The following standard reduction potentials have been determined
for the aqueous chemistry of ytterbium:
Yb3+(aq) + e- | -----> | Yb2+(aq) | E° = -1.050 V |
Yb2+(aq) + 2e- | -----> | Yb(s) | E° = -2.760 V |
Calculate the equilibrium constant (K) for the disproportionation
of Yb2+(aq) at 25 °C.
3Yb2+(aq) <----->Yb(s) + 2Yb3+(aq) |
K = |
Yb(s) --------------> Yb^2+ (aq) + 2e^- E0 = 2.760v
2Yb^3+ (aq) + 2e^- ------> 2Yb^2+ (aq) E0 = -1.050v
-------------------------------------------------------------------------------
Yb(s) + 2Yb^3+ (aq) -----------> 3Yb^2+ (aq) E0cell = 1.71v
3Yb2+(aq) <----->Yb(s) + 2Yb3+(aq) E0cell = -1.71v
n = 2
G0 = -nE0cell*F
= -2*-1.71*96500
= 330030J
G0 = -RTlnK
330030 = -8.314*298lnK
330030 = -2477.57lnK
lnK = 330030/-2477.57
lnK = -133.2
K = 1.42*10^-58 <>>>answer
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