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I don't understand, how to find the half equation for dilute sodium chloride

GCSE AQA Combined Science Chemistry
Can someone help me please?
(edited 1 year ago)
I'm confused by this table:
Screenshot 2022-04-22 230525.png
Its showing the products of the electrolysis of SOLUTIONS. It is also showing the effect of how changing the concentration of ions present can affect the products obtained.

try here it may offer a better explanation or not
I not entirely sure about discharge element, dilute and concentrate.
the discharge element is simply the element which is produced at the anode and the cathode.

Dilute just means it has lots of water in it
Concentrated just means that there is not much water present.

In the table you posted you can see for example it says that products produced at the anode and the cathode during the electrolysis of hydrochloric acid changes depending on the concentration of the acid. When its dilute according to your table water is discharged (electrolysed or oxidised) rather than chloride ions at the anode.
How to write the half equation for dilute sodium chloride?
the equations are in the link page above.

Original post by scimus63
the equations are in the link page above.


I don't understand how they had worked the half equation out?
(edited 1 year ago)
at the cathode metal ions are usually reduced to form metal atoms which can form on the surface of the cathode e.g.

Na+ + e ----------------- Na

however in solutions there is hydrogen ions present as well as metal ions so we can have:

H+ + e ----------------- H

but hydrogen is a diatomic gas so need to multiply by x2

2H+ + e ----------------- H2

Only one ion discharged at a time at the cathode, the metal lowest in the reactivity series discharges first, in this case hydrogen NOT sodium. So you will get hydrogen gas at the cathode.

At the anode non-metal ions are oxidised.

In this case the chloride ions are oxidised to form chlorine gas

Cl- -e --------------Cl

but chlorine is a diatomic element so need to multiply by x2

2Cl- -2e --------------Cl2

However according to your table if you use dilute sodium chloride then the concentration of chloride ions is obviously low, so water is oxidised instead of the chloride ions. The equation for the oxidation of water at the anode is shown below, the OH- ions come from the break up or ionization of water molecules:

4OH- 2H2O + O2 + 4e

so you get hydrogen gas at the cathode and oxygen gas at the anode

I hope this helps you a bit!!
(edited 1 year ago)
Thank you guys
(edited 1 year ago)

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