Correct Answer - a.ΔrGc-=-196.86kJmol-1,Kc=3.192×1034
b.ΔrGc-=-2.895kJmol-1,Kc=3.22
a. E
c-
cell
=E
c-
cathode
-E
c-
anode
=-0.40V-(-0.74)=+0.34V
ΔrGc-=-nFE
c-
cell
=-6mol×96400Cmol-1×0.34V
=-196860Jmol-1=-196.86kJmol-1
ΔrGc-=--2.303RTlogK
196860=2.303×8.314×298logK
logK=34.5014
K=Antilog(34.5014)=3.192×1034
b
. Ec-
_(cell)=+0.80V-0.77V=+0.03V
ΔrGc-=-nFE
c-
cell
=-(1mol)×(96500Cmol-1)×(0.03V)
=-2895Jmol-1ΔrGc-=-2.303RTlogK
-2895=-2.303×8.314×298×logK
or log K=0.5074
or K=Anitlog(0.5074)=3.22