Can anyone check if I am doing this right? Given: Pb2+ + 2e- ---> Pb(s) E standard = -0.13V Ag+ + 1e- ---> Ag(s) E standard = 0.80V [Pb2+] = 0.05M [Ag+] = 0.5 M these are non standard concentrations Temp = 298K Using the Nernst equation, find E My answer: Pb oxidized and Ag is reduced ( is this right?.... because Ag has the higher reduction potential compared to Pb) E = E standard - (RT/nF) ln Q Q = [products]^p/[Reactants]^r the reaction is: 2Ag+ + Pb(s) + 2e- ---> 2Ag(s) + Pb2 + 2e- (the electrons cancels out) therefore, Q = (0.05M) / (0.5)^2 Q= 0.20 the E standard = 0.80 - (-0.13) = 0.80 + 0.13 = 0.93V E = E standard - (RT/nF) ln Q E = 0.93 - (8.314*298K/2*9.649E4) ln (0.20) E= 207.56 V Can anyone check the calculations for me and see if I am doing this right please? Thank you!!