Question
Assume that you havea cell that has an iron (II) concentration of 0.015 M and anH+ concentration of 1.0 x 10-3 M. The celltemperature is 311 K, and thepressure of hydrogen is maintained at 0.04 bar. What would the cellpotential (E), G, and K beunder these conditions? Assume that you havea cell that has an iron (II) concentration of 0.015 M and anH+ concentration of 1.0 x 10-3 M. The celltemperature is 311 K, and thepressure of hydrogen is maintained at 0.04 bar. What would the cellpotential (E), G, and K beunder these conditions?
Explanation / Answer
From the reduction potential values, we have - Fe+2/Fe = -0.440 V Therefore for the cell reaction, Fe (s) + 2H+ (aq) ---------> Fe+2(aq) + H2 (g) The cell potential, E0cell =0 - (-0.440 V) = 0.440 V Hence - The cell potential at the given concentrations would be,Ecell = Ecell0 -0.0592/2 log [Fe+2] /[H+]2 = 0.440 - 0.0592 /2 log [(0.015) / (0.001)2] = 0.440 - 0.123 = 0.316 V Since G = - nFEcell = - 2*96500*0.316 = - 61062 J Similarly G = - RT lnK -61062 J = 8.314 J/K.mol * 311 K ln K Therefore ln K= 23.61 Thus K =e23.61 = 1.80*1010 = - 2*96500*0.316 = - 61062 J Similarly G = - RT lnK -61062 J = 8.314 J/K.mol * 311 K ln K Therefore ln K= 23.61 Thus K =e23.61 = 1.80*1010