Redox reaction exam question - oxidation and reduction

Fe(III) reacts with Sn(II) to produce Fe(II) and Sn(IV). Write a balanced redox equation for this reaction. Clearly state which element is being oxidised and which is being reduced. Show all of your working.

Reply 1

halloween123456

reduction: Fe^3+ + e- -> Fe^2+ |*2
oxidation: Sn^2+ - 2e- -> Sn^4+
_______________________________
2Fe^3+ + Sn^2+ -> 2Fe^2+ + Sn^4+

HOpe this helps :smile:

(edited 10 years ago)

Liverpool John Moores University

Liverpool

Reply 2

LauraDew

Original post by HPooveli

reduction: Fe^3+ + e- -> Fe^2+ |*2
oxidation: Sn^2+ - 2e- -> Sn^4+
_______________________________
2Fe^3+ + Sn^2+ -> 2Fe^2+ + Sn^4+

HOpe this helps :smile:

Why do you multiply the Fe line by two?

Reply 3

honeywhite

Original post by LauraDew

Fe(III) (that is, the ferric cation) has an oxidation state of plus three; Sn(II) (the stannous cation) has an oxidation state of plus two. Fe(II) (ferrous) has an oxidation state of plus two, while Sn(IV) has an oxidation state of plus four.

Remember, LEO the lion says GER: Loss of Electrons is Oxidation, and Gain of Electrons is Reduction. Since oxidation state is essentially the number of "missing" or "additional" electrons, Sn(II) loses two electrons to become Sn(IV)... it becomes oxidised. Fe(III) gains an electron to become Fe(II) and so it becomes reduced.

I need a few minutes to calculate the actual equation but there's your theory.

Reply 4

halloween123456

Original post by LauraDew

Why do you multiply the Fe line by two?

So the electrons cancel each other out: (For Sn you need 2 e-)
You always multiply a half reaction with the number of electrons you need for the other half reaction. I could also multiply the Sn-Half reaction with 1, but because it's the same I left it out.

reduction: Fe^3+ + e- -> Fe^2+ |*2
oxidation: Sn^2+ - 2e- -> Sn^4+ |*1
_______________________________
2Fe^3+ + Sn^2+ + 2e- - 2e- -> 2Fe^2+ + Sn^4+

Reply 5

LauraDew

Original post by HPooveli