1)1. You are given the following cell at 25°C: Pb(s) | Pb2+(aq,
0.60M) || Sn2+(aq, 0.1 M), Sn4+(aq, 0.15 M) |Pt(s) Write the half
reactions for the cell and using Appendix D in your text book,
calculate the E°cell and the Ecell
oxidation half reaction = Pb(s) ------------>
Pb+2(aq) + 2e-
reduction half reaction = Sn+4(aq) +
2e- --------> Sn+2(aq)
E0cell =
E0Sn+4/sn+2
E0Pb+2/Pb
E0cell = 0.15 - (-0.126)
E0cell = 0.276 V ( i'm not
sure if it correct...