1. Daniel Cell
Zn + Cu2+  Zn2+ + Cu We used 1.0 M Zn2+ and 1.0 M Cu2+ as electrolyte solutions. These are standard conditions (Eo) For the standard Daniel Cell, Eocell = 1.10 V If we did not use 1.0 M, non standard conditions, E

2. Relate E to Eo using Q Nernst Equation E = Eo – (0.0592/n) log Q
Recall, Q = P/R
As reaction proceeds, ↑P, which ↑Q (>1)
As ↑Q, ↓E. As E → 0, the cell dies (voltage ↓)
If Q < 1 (R > P), Ecell > E0cell
If we used 1.0 M Zn2+ and 2.0 M Cu2+, E > 1.10 V
So Ecell depends on half reactions and [R] vs. [P]

3. Pg 906 #73 (brown and lemay)
A voltaic cell employs the following redox reaction
Sn2+ + Mn → Sn + Mn2+
Calc the cell potential at 25oC under
Standard conditions
[Sn2+] = M; [Mn2+] = 2.00 M
Eo Sn/Sn2+ = V Eo Mn/Mn2+ = V

4. Concentration Cells
The half reactions are the same (same material in both beakers)
That means there is no reduction potential driving the e-, Eocell = 0 V
To drive the current, use concentration gradient
Cell operates by transferring e- from less conc side to more conc side until the [Mn+] on both sides is equal.
Once they are equal, the cell is dead

5. Pg 906 #80 Consider the conc cell cell. Label the anode and cathode.
Indicate the direction of e- flow.
Indicate what happens the [Pb2+] in each half

6. Pg 906 #81
A conc cell consists of two Sn/Sn2+ half cells. The cell has E = 0.10 V at 25oC. What is the ratio of the Sn2+ conc in each half cell?