experiment gf24d29202 1280 Factors That Determine The Products Of Electrolysis

Factors That Determine The Products Of Electrolysis


The products of electrolysis are the elements that make up the electrolyte. For example, sodium chloride is made up of sodium and chlorine, which means the electrolysis of sodium chloride will yield sodium and chloride. However, predicting the products of electrolysis is not that straightforward, especially if the electrolyte is in aqueous solution.

There are basically 3 ways of determining the products of electrolysis:

Order of preferential discharge of ions

Some ions are more easily discharged than others. The order of preferential discharge of ions is the order in which ions are more likely to be discharged. The order of discharge of cations (negatively charged ions) is the same as the order of the ions on the electrochemical series. Ions of less reactive metals are more easily discharged than ions of more reactive metals.

Order of discharge of cations in order of increasing chance of discharge

Cation Chance of bieng discharged
K+ least likely
Na+ โ†“
Mg2+ โ†“
Zn2+ โ†“
H+ โ†“
Pb2+ โ†“
Cu2+ โ†“
Ag+ most likely

Order of discharge of anions in order of increasing chance of discharge

Anion Chance of bieng discharged
SO42- and NO3โ€“ least likely
Clโ€“ โ†“
Brโ€“ โ†“
Iโ€“ โ†“
OHโ€“ most likely

Example 1: Electrolysis of lead(II) nitrate solution

In the electrolysis of lead(II) nitrate solution, Pb(NO3)2(aq):

  • the cations present are lead ions (Pb2+) from the lead(II) nitrate and H+ from the water, since it is in solution.
  • the anions present are nitrate ions (NO3โ€“) from the lead(II) nitrate and OHโ€“ from the water.

During electrolysis, the cations Pb2+ and H+ moved to the cathode where they compete for discharge. Since Pb2+ is lower than H+ in the order of discharge, Pb2+ is discharge instead of H+. Each Pb2+ ion gains two electrodes and is discharged as a Pb atom at the cathode.

Pb2+(aq) + 2e โ†’ Pb(s)

The anions, NO3โ€“ and OHโ€“ move to the anode. At the anode OHโ€“ is discharged instead of NO3โ€“ because OHโ€“ is lower in the order of discharge, hence more easily discharged. During discharge, OHโ€“ lose electrons to form water and oxygen gas.

4OHโ€“(aq) โ€“ 4e โ†’ 2H2O(l) + O2(g)

Concentration of solution

Concentration of the electrolyte in solution also determines the products at the electrodes. An ion is more easily discharged if its concentration is higher. Concentration usually determines the anode product.

Example 2 โ€“ Electrolysis of dilute sodium chloride solution

Dilute sodium chloride, NaCl(aq) contains the following ions, Na+, Clโ€“, H+, and OHโ€“. However, the fact that is a dilute solution means that there is more water in the solution and the concentration of OHโ€“ ions is higher than the concentration of Clโ€“ ions.

The cations, Na+ and H+ move to the cathode. H+ is lower than Na+ in the order of discharge, so the H+ ions are discharged first.

2H+(aq) + 2e โ†’ H2(g)

Now, we use concentration of the ions to predict the anode product. The solution is dilute, which means there are more OHโ€“ ions than Clโ€“ ions in the solution. Therefore, OHโ€“ ions are discharged first.

4OHโ€“(aq) โ€“ 4e โ†’ 2H2O(l) + O2(g)


Example 3 โ€“ Electrolysis of concentrated sodium chloride solution

Concentrated sodium chloride, NaCl(aq) contains the following ions, Na+, Clโ€“, H+, and OHโ€“. However, the fact that is a concentrated solution means that there is less water in the solution and the concentration of OHโ€“ ions is less than the concentration of Clโ€“ ions.

The cations, Na+ and H+ move to the cathode. H+ is lower than Na+ in the order of discharge, so the H+ ions are discharged first.

2H+(aq) + 2e โ†’ H2(g)

Again, we use concentration of the ions to predict the anode product. The solution is concentrated, which means there are less OHโ€“ ions than Clโ€“ ions in the solution. Therefore, Clโ€“ ions are discharged first.

2Clโ€“(aq) โ€“ 2e โ†’ Cl2(g)

Nature of electrodes

The nature of electrodes also affect the products of electrolysis. If the anode is made up of the same metal as the ions in the electrolyte the anode dissolves during electrolysis.

Example 4 โ€“ electrolysis of copper(II) sulphate solution using a copper anode

Copper(II) sulphate solution contains the following ions, Cu2+, H+, SO42-, OHโ€“.

If the anode is copper, the atoms in the anode loses electrons and dissolve to form Cu2+ ions.

Cu(s) โ€“ 2e โ†’ Cu2+(aq)

The copper ions travel to the cathode where they gain electrons to form copper atoms.

Cu2+(aq) + 2e โ†’ Cu(s)

Example 5 โ€“ electrolysis of copper(II) sulphate solution using a copper anode

Copper(II) sulphate solution contains the following ions, Cu2+, H+, SO42-, OHโ€“.

If the anode is inert (i.e. carbon or platinum), the OHโ€“ are discharged instead of SO42- due to preferential discharge.

4OHโ€“(aq) โ€“ 4e โ†’ 2H2O(l) + O2(g)

The copper ions in solution travel to the cathode where they gain electrons to form copper atoms.

Cu2+(aq) + 2e โ†’ Cu(s)