1. Use the Electromotive chart below to answer parts A&B. Please show all work, thank you!!:) Electromotive Reaction TableElementE'(volts)Half ReactionLithiumPotassiumBarium.StrontiumCalcium.Sodium ..3.0452.925.2.90.2.89:2.87.2.714.2.37.1.661.18.0.763.0.74.0.51..0.440.0.403.0.3363.0.2770.2500.136.0.1260.000-0.152.-0.2475-0.32-0.337.-0.45:-0.521.-0.5355.-0.771'.-0.7991-0.920Li Li + le-.K = K* + 1e-Ba = Ba² + 2e-.Sr = Sr² + 2e-Ca = Ca³ + 2e-Na = Na + le-MagnesiumAluminum.Manganese.Zinc ...Chromium ....PhosphorusIron IICadmiumThailiumCobalt.NickelTinLead.HydrogenAntimony..Arsenic .Bismuth..Mg = Mg- + 2e-Al = Al+3 + 3e-Mn = Mn +2e-Zn = Zn² + 2e-Cr = Cr + 3e-P+ 2H,0 = 3H,PO, + H* + 1e-Fe = Fe + 2e-Cd = Cd2 + 2e-TI = TI + 1e-Co = Co +2e-Ni = Ni +2e-Sn Sn + 2e-Pb = Pb•² + 2e-Copper II.Gulfur/Sulfurous Acid.Copper I.Iodide.Iron IIISilverMercuryPalladiumBromine.Oxygen/WaterChlorineGold ..Oxygen/OzoneFlourineH = 2H* + 2e-2Sb + 3H,0 = Sb,O, +6H* + 6e-As + 2H,Ó = HASO, +3H¯ + 3e-Bi + H,Ó = BiO* + 2H* + 3e-Cu = Cu² + 2e-.S+3H,0 = H,SO; + 4H: +4e-Cu = Cu* + 1e-.21- = L, + 2e-Fe = Fe + 1e-Ag = Ag* +1e-.Hg,² = 2H&² + 2e-Pd = Pd+2 + 2e-2Br- = Br, + 2e-2H,0 = 0, + 4H* + 4e-2CI = CL + 2e-Au = Au + 3e-O, + H,O = 0,+ 2H* + 2e-.2F = F, + 2e-.-0.987-1.066.-1.229-1.3595:-1.50-2.07-2.85Turch 1)a) Calculate the K for this reaction:2Fe*2(aq)+ Pd*2,(aq) 2Fe"(aq)+3+ Pdb) Calculate the equilibrium concentrations,if the above all start at 2.00M.

Interpretation : We have to calculate K and equilibrium concentration of all the species.Calculations: A)A)2Fe2+ (aq)---> 2Fe3+ (aq)+2 e- E0 = -0.771VPd2+ (aq)+ 2e----> Pd(s) E0 =0.987VE0 = E0cathode + E0anodeE0 =0.987V -0.771V= 0.216V-nFE0 =-RTlnK2 x 96500 x 0.216V = 8.314 x 298 x 2.303 logKK = 2.02 x 107B) 2Fe2+ (aq) + Pd2+ (aq)⇔ 2Fe3+ (aq) + Pd(s) Initial : 2 2 2 2At equi: 2-2x 2-x 2+2x 2+xK = [Products][Reactants] 2.02 x 107 = [ 2+2x ]2[ 2+x][ 2-2x ]2[ 2-x]Since K is very large we can neglect denomenator .2.02 x 107 =[ 2+2x ]2[ 2+x]x = -0.1105[Fe2+] = 2-2x = 2-2x(-0.1105)=2.221M[Pd2+] = 2-x =2-(-0.1105)=2.11MFe3+ =will be almost 2M