Question

Use the appropriate standard reduction potentials to calculate the value of K at 25.0 ^oC

NiO₂(s) + 2Cl^-(aq) + 4H⁺(aq) ----> Cl₂(g) + Ni²⁺(aq) + 2H₂O

Answer 1

The overall reaction is:

NiO₂(s) + 2Cl^-(aq) + 4H⁺(aq) ----> Cl₂(g) + Ni²⁺(aq) + 2H₂O

Standard reduction potentials for the half reactions of the cell:

NiO₂(s) + 4H⁺(aq) + 2e^- ----> + Ni²⁺(aq) + 2H₂O………………………E^o_red = 1.678 V

2Cl^-(aq) ----> Cl₂(g) + 2e^-………………………………………………. E^o_{oxi =} -1.358 V

For calculating standard potential, E^o_cell, of the overall reaction we need to add standard reduction and oxidation potentials of the half reactions. For obtaining Cl^- oxidation potential the reduction potential of Cl₂ is taken and the sign is reversed. Thus

Eo _{cell =} E^o_red +E^o_oxi ………………………………………………….eq.1

           = 1.678 + (-1.358) = 0.32 V

Standard potential of reaction, E^o_cell = 0.32V

Having found E^o_cell of reaction, in order to find equilibrium constant of the reaction we need to use Gibbs free energy equations in relation with standard potential of reaction, E^o_cell. Accordingly,

The following two equations will be used to arrive at equilibrium constant of the reaction. Thus

G^o = -nF E^o_cell …………………………………….eq.2

and  G^o = -RTlnK_eq……………………………….eq.3

Hence to find G^o from E^o_cell, the corresponding values are substituted in eq.2

Thus

For G^o = -nF E^o_cell

where

n = number of electrons in reaction = 2

F = Faraday constant = 96,485 J (96.485 kJ) per volt gram equivalent

Thus G^o = -2 x 96,485 x 0.32 = - 61,450 J/mol

Now knowing G^o of the reaction it’s easy to find equilibrium constant of the reaction using

G^o = -RTlnK_eq……………………………….eq.3

K_eq = exp (-G/RT) …………………………………… eq.4

Where

R = gas constant = 8.314 J K⁻¹ mol⁻¹

T = temperature in K = 298 K

K_eq = equilibrium constant

Thus from eq.4

RT = 8.314x298 = 2477.572

K_eq = e^{-(-61450/2477.572)}

K_eq = e^-(-24.80) = e^24.80= 5.8952 x 10¹⁰

Thus the equilibrium constant, K_eq, of the above reaction = 5.8952 x 10¹⁰