Question

Calculate the cell potential and the equilibrium constant for the following reaction at 298 K:

I₂(s) + Hg(l) = 2I^-(aq) + Hg²⁺(aq)

Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.

Equilibrium constant: -------

Answer 1

I₂ is reduced to I^- while Hg is oxidized to Hg²⁺. The standard half cell reactions and the standard reduction potentials are listed below.

I₂ (s) + 2 e^- -------> 2 I^- (aq); E⁰₁ = +0.535 V …..(1)

Hg²⁺ (aq) + 2 e^- ------> Hg (l); E⁰₂ = +0.855 V….(2)

The listed values are obtained from the following internet source (http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm).

The given cell reaction is obtained by subtracting (2) from (1). The standard cell potential is given as

E⁰_cell = E⁰₁ – E⁰₂ = (+0.535 V) – (+0.855 V) = -0.32 V

The given reaction takes place at 298 K (standard conditions) and hence we can write the standard free energy change of the reaction as

ΔG⁰ = -n*F*E⁰_cell where n = number of moles of electrons transferred; F = 1 Faraday of electricity = 96485 J/V.mol

ΔG⁰ = -(2 mole)*(965485 J/V.mol)*(-0.32 V) = +61750.4 J = (617504 J)*(1 kJ/1000 J) = 61.7504 kJ

The equilibrium constant is given as

ΔG⁰ = -R*T*ln K_eq

===> ln K_eq = -ΔG⁰/RT = -(61750.4 J)/(8.314 J/mol.K)(298 K) = -24.923756

===> K_eq = e^(-24.923756) = 1.49882*10^-11 ≈ 1.500*10^-11 (ans).