Ionic Equation for Nickle Sulphate

OP

Original post by EierVonSatan

That's an oxidation half equation?

If you want the ionic equation for that, then all you have to do is split up the aqueous ionic compounds into their separate ions and cancel the ions which appear on both sides :smile:

Sorry to be a pain but Im still struggling ? Mind going through the steps ? :smile:

Reply 3

Start on the left. Is NiCO₃ soluble? If it is - write it as ions, if it is not, write it as NiCO₃(s). Is sulfuric acid dissociated? If so, write it as ions, if not - write it as molecule.

Now, move to the right hand side and repeat the same.

Balance.

Cancel out.

Done.

Reply 4

OP

okay so NiCO3 doesnt dissolve in water so I can ignore that -
Then you have H2 (2+) and S04 (2-) on the left
and I know Ni ---> Ni 2+ + 2e-

So Ni 2+ + SO4 2- - --> NiSO4 ?

Reply 5

You were already told there is no need for redox here - so your half reaction with electrons is of no use. Scratch it.

NiCO₃ is insoluble - but it doesn't mean you can ignore that. Please reread what I already wrote in my previous post.

NiSO₄ is not insoluble, so it can't be written on the right as NiSO₄, but it must be written in dissociated form.

Reply 6

Just ZK

Original post by oriane89

Okay so I'm making nickle sulphate following the below equation :

NiCO3 +H2S04 ---> NiSO4 + CO2 + H20

I think the first bit of the ionic equation would be this but I've no idea as to the rest ????

Ni ---> Ni 2+ + 2e-

What i got after doing what Borek said is this: CO3(2-)---------->CO2 +O(2-)
I dont know exaxtly whether this is correct or not!

Reply 7

Just ZK

I got it
First u will seperate the ions:
Ni(2+) + CO3(2-) + 2H+ + SO4(2-)------> Ni(2+) SO4(2-) + CO2 + H2O
Then cancel the ones that appear on both sides and u get the final equation as:
CO3(2-) + 2H+ -----> CO2 + H2O
Which is your ionic equation!!

Reply 8

OP

Original post by Just ZK

I got it
First u will seperate the ions:
Ni(2+) + CO3(2-) + 2H+ + SO4(2-)------> Ni(2+) SO4(2-) + CO2 + H2O
Then cancel the ones that appear on both sides and u get the final equation as:
CO3(2-) + 2H+ -----> CO2 + H2O
Which is your ionic equation!!

brill thank you ! :biggrin:

Reply 9

Original post by Just ZK

I got it
First u will seperate the ions:
Ni(2+) + CO3(2-) + 2H+ + SO4(2-)------> Ni(2+) SO4(2-) + CO2 + H2O
Then cancel the ones that appear on both sides and u get the final equation as:
CO3(2-) + 2H+ -----> CO2 + H2O
Which is your ionic equation!!

No, it is not.

You can't list NiCO₃ as ions when it is an insoluble solid.

Net ionic equation can contain no-ionic species, if that's how they enter (or leave) the reaction. You had no problems putting H₂O and CO₂ on the right, didn't you?

In this case it means there should initially be

NiCO₃(s) + 2H⁺ + SO₄²⁺

on the left, and NiCO₃(s) will be present in the final reaction.

(edited 10 years ago)

Reply 10

OP

Original post by Borek

No, it is not.

You can't list NiCO₃ as ions when it is an insoluble solid.

Net ionic equation can contain no-ionic species, if that's how they enter (or leave) the reaction. You had no problems putting H₂O and CO₂ on the right, didn't you?

In this case it means there should initially be

NiCO₃(s) + 2H⁺ + SO₄²⁺

on the left, and NiCO₃(s) will be present in the final reaction.

okay so that means the ions I have present on the left are H2+ and SO4(2-) and on the right the only ions I have is the ionic salt ?

Reply 11