Cd(s) ----------------> Cd^2+ (aq) + 2e^- E0 = 0.40v
2Cu^2+ (aq) + 2e^- ------> 2Cu(s) E0 = 0.16v
--------------------------------------------------------------------------------
Cd(s) + 2Cu^2+(aq) ------. Cd^2+ (aq) + 2Cu(s) E0cell = 0.56v
n = 2
G0 = -nE0cell*F
= -2*0.56*96500
= -108080J
G0 = -RTlnK
-108080 = -8.314*298lnK
lnK = -108080/(-8.314*298)
lnK = 43.62
K = 8.8*10^18 >>>answer
G0 for this reaction would be less than zero
b.
Hg(l) ------------> Hg^2+ (aq) + 2e^- E0 = -0.85g
2Cu^2+ (aq) + 2e^- ------> 2Cu(s) E0 = 0.16v
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Hg(l) + 2Cu^2+(aq) -----> Hg^2+ (aq) + 2Cu(s) E0cell = -0.69v
n = 2
G0 = -nE0cell*F
= -2*-0.69*96500
= 133170J
G0 = -RTlnK
133170 = -8.314*298lnK
lnK = 133170/(-8.314*298)
lnK = -53.75
K = 4.5*10^-24 >>>answer
G0 for this reaction would be greater than zero
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cu-+ (aq) + Cd(s)...
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