20)
Lets find Eo 1st
from data table:
Eo(Fe2+/Fe(s)) = -0.44 V
Eo(Au3+/Au(s)) = 1.52 V
here:
cathode is (Au3+/Au(s))
anode is (Fe2+/Fe(s))
The chemical reaction taking place is
2Au3+(aq) + 3Fe(s) --> 2Au(s) + 3Fe2+(aq)
Eocell = Eocathode - Eoanode
= (1.52) - (-0.44)
= 1.96 V
Number of electron being transferred in balanced reaction is 6
So, n = 6
Use:
E = Eo - (2.303*RT/nF) log {[Fe2+]^3/[Au3+]^2}
2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591
E = Eo - (0.0591/n) log {[Fe2+]^3/[Au3+]^2}
E = 1.96 - (0.0591/6) log (1.5^3/0.004^2)
E = 1.96-(0.052)
E = 1.91 V
Answer: E
I am allowed to answer only 1 question at a time
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