[Fablowable]

Hello folks,

this is my first post and I have a question about a redox reaction.

Citric acid reacts with Zinc ---> Zinc(II)citrate + Hydrogen

I have come this far but I am totally confused as to how to proceed.

                          h8c6o7 + Zn ---->Zn(h6c607) + H2

Oxidation numbers: 1/6/-2     0         +2 for Zn       0

Can you please tell me how to continue?

Thanks

[AWK]

This is a single displacement reactionin which citrate^3-, hydrogen citrate^2- or dihydrogen citrate^- (3 reaction are possible depending on stoichiometry) do not change oxidation state of any atom.
There is much better balancing this reaction:
H⁺ + Zn = Zn²⁺ + H₂
 then adjusting coefficients of this reaction to stoichiometry of zinc reaction with citric acid.

[Fablowable]

Thanks for the explanation!

This is the solution presented, however it just confuses me and I dont seem to understand how I can get there.


4 C3H4OH(COOH)3 + 6 Zn ⇄   2  Zn3(C3H4OH(COO)3)2  + 3 H2

[Borek]

Thanks for the explanation!

This is the solution presented, however it just confuses me and I dont seem to understand how I can get there.


4 C3H4OH(COOH)3 + 6 Zn ⇄   2  Zn3(C3H4OH(COO)3)2  + 3 H2

C3H4OH(COOH)3 is just a complicated (although perfectly correct) way of presenting the citric acid. Would it be easier if I told you to write it as H₃Citrate (which reflects the fact citric acid is triprotic)?

[AWK]

4 C3H4OH(COOH)3 + 6 Zn ⇄   2  Zn3(C3H4OH(COO)3)2  + 3 H2

Something wrong with coefficients (almost all)

[Vidya]

only ionizable H will be reduced and one mole of citric acid has there moles of ionizable H which can give you 3/2 moles of H2
3H⁺ +3e ------> 3/2H2
It means
6H⁺ + 6e- -----> 3H2
Zn ----> Zn²⁺ + 2 e -
3Zn ----> Zn²⁺ + 6 e -

now you can add these reactions to get final equation