Question

An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO3. The cell uses the half reactions shown below to produce the overall dissociation reaction. cathode reaction: anode reaction: overall reaction: PbCO3(s) + 2e — Pb(s) + Coz- (aq) Ph(s) — Pb2+ (aq) + 2 e- PbCO,(s) — Pb2+ (aq) + CO2 (aq) Which equality correctly represents the relationship between the equilibrium constant expression for the overall reaction and the solubility product constant for PbCO,? OK = -Ksp = [Pb2+] [co] O K = Ksp = [Pb2+] [CO3-) K = (Ksp)' = po [Pb2+][003] O K = Ksp = PbCO] A student constructs this cell using solutions that are 1 M. The potential of the cell at 298 K measures -0.379 V. Calculate the experimental Ksp of PbCO3 based on the measured potential. Ksp =

Answer 1

po coz (s) → pot ac) + coz (as) K= ksp = {p.627] [co ] GIVEN E call = -0.379 16° = – RT Iu Ksp = - Of Eisle n = 2, No. of é's TamUS FERRED IN S THE CELL REACTION F IS FARADAY CONSTANT • - – 8T A Ke - OFFeu du ksp = 2x96485 C moet x(-0.3791) 8.314 JK moet x 298k - - 29.52 Ksp = 8 29:52 ..kep = (1.46 x 1013]