Question

The Ksp of Cd(OH)2 is 7.2×10−15. Cd2+(aq)+2e−→Cd(s), E∘=−0.40V

Part A Calculate the standard electrode potential for half-reaction Cd(OH)2(s)+2e−⇌Cd(s)+2OH−(aq)

Answer 1

Sol :-  

The given reaction is :

Cd(OH)₂ (s) <--------------------> Cd²⁺ (aq) + 2 OH^- (aq) , K_sp = 7.20 x 10^-15

We know that ,

ΔG = - nFE⁰_cell = - 2.303 RT log K_sp

- 2 x 96500 x E⁰_cell = - 2.303 x 8.314 J K^-1mol^-1 x 298 K x log (7.20 x 10^-15)

On solving, we have

E⁰_cell = - 0.418 V

Now,

Cd(OH)₂ (s) <--------------------> Cd²⁺ (aq) + 2 OH^- (aq) , E₁⁰ = - 0.418 V ............(1)

Cd²⁺ (aq) + 2e^- -------------------> Cd (s) , E₂⁰ = - 0.40 V ............(2)

Adding both the equation, we have the required equation i.e.

Cd(OH)₂ (s) + 2e^- ------------------> Cd (s) + 2 OH^- (aq) , E⁰ = E₁⁰ + E₂⁰ = - 818 V = - 0.82 V

Hence, E⁰ of the required equation = - 0.82 V