# Balancing redox reactions help!

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#1
I am doing a worksheet on balancing redox equations and I'm a bit stuck on one of them. The question is: in acid solution dichromate(VI) ion, Cr2O7(2- ) oxidise tin(II) to tin (IV) ions. Chromium (III) ions are also produced.
I have tried using the normal method where you work out the oxidation state change for chromium and tin.

For chromium I got + 6 to +3 so a change of -3. For tin I got +2 to +4 so a change of +2. I multiplied chromium compounds by 2 and tin by 3 and balanced using h2o p, e- and h+ In the normal way. My answer is:
2Cr2O7(2-) + 3Sn(2+) + 28H+ + 6e- ==> 4Cr3+ + 3Sn4+ +14 H2O

Cr2O7(2-) + 3Sn(2+) + 14H+ ==> 2Cr3+ + 3Sn4+ 7 H2O
I understand how to get this by first writing half equations, the making e- equal and cancelling. I don't understand why the normal method doesn't work for this question, and why there are still electrons in the final answer. Please help!!
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3 years ago
#2
(Original post by PuffyPenguin)
For chromium I got + 6 to +3 so a change of -3.
There are 2x Cr atoms in each Cr2O72-, which is a total change of -6 per Cr2O72-.
1
#3
(Original post by Pigster)
There are 2x Cr atoms in each Cr2O72-, which is a total change of -6 per Cr2O72-.
Is the change in oxidation state counted per all atoms of the element, not per each atom then? I didn't know that - thanks!
0
3 years ago
#4
(Original post by PuffyPenguin)
I am doing a worksheet on balancing redox equations and I'm a bit stuck on one of them. The question is: in acid solution dichromate(VI) ion, Cr2O7(2- ) oxidise tin(II) to tin (IV) ions. Chromium (III) ions are also produced.
I have tried using the normal method where you work out the oxidation state change for chromium and tin.

For chromium I got + 6 to +3 so a change of -3. For tin I got +2 to +4 so a change of +2. I multiplied chromium compounds by 2 and tin by 3 and balanced using h2o p, e- and h+ In the normal way. My answer is:
2Cr2O7(2-) + 3Sn(2+) + 28H+ + 6e- ==> 4Cr3+ + 3Sn4+ +14 H2O

Cr2O7(2-) + 3Sn(2+) + 14H+ ==> 2Cr3+ + 3Sn4+ 7 H2O
I understand how to get this by first writing half equations, the making e- equal and cancelling. I don't understand why the normal method doesn't work for this question, and why there are still electrons in the final answer. Please help!!
Your full equation should not have electrons included on either side; all electrons should have been transferred from the reducing agent (dichromate in this case) to the entity being reduced (Sn2+ in this case). If you have electrons remaining in your full equation that should ring alarm bells.

(You would always see electrons in a redox half equation, but never in a full redox equation).
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