[Shaun1979]

Hi,

I understand redox reactions mostly, but I cannot seem to get this:

Question: Balance the equation for the reaction of dichromate ion with lead(II) ion to produce chromium(III) and other products.
Hint: The reaction takes place in acidic medium.

This is what I came up with for one of the half reactions:

Cr2O7(2-) = Cr(3+)

Cr2O7(2-) = 2Cr(3+)

Cr2O7(2-) = 2Cr(3+) + 7H2O

14H(+) + Cr2O7(2-) = 2Cr(3+) + 7H2O

6e- + 14H(+) + Cr2O7(2-) = 2Cr(3+) + 7H2O


That seems to make sense to me (unless I am wrong) .  but what I cannot seem to fathom is what these "other products" are that they are referring to in the question? Without which I can't do the other half? Maybe I am reading too much into it, but my textbook doesn't mention anything of the sort.

Any help would be greatly appreciated.

[Borek]

Water is already other product

Chromium got reduced, it reacted with Pb(II) - so the Pb(II) had to be oxidzed...

[Shaun1979]

Thanks Borek!

That makes sense to me, I just don't understand how to go about writing that as a half reaction?

[AWK]

As far as I know this reaction is impossible but oxidation of Pb(II) to Pb(IV) can be performed in strongly alkaline solution with H₂O₂ or Br₂.