Consider the following reaction and its Δ𝐺∘ at 25.00 °C. 2Ag+(aq)+Cu(s)⟶2Ag(s)+Cu2+(aq)Δ𝐺∘=−88.66 kJ/mol

Question

Calculate the standard cell potential, 𝐸∘cell, for the reaction.

𝐸∘cell=V?

Calculate the equilibrium constant, 𝐾, for the reaction.

𝐾=?

Answer 1

Ag⁺ +e- ==> Ag .. Eº = 0.80 V (cathode - reduction)

Cu²⁺ + 2e- ==> Cu .. Eº = 0.34 (anode - oxidation)

2Ag⁺ + Cu ==> 2Ag + Cu²⁺ .. balanced redox

Eºcell = cathode - anode = 0.80 - 0.34

Eºcell = 0.46 V

∆Gº = -RT ln K

∆Gº = -88.66 kJ /mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (note to change units to agree with those of ∆Gº)

T = temp in Kelvin = 25 + 273 = 298K

K = ?

-88.66 = - (0.008314)(298) ln K

ln K = 35.79

K = 3.5x10¹⁵