Electrolysis - Competing Half-reactions in Solution Electrolysis

Competing Half-reactions in Solution Electrolysis

Using a cell containing inert platinum electrodes, electrolysis of aqueous solutions of some salts leads to reduction of the cations (e.g., metal deposition with, e.g., zinc salts) and oxidation of the anions (e.g. evolution of bromine with bromides). However, with salts of some metals (e.g. sodium) hydrogen is evolved at the cathode, and for salts containing some anions (e.g. sulfate SO42−) oxygen is evolved at the anode. In both cases this is due to water being reduced to form hydrogen or oxidised to form oxygen. In principle the voltage required to electrolyze a salt solution can be derived from the standard electrode potential for the reactions at the anode and cathode. The standard electrode potential is directly related to the Gibbs free energy, ΔG, for the reactions at each electrode and refers to an electrode with no current flowing. An extract from the table of standard electrode potentials is shown below.

Half-reaction (V) Ref.
Na+ + e− Na(s) −2.71
Zn2+ + 2e− Zn(s) −0.7618
2H+ + 2e− H2(g) ≡ 0
Br2(aq) + 2e− 2Br− +1.0873
O2(g) + 4H+ + 4e− 2H2O +1.23
Cl2(g) + 2e− 2Cl− +1.36
S2O2–
8 + 2e− 2SO2−
4
+2.07

In terms of electrolysis, this table should be interpreted as follows

  • oxidised species (often a cation) nearer the top of the table are more difficult to reduce than oxidised species further down. For example it is more difficult to reduce sodium ion to sodium metal than it is to reduce zinc ion to zinc metal.
  • reduced species (often an anion) near the bottom of the table are more difficult to oxidise than reduced species higher up. For example it is more difficult to oxidise sulfate anions than it is to oxidise bromide anions.

Using the Nernst equation the electrode potential can be calculated for a specific concentration of ions, temperature and the number of electrons involved. For pure water (pH 7):

  • the electrode potential for the reduction producing hydrogen is −0.41 V
  • the electrode potential for the oxidation producing oxygen is +0.82 V.

Comparable figures calculated in a similar way, for 1M zinc bromide, ZnBr2, are −0.76 V for the reduction to Zn metal and +1.10 V for the oxidation producing bromine. The conclusion from these figures is that hydrogen should be produced at the cathode and oxygen at the anode from the electrolysis of water which is at variance with the experimental observation that zinc metal is deposited and bromine is produced. The explanation is that these calculated potentials only indicate the thermodynamically preferred reaction. In practice many other factors have to be taken into account such as the kinetics of some of the reaction steps involved. These factors together mean that a higher potential is required for the reduction and oxidation of water than predicted, and these are termed overpotentials. Experimentally it is known that overpotentials depend on the design of the cell and the nature of the electrodes.

For the electrolysis of a neutral (pH 7) sodium chloride solution, the reduction of sodium ion is thermodynamically very difficult and water is reduced evolving hydrogen leaving hydroxide ions in solution. At the anode the oxidation of chlorine is observed rather than the oxidation of water since the overpotential for the oxidation of chloride to chlorine is lower than the overpotential for the oxidation of water to oxygen. The hydroxide ions and dissolved chlorine gas react further to form hypochlorous acid. The aqueous solutions resulting from this process is called electrolyzed water and is used as a disinfectant and cleaning agent.

Read more about this topic:  Electrolysis

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