Question

Predict if the following reactions are spontaneous (e.g. using the Z method) and calculate E⁰, DG⁰ and K for each reaction

Reduction of H⁺ by Zn(s) (to form H₂ and Zn²⁺)

Oxidation of Cu by Cl₂(g) (to form Cu²⁺ and Cl^-)

Oxidation of Cl^- by F₂(g) (to form Cl₂ and F^-)

Reduction of Cu²⁺ by Fe²⁺ (to form Cu(s) and Fe³⁺)

Answer 1

Use E^o = E^ocathode – E^oanode

If E^o > 0, the reaction is spontaneous.

Reduction of H⁺ by Zn(s) (to form H₂ and Zn²⁺)

E^o = 0.00 V– ( -0.76V ) = 0.76 V >0

n=2   

Note:1J = 1Cx1V

dG^o = - nFE^o = - 2 val/mol x 96500 C/val x 0.76 V = - 146.7 kJ/mol

ln K = - dG^o / RT = - dG^o / (8.314 J.mol.K x 298 K) =

        = - 4.036 x 10^-4 J^-1.mol^-1 · dG^o

K = e ^{- 4.036 x 10-4 dGo} = e ^{- 4.036 x 10-4 x (- 146.7 x 10^3)} = e^59.208 = 5.2x10²⁵

(For dG use the value in J/mol for calculation)

Oxidation of Cu by Cl₂(g) (to form Cu²⁺ and Cl^-)

E^o = 1.358 V– ( 0.153V ) = 1.205 V >0

n=2   

dG^o = - nFE^o = - 2 val/mol x 96500 C/val x 1.205 V = - …….. kJ/mol

ln K = - dG^o / RT = - dG^o / (8.314 J.mol.K x 298 K) =

        = - 4.036 x 10^-4 J^-1.mol^-1 · dG^o

K = e ^{- 4.036 x 10-4 dGo} =………

Oxidation of Cl^- by F₂(g) (to form Cl₂ and F^-)

E^o = 2.87 V– ( 1.35V ) = 1.52 V      >0

n=2   

Note:1J = 1Cx1V

dG^o = - nFE^o = - 2 val/mol x 96500 C/val x 1.52 V = - …… kJ/mol

ln K = - dG^o / RT = - dG^o / (8.314 J.mol.K x 298 K) =

        = - 4.036 x 10^-4 J^-1.mol^-1 · dG^o

K = e ^{- 4.036 x 10-4 dGo} =…..

Reduction of Cu²⁺ by Fe²⁺ (to form Cu(s) and Fe³⁺)

E^o = 0.153– ( 0.771 ) = - 0.618 V     <0

n=2   

Note:1J = 1Cx1V

dG^o = - nFE^o = - 2 val/mol x 96500 C/val x (-0.618) V = + ….. kJ/mol

ln K = - dG^o / RT = - dG^o / (8.314 J.mol.K x 298 K) =

        = - 4.036 x 10^-4 J^-1.mol^-1 · dG^o

K = e ^{- 4.036 x 10-4 dGo} =…..