1. Calculate the standard electrode potential Eθcell for a cell
constructed using the following
Al3+(aq) + 3e 
Al(s) E = -1.66V
Pb2+(aq) + 2ePb(s) E = -0.13V
Al(s) E = -1.66V
Pb2+(aq) + 2ePb(s) E = -0.13V| Step I: The Pb2+ /Pb half cell has the most positive Eθ value. It forms the right cell electrode. The Al3+/Al forms the left cell electrode. | |
| Step II: Eθright = E (Pb2+/Pb) = -0.13V Eθleft = E (Al3+/Al) = -1.66V Eθcell = Eθright Eθleft | |
| (-) | |
| =-0.13 = 1/ | |
| = |
|