Chemistry - Electrochemistry Question with Solution | TestHub

Concept:

• Frost diagram construction

• Standard reduction potential

• Thermodynamic stability 

Solution:

Frost diagram ordinate: Δ$G^0$/F =-nE^₀

The stability is determined by the value of nE^₀ (which is proportional to Gibbs Free Energy).

The lower the value, the more stable the state.

For Mº: nEº =0 (Reference point)

For M²⁺ :n=2 , E^₀=-0.14 V

nE^₀ = 2(-0.14) =-0.28

For M³⁺ : n=3 . We add the energy of M²⁺ to the energy of the M³⁺/M²⁺step:

 nE°= -0.28 +(0.76) =+ 0.48

Oxidation State                       nE° Value        

 M°                                        0        

 M²⁺ –0.28 (Lowest)          

M³⁺ +0.48

The most stable oxidation state is M²⁺ because it represents the minimum point on the Frost diagram. Final Answer:  Option(B)