Electrode Potential and Electrolysis:
When an electrolyte is dissolved in water or melted and an electric current is passed through it, the Cations (positive ions) move towards the cathode and Anions (negative ions) towards the anode. The formation of products at the respective electrodes is due to oxidation (loss of electrons) at the Anode and reduction (gain of electrons) at the cathode. For example- in the electrolysis of molten NaCl, the electrode reactions are-
NaCl ⇌ Na+ + Cl– (Ionisation)
At Cathode: Na+ + e– ———–> Na (Reduction)
At Anode: Cl– ———> 1/2 Cl2 + e–
But, if the solution contains more than one reducible or oxidizable species then the products of electrolysis depends upon their respective electrode potentials as described below-
At the Cathode- When two or more cations move towards the cathode, the one with higher reduction potential will be discharged at the cathode i.e. for different competing reduction processes, the one with higher reduction potential will takes place. Example- during the electrolysis of an aqueous solution of NaCl, the following two reactions are possible at cathode i.e.
- Reduction of Na+ ions.
|Na+(aq) + e– ———–> Na (s) ; E0red = -2.71 V|
- Reduction of H2O molecules.
|H2O (l) + e– ———–> 1/2 H2 (g) + OH– (aq) ; E0red = -0.41 V|
From the electrode (reduction) potential values, it is clear that water shows more reduction tendency than Na+ ions and hence, H2O is preferentially reduced. Thus, the product of electrolysis at the cathode will H2 gas and not Na-metal.
At the Anode- When two or more anions move towards the Anode, the one with higher oxidation potential (or lower reduction potential) will be discharged at the Anode i.e. for different competing oxidation processes, the one with lower reduction potential will take place during the electrolysis of an aqueous solution of CuSO4, the two competing oxidation processes at the Anode are-
- Oxidation of SO4-2 ions.
|2SO4-2 (aq) ————–> S2O8-2 + 2e– ; E0red = 2.01 V or E0oxd = -2.01 V|
- Oxidation of H2O molecules.
|H2O (l) ———–> 1/2 O2 (g) + 2H+ (aq) + 2e– ; E0red = 1.23 V or E0oxd = -1.23 V|
From the electrode potential values, it is clear that the oxidation potential of H2O is higher than SO4-2 ions or the reduction potential of H2O is lower than SO4-2 ions and hence, water is preferentially oxidized, i.e. the product formed at the Anode will be O2 gas instead of S2O8-2 ion.
However, in some cases unexpected results are obtained due to over-voltage i.e. sometimes higher voltage is required to carry out a change, and if the two potentials are close to each other, it will not be possible to predict with certainty the species that will undergo oxidation or reduction. Example- consider the following two changes-
|H2O (l) ———–> 1/2 O2 (g) + 2H+ (aq) + 2e– ; E0oxd = -1.23 V|
Cl– (aq) ———> 1/2 Cl2 (g) + e– ; E0oxd = -1.36 V
Though the oxidation potential of H2O is more than that of Cl– ions, but during the electrolysis of an aqueous NaCl solution, Cl– ions oxidize in preference to H2O molecules at the anode giving Cl2 gas.