Electrolysis - ionic equations

[Ozzy]

Hi,

 

I'm stuck with some homework making ionic equations. Here's what I think:

 

Sodium Chloride

 

Chlorine made at anode

Sodium made at cathode

 

Chlorine: 2cl- -2e --> cl2

 

Sodium: 2Na+ +2e --> Na

 

Is that right? It probably isn't.

 

I also need help with:

 

Aluminium oxide

Lithium fluoride

Copper bromide

Magnesium sulphide

 

Please explain it in steps because I'm really dumb.

 

Cheers

[LazerFazer]

I think the half-equations for the formation of NaCl would be:

 

Na --> Na+ + e-

Cl + e- --> Cl-

 

And the same for the others. (I might be wrong tho... havent taken chem now for about half a year)

 

cheers,

LF

[RyanJ]

Your first one looks correct

 

I'll show you #2 and lets se if you can do the rest!

 

Aluminium(III) Oxide, [ce]Al2O3[/ce]

 

Splits into...

 

[ce]2Al2O3 <=> 4Al^{+3} + 6O^{-2}[/ce]

 

Anode = positive, cathode = negative.

 

Anode: [ce]6O^{-2} - 12e^{-} -> 3O2 + 12e^{-}[/ce]

Cathode: [ce]4Al^{+3} + 12e^{-} -> 4Al[/ce]

 

I ballanced the equation also

 

You may need a little help with the ions that are formed so I'll help you tere too and you can work the equations out:

 

Lithium forms [ce]Li^{+}[/ce] ions

Fluoride ions are always: [ce]F^{-}[/ce]

Bromide ions are normally: [ce]Br^{-}[/ce] (Unless with something more electronegative)

Copper is [ce]Cu^{+}[/ce] in this case.

Magnesium is always [ce]Mg^{+}[/ce]

Sulphide ions are [ce]S^{-2}[/ce]

 

Good luck!

 

Cheers,

 

Ryan Jones