Electrolysis

Voltaic cells are based upon the spontaneity of a particular redox reaction

• This in turn is based upon the relative reduction potentials for the two half-reactions
• The half-reaction with the greater reduction potential will drive the other half-reaction in the direction of oxidation
• The relative difference between the two reduction potentials provides a quantitative description of the electromotive force of the voltaic cell

However, we have seen that some commercial voltaic cells can be "recharged"

• This is achieved by the application of current in the opposite direction to that of the spontaneous direction of current flow in the voltaic cell
• This application of current in the opposite direction will cause the redox reaction to proceed in the opposite direction
• The half cell where oxidation occurred will now undergo a reduction reaction
• The half cell where reduction occurred will now undergo an oxidation reaction

Sodium metal and chlorine gas spontaneously react to form the ionic compound Sodium Chloride:

2Na(s) + Cl₂(g) ® 2NaCl(s)

• This reaction is a redox reaction:

Anode: 2Na(s) ® 2Na⁺(s) + 2e^-

Cathode: Cl₂(g) + 2e^- ® 2Cl^-(s)

• The reverse reaction, the decomposition of ionic NaCl into Na(s) and Cl₂(g), can be achieved by the application of an EMF (i.e. external current)

2NaCl(l) ® 2Na(l) + Cl₂(g)

The electrolytic cell contains two electrodes and either a molten salt solution (as with NaCl) or some other type of solution (e.g. aqueous)

• The external EMF is provided by a source of direct current (sometimes another voltaic cell, or battery)
• This external current acts as an electron pump to push electrons onto one electrode (and drive a reduction reaction) and to withdraw electrons from the other electrode (and drive an oxidation reaction)
• The electrode where oxidation occurs is still called the anode. However, for an electrolytic cell it is labeled as "+" to indicate that this is the electrode where electrons are being withdrawn by the external EMF
• The electrode where reduction occurs is stall called the cathode. However, for an electrolytic cell it is labeled as "-" to indicate that this is the electrode where electrons are being pumped into by the external EMF

With aqueous solutions of salts it must be considered that the electrolysis might end up driving the reduction or oxidation of H₂O(l)

• H₂O(l) can be oxidized to produce O₂(g)
• H₂O(l) can be reduced to produce H₂(g)
• If a salt is present, the cation of the salt may or may not be preferentially reduced compared to H₂O(l). Likewise, the anion may or may not be preferentially oxidized compared to H₂O(l)

Here are the reduction potential data for the reduction of H₂O(l) and Na⁺(aq):

2H₂O(l) + 2e^- ® H₂(g) + 2OH^-(aq) E⁰_red = -0.83 V

Na⁺(aq) + e^- ® Na(s) E⁰_red = -2.71 V

• When comparing any two reduction reactions the reaction that is favored is the one with the more positive reduction potential. In this case, the reduction of H₂O(l) is the favored reduction reaction

Here are the reduction potential data for the oxidation of H₂O(l) and Cl^-(aq). Remember that the reduction potential provided is referencing the reduction (i.e. reverse) reaction

2H₂O(l) ® 4H⁺(aq) + O₂(g) + 4e^- E⁰_red = 1.23 V

2Cl^-(aq) ® Cl₂(g) + 2e^- E⁰_red = 1.36 V

• When comparing the reduction potentials, the reduction of Cl^- is favored. However, we are interested in the reverse reactions, i.e. the oxidation reactions. In this case, remember that a substance that is readily reduced is difficult to oxidize, and vice versa. Thus, if the Cl^- is favored for reduction, then the water reaction is favored for oxidation
• What this analysis suggests is that electrolysis of an aqueous solution of Na⁺(aq) and Cl^-(aq) ions will not result in the reduction of Na⁺ or oxidation of Cl^-, but rather, will result in the reduction of H₂O to produce hydrogen, and the oxidation of H₂O to produce oxygen.

• The redox reactions for the electrolysis of an aqueous solution of NaCl are therefore:

(reduction) 2H₂O(l) + 2e^- ® H₂(g) + 2OH^-(aq)

(oxidation) 2Cl^-(aq) ® Cl₂(g) + 2e^-

• And the overall reaction is:

2H₂O(l) + 2Cl^-(aq) ® Cl₂(g) + H₂(g) + 2OH^-(aq)

• What external EMF is required do drive this particular electrolysis?

(reduction; cathode) 2H₂O(l) + 2e^- ® H₂(g) + 2OH^-(aq) E⁰_red = -0.83 V

(oxidation; anode) 2Cl^-(aq) ® Cl₂(g) + 2e^- E⁰_red = 1.36 V

• Inert electrodes do not enter into the chemical reaction, and just serve as a surface upon which electron transfers can occur (and be conducted to the other half-cell)
• Active electrodes chemically participate in the redox reaction

If an aqueous solution contains solid metal electrodes, the electrodes may be oxidized to the ionic form if the metal is easier to oxidize than water. Since we think in terms of reduction potentials and not ease of oxidation, this means that the metal electrodes will be oxidized to ionic form if the reduction potential of the metal is less than that of water

Ni(s) ® Ni²⁺(aq) + 2e^- E⁰_red = -0.28 V

2H₂O(l) ® 4H⁺(aq) + O₂(g) + 4e^- E⁰_red = 1.23 V

• In this case the reduction potential of H₂O(l) is greater than that for Nickel metal. Thus, H₂O is easier to reduce, and therefore more difficult to oxidize, than Nickel. Thus, at the anode, Nickel will preferentially be oxidized compared to H₂O(l) if we provide an external EMF that "sucks" electrons out at this electrode

What about the cathode (reduction) reaction involving Nickel ion and water?

Ni²⁺(aq) + 2e^- ® Ni(s) E⁰_red = -0.28 V

2H₂O(l) + 2e^- ® H₂(g) + 2OH^-(aq) E⁰_red = -0.83 V

• In this case, the half-reaction with the greater reduction potential is the reduction of Nickel. Therefore, at the cathode, Nickel will preferentially be reduced if we provide an external EMF that forces electrons onto this electrode

• Thus, a metallic cathode may be coated with solid nickel metal in a process known as "electroplating"