Question

A galvanic cell consists of a silver electrode in 1 M AgNO3 and a chromium electrode in 1 M Cr(NO3)3. What is the equilibrium constant for this reaction at 25oC? Enter you answer with 2 significant digits, using the syntax of "1.0x10(22)" for "1.0x1022"

Answer 1

Lets find Eo 1st

from data table:

Eo(Cr3+/Cr(s)) = -0.74 V

Eo(Ag+/Ag(s)) = 0.7996 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ag+/Ag(s))

anode is (Cr3+/Cr(s))

The chemical reaction taking place is

3 Ag+(aq) + Cr(s) --> 3 Ag(s) + Cr3+(aq)

Eocell = Eocathode - Eoanode

= (0.7996) - (-0.74)

= 1.5396 V

here, number of electrons being transferred, n = 3

Eo = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, Eo = (0.0592/n)*log Kc

1.5396 = (0.0592/3)*log Kc

log Kc = 78.0203

Kc = 1.048*10^78

Answer: 1.0*10^78