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Year 13 chem question

The question says “the redox reaction in aqueous solution between acidified potassium manganate(VII) and sodium ethanedioate is autocatalysed. Write an equation for this redox reaction”

I checked the mark scheme and I don’t understand it.
Original post by Anonysus
The question says “the redox reaction in aqueous solution between acidified potassium manganate(VII) and sodium ethanedioate is autocatalysed. Write an equation for this redox reaction”

I checked the mark scheme and I don’t understand it.


which bits do you not understand?

some points to consider:
"redox reaction" some things will be reduced, some will be oxidised. what do you think will happen to the manganate (VII) ions?
"acidified" H+ present think about the role H+ can have in redox reactions
"potassium manganate(VII) and sodium ethanedioate" can you work out which ions will be involved in the redox reaction? can you write formulae for the species?
"autocatalysed" what is autocatalysis?

Spoiler

Reply 2
Original post by bl0bf1sh
which bits do you not understand?

some points to consider:
"redox reaction" some things will be reduced, some will be oxidised. what do you think will happen to the manganate (VII) ions?
"acidified" H present think about the role H can have in redox reactions
"potassium manganate(VII) and sodium ethanedioate" can you work out which ions will be involved in the redox reaction? can you write formulae for the species?
"autocatalysed" what is autocatalysis?

Spoiler



How do I know what gets reduced and what gets oxidised and what they form when they are reduced/oxidised?
(edited 1 year ago)
Original post by Anonysus
How do I know what gets reduced and what gets oxidised and what they form when they are reduced/oxidised?


Well, you can make a good guess using oxidation numbers. If you find a species with a very positive oxidation number, it generally is what is being reduced.

If you work out the oxidation number of Mn in MnO4^-, it should be +7, meaning it’s already pretty oxidised and ought to get reduced.

As MnO4^- has all these oxygens bound to it, which it would have to get rid of to be reduced, the oxygens must be removed in the form of water molecules (the acidification of the solution provides the H^+ ions required for this). You have to memorise the fact that the manganese is reduced to Mn^2+ at A level.

C2O4^2- is a little more tricky. Because you’ve worked out that the MnO4^- is getting reduced, then C2O4^2- must therefore be oxidised. That means it must form electrons as one of the products. Although C2O4^2- has several oxygens on it, it doesn’t need to remove them as water molecules (because this isn’t reduction). It can instead break down to form CO2.

Can you now construct some balanced ionic equations for each process?
(edited 1 year ago)

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