Question

1A: Calculate E for the cell Cu(s)+2Ag⁺(aq)--->Cu⁺²(aq)+2Ag(s) at 298K when [Ag⁺]=0.025M and [Cu²⁺]=0.068 M

1B: Consider the cell Cd(s)+Cu²⁺(aq)-->Cu(s)+Cd²⁺. Calculate K for this reaction.

Answer 1

1A)

Oxidation half reaction (at anode)

Cu(s) ------> Cu²⁺(aq) + 2e E°_red = +0.337 V

Reduction half reaction(at cathode)

2Ag⁺(aq) + 2e ------> 2Ag(s) E°_red = +0.800V

Overall reaction

Cu(s) + 2Ag⁺(aq) --------> Cu²⁺(aq) + 2Ag(s)  

E°_cell = E°_red,cathod - E°_{red, anode}

E°_cell = +0.800V - (+0.337V)

E°_cell = 0.463V

reaction quotient ,Q= [ Cu²⁺]/[Ag⁺]² = 0.068M/(0.025M))² = 108.8

number of elecron transfer , n = 2

At 25℃ , Nernst equation is as follows

E = E°_cell - (0.0592V/n)logQ

E = 0.463V - (0.0592V/2)log(108.8)

E = 0.463V - 0.06V

E = 0.403V

1B)

Oxidation half reaction

Cd(s) ------> Cd²⁺(aq) + 2e E°_red = -0.400V

Reduction half reaction

Cu²⁺(aq)   + 2e -------> Cu(s) E°_red = +0.337V

E°_cell = +0.337V - (-0.400V) = 0.737V

n = 2

∆G° = -nFE°_cell

∆G° = -(2× 96485C/mol × 0.737V)

∆G° = -142.22kJ/mol

∆G° = - RTlnK

K = 1×10^{-(∆G°/2.303RT)}

K = 1× 10^{-(-142.22kJ/mol/(2.303× 0.008314(kJ/mol K) × 298.15K)}

K = 1× 10^24.91

K = 8.13 ×10²⁴