At Anode: oxidation: Sn(s) ------> Sn2+ + 2e-
At cathode reduction Pb2+ + 2e- -------> Pb(s)
Overall reaction: Sn + Pb2+ -----> Pb + Sn2+
Eocell = Eocathode - Eoanode
Given Eocathode = -0.131 V and Eoanode = -0.143V
Eocell = Eocathode - Eoanode = -0.131 - (-0.143) = 0.012 V
Nernest equation
Ecell = Eocell - (0.0592/n) * log([Sn2+]/[Pb2+])
At the point when [Sn2+]=0.355 mol/L Means total reacted concentration of Sn is = 0.871 - 0.355 = 0.516 M
Same concentration of Pb2+ must be reacted. Hence [Pb2+] = 0.927-0.516 = 0.411 M
Ecell = Eocell - (0.0592/n) * log([Sn2+]/[Pb2+]) = 0.012 - (0.0592/2)*log(0.355/0.411) = 0.0139 V
HHelp ASAP Consider the cell described below at 283 K: Sn | Sn2+ (0.871 M) || Pb2+ (0.927 M) | Pb Given E°Pb2+>Pb = -...
Consider the cell described below at 253 K: Sn | Sn2+ (0.979 M) || Pb2+ (0.943 M) | Pb Given E°pb2+_Pb = -0.131 V, Eºsn2+_Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed 0.355 mol/L. .00893 V Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previous Tries
I am not sure how to answer this question
Consider the cell described below at 285 K: Sn | Sn2+ (1.01 M) || Pb2+ (2.07 M) Pb Given EºPb2+ + 2e-Pb = -0.131 V and Eºsn2+ + 2 e-Sn = -0.143 V. What will the concentration of the Pb2+ solution be when the cell is dead? Submit Answer Tries 0/45
A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). a. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment? b. If the anode compartment contains [SO2−4]= 1.50 M in equilibrium with PbSO4(s), what is the Kspof PbSO4?
1. A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). A. If the concentration of Sn2+ in the cathode compartment is 1.30 M and the cell generates an emf of 0.25 V , what is the concentration of Pb2+ in the anode compartment? B. If the anode compartment contains [SO2−4]= 1.30 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4? 2. A voltaic cell utilizes the following reaction: 2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). A. What is the emf...
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M) | Fe(s) given the following reduction half-reactions. Pb2+(aq) + 2 e– ® Pb(s) E° = –0.126 V Fe2+(aq) + e– ® Fe(s) E° = –0.44 V
Consider the cell described below at 253 K: Fe | Fe2+ (0.867 M) || Cd2+ (0.945 M) | Cd Given EoCd2+Cd -0.403 V, E°Fe2+_Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.311 mol/L. Tries 0/45 Submit Answer
Consider the cell described below at 275 K: Fel Fe2+ (0.935 M) || Cd2+ (0.939 M) 1 Cd Given E°Cd2+--Cd = -0.403 V, EºFe2+-Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.399 mol/L.