Standard Potentials

We will demonstrate how to use a table of standard reduction potentials in the following segment. We will use a shortened table:

Reduction Half-Reaction			E° (V)
F2(g) + 2e-		2 F-(aq)	+2.87
Ag+(aq) + e-		Ag(s)	+.80
2H3O+(aq) + 2e-		H2(g) + 2H2O(l)	0.00
Zn2+(aq) + 2e-		Zn(s)	-0.763
Li+(aq) + e-		Li(s)	-3.045

You can also see the larger table: Standard Reduction Potentials

Note:

• The voltages are for reactions written as oxidized form and electrons ==> reduced form. If your reaction is reduced form ==> oxidized form and electrons reverse the sign of E°
• All of the listed reactions are reversible.

This table gives us a considerable amount of information very compactly. This table tells us:

• The more positive the value of E° is, the better the oxidizing agent. In our sample table, F₂(g) is the best oxidizing agent, and correspondingly, the worst reducing agent.
• The more negative the value of E° is, the better the reducing agent. In our sample Li⁺(aq) is the best reducing agent, and correspondingly, the worst oxidizing agent.
• The sign of the E° of a redox reaction is the sign of the electrode when it is attached to a H₂/H₃O⁺ standard cell.
• Any substance on the left side of the reaction arrow will spontaneously oxidize any substance on the right side of the arrow, if the second substance is lower in the table.
• The potentials are constant, even if the amount of each molecule is increased or decreased, provided the ratios remain the same. (So that the reaction is balanced)

To calculate the net potential of a redox reaction we follow the following steps.
First say our problem is: calculate the net potential of the below redox reaction:
Zn(s) + Sn²⁺(aq) ==> Sn(s) + Zn²⁺(aq)

1. First we separate the redox equation into half reaction as follows:
   Zn(s) ==> Zn²⁺(aq) + 2e^-
   Sn²⁺(aq) + 2e^- ==> Sn(s)
2. We find the potentials for these reactions is the table of standard reduction potentials.
   We find:
   Zn(s) ==> Zn²⁺(aq) + 2e^- ......... E° = +.763 V (remember, polarity is reversed)
   Sn²⁺(aq) + 2e^- ==> Sn(s) ......... E° = -.14 V
3. We add the two half reaction potentials, -.763 V + -.14 V = .623 V
   Thus the net potential of the redox reaction, and our answer is: .623 V