Question

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu?" (aq) + Pb(s)—+2Cu(aq) + Pb2+ (aq) Answer: K for this reaction would be than one.

Answer 1

o Solve' Given; Reaction 2.642+(aq.) + Pb(s) — 2Cut(aq.) + Pb2+(99.) Now; E0924%cut (standard reduction potenital) = 0-16V - E 132+/Pb (standard reduction potential) = -0.13V. _Oxidation-half cell deaction: Pb(s) > Pbatint 2é OG, = nFEP87 px teen Reduction - half cell reaction: 2cument 2é 264+ (aq.) OG E-NFEO (4²t/ut Scanned with CamScanner

COverall reaction: 2 cent Pb(s) → 2Cut (aq.) + Potcaq.) Onyx o

og So; AGO--M FEO Where; G = Gibbs fore enorgy (7) ne number of electrons tromsferred. F = Faraday constant = 96500 Cmoſ: Eº - Standard cell potential. 80; OGSXN = OG + olio Giova 2 - FE Beyppet - DF Fettycent OGON = -2 x 96500 Cmoh' x (0:13V) – 2x96500(moíx(0-16v e agorn = -55970 J mot fo; OGöxne – 55.970 kg/mol. _Thus; Standard Free energy change per mol = -55.9 70 kg: Now; Gixxe = - RT.mk Raumiveral gas constant = 8:3147k moí' T: Temperature = 25°C -25+2736 = 2984 (Standard ke Equilibrium constant : So; -55970 ]moí' = -8-314_7k!mui'x 238k xhams on solving, we get k = 6-46. fos ScanJhuajin k for this reaction [greatex/more then she CamScanner