Factors that decide the ion to be eliminated or discharged first during electrolysis.

Factors that decide the ion to be eliminated or discharged first during electrolysis.

• Position in electro-chemical series:

Series for cations (high to low) K⁺, Ca²⁺, Na⁺, Mg²⁺, Al³⁺, [C], Zn²⁺, Fe²⁺, Pb²⁺, Cu²⁺, Ag⁺

Series for anions (high to low) SO₄^2-, NO₃^–, Cl^–, Br^–, I^–, -OH

The ion lower in electro-chemical series is discharged first. Consequently, during electrolysis of dilute sodium chloride, H⁺ which is low in the reactivity series than Na⁺ is discharged at the cathode whereas, ^–OH ions are discharged at the anode.

At the cathode                                                          At the anode

2H⁺(aq) + 2e →         H₂(g)                                      4^–OH (aq) – 4e   →           2H₂O(l) + O₂(g)

 

• Concentration:

When Cl^–, Br^– or I^– are concentrated, then they will be discharged in preference ^–OH.

In this case electrolysis of concentrated sodium chloride solution liberates chlorine at the anode and hydrogen gas at the cathode (concentration does not affect Na⁺)

 

At the cathode                                                          At the anode

2H⁺(aq) + 2e →         H₂(g)                                      2Cl^– (aq) – 2e      →           Cl₂(g)

 

• Nature of electrode

When mercury cathode is used Na⁺ is discharged with preference to H⁺.

Therefore, electrolysis concentrated sodium chloride using mercury cathode liberate Na at the cathode and chlorine gas at the anode (due to high concentration of Cl^–).

At the cathode                                                          At the anode

Na⁺(aq) + e                      →    Na                                             2Cl^– (aq) – 2e      →  Cl₂(g)

 

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