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Electrochemistry and Corrosion

Electrolysis of Mixtures of Ions

Let us consider an electrolytic cell containing a mixture of ions, Ag+, Cu2+, Zn2+, and Mg2+, all with a concentration of 1.0 M. Pt | H2 | H+ || Ag+, Cu2+, Zn2+, Mg2+ | M <= electrons For such a mixture, we construct a table below using standard reduction potentials.
Anode DE° Cathod
Half-reaction E° /V E° /V Half-reaction
H2 = 2 H+ + 2e 0.00 0.80 V 0.80 Ag+ + e = Ag
H2 = 2 H+ + 2e 0.00 0.34 V 0.34 Cu2+ + 2 e = Cu
H2 = 2 H+ + 2e 0.00 -0.76 V -0.76 Zn2+ + 2 e = Zn
H2 = 2 H+ + 2e 0.00 -2.36 V -2.36 Mg2+ + 2 e = Mg
The first two couples has a positive voltage, indicating that the reactions for the first two cells are spontaneous and exothermic. The Ag+ ions have the strongest oxidizing power, and they create the highest voltage for the cell. The third and fourth reactions are not spontaneous, and require a voltage to force the reaction.

Let us assume that a Mg2+ ion zaps by and is electicuted, (reduced). But Mg is more reactive than Ag, and in the fight for electrons, the stongest oxidizing ion Ag+ wins. So goes the theory, but the reality is often different. Thus, apply electrolysis for the separation of metals still require engineering research and development.

The argument provided earlier appears to be sound, but the reality can be different. Consider the the case when a salt solution is electrolyzed. The oxidation potentials are:

Half-reaction E° E° Half-reaction
Cl2 + 2 e = 2 Cl- 1.36 V 0.00 2 H+ + 2e = H2
O2 + 2 e + 4 H+ = 2 H2O 1.23 V
Since water has less negative (similar to more positive) oxidation potential, the above argument suggests that oxygen gas will form when a voltage is applied to the electrode. The reality is that we get chlorine gas under this circumstance. the mechanism of the process is complicated. An explanation suggests that its easier for the chloride to transfer electrons to the electrode than water.