HHelp ASAP Consider the cell described below at 283 K: Sn | Sn2+ (0.871 M) || Pb2+ (0.927 M) |

At Anode: oxidation: Sn(s) ------> Sn2+ + 2e-

At cathode reduction Pb2+ + 2e- -------> Pb(s)

Overall reaction: Sn + Pb2+ -----> Pb + Sn2+

E^ocell = E^ocathode - E^oanode

Given E^ocathode = -0.131 V and E^oanode = -0.143V

E^ocell = E^ocathode - E^oanode = -0.131 - (-0.143) = 0.012 V

Nernest equation

Ecell = E^ocell - (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 = E^ocell - (0.0592/n) * log([Sn2+]/[Pb2+]) = 0.012 - (0.0592/2)*log(0.355/0.411) = 0.0139 V