Question

Given that the solubility constants of HgSO4 and PbSO4 are 6.6x10-7 and 1.6x10-8 respectively, determine the cell potential when the electrolyte is saturated with both
salts at 298 K.

Answer 1

The overall reaction is Pb_(s) + Hg₂SO_4(s) ------> PbSO_4(s)  + 2Hg_(I)

An electrochemical cell can be arranged by considering the two half reactions. By fact, that Oxidation reaction takes place in anode and the reduction reaction takes place at cathode. By using the reduction half reactions for the description of both anodic and cathodic reactions.

Cathodic reaction : Hg₂SO₄(s)  + 2e^- -------> 2Hg(l) + SO₄^2- (aq)

Anodic reaction : PbSO₄(s)  + 2e^- -------> Pb(s) + SO₄^2- (aq)

These reaction indicates the presence of aqueous solution which consists of SO₄^2-

As a result the electrochemical cell is

Pb(s) | PbSO₄(s) | H₂SO₄ (aq) || H₂SO₄ (aq) | Hg₂SO₄ (s) | Hg(l)

(or)

Pb(s) | PbSO₄(s) | H₂SO₄ (aq) | Hg₂SO₄ (s) | Hg(l)

As it is mentioned the electrolyte is saturated with both salts, the above two half reaction will be

Cathodic reaction : Hg₂²⁺ (aq) + 2e^- -------> 2Hg(l)

Anodic reaction : Pb²⁺ (aq) + 2e^- --------> Pb(s)

Cell reaction is

Hg₂²⁺(aq) + Pb(s) --------> 2Hg(l) + Pb²⁺(aq)

By applying Nernst equation

E = E - RT / 2F * ln (a _Pb²⁺  / a _Hg2²⁺)

Molalities of Hg₂²⁺ and Pb²⁺ can be estimated with their solubility constants by considering the mean activity coefficients as 1.

b(Hg₂²⁺) = 2.743 * 10^-3 mol kg^-1

b(Pb²⁺) = 1.265 * 10^-4 mol kg^-1

E = E(Hg₂²⁺ , Hg) - E(Pb²⁺, Pb) = 0.79 V - (-0.13) V = 0.92 V

As a result

E = 0.92 - (- 0.049) = 0.97V