Question

5. A lot of oxidation-reduction reactions have very high Keq values. This is easy to see by relating E0 values to Keq through the Nernst Equation written at equilbrium: E0cell = (nF/ RT)lnKeq , in which n = the moles of electrons transferred in the balanced reaction, F = Faraday's constant, R = gas constant in Joules and T = temp in K What is Keq for the Zn/ Zn+2 // Cu+2/Cu cell ( T = 293K)? Expalin why or why not this reaction is product favored. Show all work please!

Answer 1

oxidation:	Zn(s) Zn2+(aq) + 2 e-	Eoox. = - Eored. = - (- 0.762 V) = + 0.762 V
reduction:	Cu2+(aq) + 2 e- Cu(s)	Eored. = + 0.339 V

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Eocell = + 1.101 V

1.101 = 8.314*293/2*96500lnKeq

lnKeq = 87.230

keq= 1.69 * 10^(87)

the reaction is favoured because we have huge keq and G= -ve so it is spontaneous