How do you get this answer? A concentration cell is constructed using two Fe3+/Fe half-cells. What is the cell potential, Ece for this concentration cell given the following two half-cell Fe3+ion concentrationsFeau 5.87 x 106M and [Fe3+]c3.74 x 10-2M? 14. a) 0.173 V ) 0.075V b0103 V c) 0.225 v Solution Following are the half cell reactions Fe(s)(dil) - - - - - - > Fe 3+ (aq) + 3e Fe 3+ (aq)(conc) + 3e - - - - - - > Fe(s) standard reduction potential for both half reaction is same and E° cell is zero Therefore Nernst equation for concentration cell is E cell = (0.0592/n)log([Fe 3+ ]conc/[Fe 3+ ]dil) where, n = number of electron transfer, 3 Q=[Fe 3+ ]conc/[Fe 3+ ?]dil = 3.74×10 -2 M /5.87×10 -6 M = 6.37×10 3 Therefore E cell = (0.0592/3)log(6.37×10 3 ) E cell = 0.075V .