# 2.3 Standard Electrode Potentials

One question you may have had when we set up our zinc and copper cell was why was zinc oxidized and copper reduced, and not the other way around? In fact, for any redox reaction what determines which element is oxidized and which is reduced?

Good question!

Metals, because they only have a few valence electrons, like to lose electrons. In other words they tend to be easily oxidized. But metals differ in how easily they lose electrons. A list of metals arranged in order of how easily the metal is oxidized is known as an activity series. Activity series are covered in more detail in Section 2.5 (an optional section).

The fact that different substances are oxidized more readily than others is the driving force behind electrochemical cells, and it is this force that forces electrons through the external circuit from the anode (site of oxidation) to the cathode (site of reduction). This force is known as the potential difference or electromotive force (emf or E). Potential difference is measured in volts (V), and thus is also referred to as the voltage of the cell. Voltage is a measure of the tendency of electrons to flow. The higher the voltage, the greater the tendency for electrons to flow from the anode to the cathode.

Tables of Standard Reduction Potentials for Half-Reactions allow us to determine the voltage of electrochemical cells. (You'll want to print this table and keep it handy.) These tables compare the ability of different half-reactions to compete for electrons (become reduced). Since half-reactions cannot occur on their own, all values in the table are determined by comparing a half-reaction with a hydrogen half-cell.

$$\ce{2H^{+} + 2e- -> H2(g)}$$ E° = 0.00 V

the degree symbol following the E (E°) indicates standard conditions:
temperature = 25°C; pressure = 100 kPa;
concentration of aqueous solutions = 1 mol·L-1

(Find this half-reaction, and other the other half-reactions described below, in the Table.) This hydrogen half-cell has been assigned a voltage of 0.00 V. If a half-reaction is better at competing for electrons than this half-cell, that half-reaction will undergo reduction and the hydrogen will be oxidized. That other half-reaction will then be assigned a positive voltage.

For example, if copper and hydrogen half-cells are joined together we find that the copper half-cell will gain electrons from the hydrogen half-cell. Thus the copper half-cell is given a positive voltage and given a relative value of +0.34 V:

$$\ce{Cu^{2+} + 2e- -> Cu}$$ E° = 0.34 V

Since both half-reactions cannot undergo reduction, we must reverse the equation of the reaction that will undergo oxidation. This will give us an electrochemical cell voltage of 0.34 V:

 E° $$\ce{Cu^{2+} + 2e- -> Cu}$$ 0.34 V $$\ce{H2(g) -> 2H^{+} + 2e-}$$ + 0.00 V $$\ce{Cu^{2+}(aq) + H2(g) -> 2H+(aq) + Cu(s)}$$ 0.34 V

We see in the Table of Standard Reduction Potentials that zinc has a negative indicating that it is not as good at competing for electrons as hydrogen.

$$\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$$ E° = -0.76 V

Therefore if zinc and hydrogen are paired together in an electrochemical cell, the hydrogen would be reduced (gain the electrons) and zinc would be oxidized (losing electrons). To determine the net redox reaction as well as the voltage of the electrochemical cell we reverse the zinc equation, and also reverse it's sign before adding the equations and together:

 E° $$\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$$ 0.76 V $$\ce{2H^{+} + 2e- -> H2(g)}$$ + 0.00 V $$\ce{Zn(s) + 2H^{+}(aq) -> Zn^{2+}(aq) + H2(g)}$$ 0.76 V