Question

Equilibrium constant (Kc) for the given cell reaction is 10. Calculate E0cell A(s) +B2+(aq) A2+(aq) + B(s)

Answer 1

∆G=-nFE0=LnKc.
from cell reacn. n=1. F=96500.
-1 X 96500 X E0 = LnKc
96500 X E0 = 2.303 X Log(10). Log10=1
TF. E0 =. -2.303/96500

Answer 2

Answer:

The EMF of the cell is 0.029 Volts.

Explanation:

$A(s) +B^{2+}(aq)\rightarrow A^{2+}(aq) + B(s)$

n = 2

Equilibrium constant of the reaction = K = 10

(EMF) Cell potential at equilibrium:$E^o$

$E^o=\frac{2.303RT}{nF}\log K$

$E^o=\frac{0.0591}{n}\log K$

$E^o=\frac{0.0591}{2}\log [10]=0.029 V$

The EMF of the cell is 0.029 Volts.