# D40.4 Using Standard Half-Cell Potentials

Based on the methods described in the preceding section, standard half-cell potentials have been determined for many half-cells. The table below gives half-cell potentials for selected half-cells. Click on the table to open a version that can be enlarged or printed.

There are several important aspects to note:

• The oxidizing agent is on the reactant side of a reduction half-reaction equation.
• The strongest oxidizing agents (the substances most easily reduced) have the largest positive E° values and are at the top of the table. (For example, F2(g) is a very strong oxidizing agent.)
• The reducing agent is on the product side of a reduction half-reaction equation.
• The strongest reducing agents (the substances most easily oxidized) have the most negative E° values and are at the bottom of the table. (For example, Li(s) and K(s) are very strong reducing agents.)
• A redox reaction is product-favored when a stronger oxidizing agent reacts with a stronger reducing agent. This results in a positive value for E°cell when the oxidation half-reaction is combined with the reduction half-reaction.
• Half-cell reactions are reversible and the direction a half-cell reaction goes depends on the potential of the other half-cell to which it is connected in a cell.

The table of standard half-cell potentials can be used to determine the E°cell for any voltaic cell and predict whether a specific redox reaction is product-favored. For example, for the cell:

Cu(s) | Cu2+(aq, 1 M) || Ag+(aq, 1 M) | Ag(s)
E°cell = E°right half-cellE°left half-cell = E°Ag+|AgE°Cu2+|Cu = 0.7991 V – 0.340 V = 0.459 V

Because the value of E°cell is positive, the redox reaction corresponding to this cell notation is product-favored and the voltaic cell can produce electrical energy. The overall reaction can be obtained from the cell notation by writing the oxidation and the reduction half-reactions, multiplying each half-reaction by an appropriate number to balance electrons, and summing the two half-reactions:

 Oxidation (left): Cu(s) ⟶ Cu2+(aq) + 2e− E°anode = +0.340 V Reduction (right): 2 × (Ag+(aq) + e− ⟶ Ag(s)) E°cathode = +0.7991 V Overall: Cu(s) + 2Ag+(aq) ⟶ Cu2+(aq) + 2Ag(s) E°cell = +0.459 V

Note that:

• Even though the reduction half-reaction is multiplied by 2, the E°cathode is not multiplied by 2 when E°cell is calculated. This is because electrical potential is a ratio of energy per coulomb of charge transferred (V = J/C). If the half-reaction is doubled, both energy and charge transferred are doubled, leaving the ratio constant.
• The oxidation half-reaction is the reverse of the reaction as shown in the half-cell potential table, but E°anode is not multiplied by -1 . This “-1” is accounted for by the fact that E°anode is subtracted from E°cathode. (In this way, the cell potential does not depend on the direction of a half-reaction; it depends only on the composition of the half-cell.)

E° for many half-cells are included in the appendix. Tables like these make it possible to determine the E°cell for many redox reactions. Moreover, by comparing the standard potentials, we can discern which species is easier to reduce/harder to oxidize/a stronger oxidizing agent (higher/more positive E°) and which species is easier to oxidize/harder to reduce/a stronger reducing agent (lower/more negative E°).

Exercise: Using Standard Half-cell Potentials

Exercise: Predicting whether a Redox Reaction is Product-favored