm{Hg}\left(l\right)+2{\mathrm{Cl}}^{-}$

Second half-reaction (oxidation, reversed): $2\mathrm{Hg}\left(l\right)\to {\mathrm{Hg}}_2^{2+}+2e^{-}$

Adding them gives the net cell reaction: ${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)\rightleftharpoons {\mathrm{Hg}}_2^{2+}+2{\mathrm{Cl}}^{-}$

This is exactly the dissolution reaction for which the equilibrium constant is Ksp.

Step 4: Relate the standard cell potential to the equilibrium constant (Ksp)

The standard cell potential for this reaction is:

$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.27\text{ V}-0.81\text{ V}=-0.54\text{ V}$

The relationship between standard cell potential and the equilibrium constant K is given by the Nernst equation at equilibrium:

$0=E_{cell}^{\circ }-\frac{RT}{nF}\ln K$

Rearranging for ln K:

$\ln K=\frac{nF{E}_{cell}^{\circ }}{RT}$

Here, K = Ksp = X, n = 2 (number of electrons transferred), and at 298 K, RT/F = 0.059 V (approximately).

Therefore, the formula becomes:

$\log X=\frac{n{E}_{cell}^{\circ }}{0.059}=\frac{2\times (-0.54)}{0.059}$

Step 5: Calculate log X and then pX

First, calculate log X:

$\log X=\frac{-1.08}{0.059}\approx -18.305$

By definition, pX = -log X. Therefore:

$pX=-(-18.305)=18.305$

Step 6: Find the value of pX/3

$\frac{pX}{3}=\frac{18.305}{3}\approx 6.1017$

Final Answer: The value of pX/3 is approximately $6.10$.

Related Topics & Formulae:

1. Solubility Product Constant (K_sp):

For a general dissolution reaction: $A_a{B}_b\left(s\right)\rightleftharpoons a{A}^{n+}+b{\mathrm{B</mi}}^{}$