Question

What is the cell potential for the spontaneous concentration cell with the reaction Fe3+(aq) + Fe(s) → Fe(s) + Fe3+(aq), with concentrations of Fe3+ of 0.873M and 0.342M at a temperature of 310. K? Ecell = E°cell −(RT/nF)lnQ

Answer 1

For concentration cell, cathode and anode are same electrode
So, Eo = 0
Number of electron being transferred in balanced reaction is 3
So, n = 3
If E is positive anode will be the one with lower concentration

we have below equation to be used:
E = Eo - (2.303*RT/nF) log {[Fe3+] at anode/[Fe3+]at cathode}

Here:
2.303*R*T/F
= 2.303*8.314*310.0/96500
= 0.0615

So, above expression becomes:
E = Eo - (0.0615/n) log {[Fe3+] at anode/[Fe3+]at cathode}
E = 0 - (0.0615/3) log (0.342/0.873)
E = 8.344*10^-3 V
Answer: 8.344*10^-3 V

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know