Question

Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the ions in these reactions is 0.0005 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-K, and F = 96,485 C/mol):
Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the ions in these reactions is 0.0005 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-k, and F = 96,485 C/mol): (a) 2 Cr3+ (aq) + 1 Ca(s) --> 2 Cr2+(aq) + 1 Ca2+(aq) E = v (b) 1 Ni2+(aq) + 1 Pb(s) --> 1 Ni(s) + 1 Pb2+ (aq) E = (c) 1 Fe2+ (aq) + 1 Cd(s) --> 1 Fe(s) + 1 Cd2+(aq) (d) 2 A13+(aq) + 6 1(aq) --> 2 Al(s) + 3 12(s) E = v

Answer 1

solution a) 2012 ap) +1268) -> 207*@g) + latav) Cathode (reduction) -> Reght the 2017 Anode (oxidation) -> ca cat the Echogy any Canody 2.87 E cell = Ecathodi Eanode -0.41 – 2.87 Fill = -3.28 V n=2 (2 electrons are transferred) Ecell = E - RT ln (anode] - cell nf. - Cathode] - -3.28 – 8-314 X 298 e [0005] 2 x 96485 ods] = -3.28 - 0.07568 = -3.28-0.02568 6) Ecell = -3.28'V i IPb (S) -> I Night 1 PbCall) cathode (reduction) Nit the Ni Anode oxidation) P6 -> P6²t the -0.28 0.126 Scanned with CamScanner

E cell = Ecathode - Fanoche = -0.28 -0.126 = -0.400V co RT , Ganode] Ecell = { cell [Cathode] = -0.46 - Phare to Factory = -0.406 - BI beni = -0.405-fat xo HE -0.406 (-0.440) (0-403) Ecell = -0.400 I Feest lcd(s) > Fe (s) + 1 cd?tag) cathode (reduction) Fellyt ze ? Fews Anode (oxidations' cdo (dentre Eu - Ecathode – Eanode = -0.440 - 0.403 = -0.843 v Ecell = E cell - In Cathode - RT e Anode] = -0.843-21 txo CS Scanned Evil = -0.843 v