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Worked example: Balancing a redox equation in acidic solution

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)
When balancing equations for redox reactions occurring in acidic solution, it is often necessary to add H⁺ ions or the H⁺/H₂O pair to fully balance the equation. In this video, we'll walk through this process for the reaction between dichromate (Cr₂O₇²⁻) and chloride (Cl⁻) ions in acidic solution.. Created by Jay.

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• How do we balance the reactions if "Spectator ions" are also present in the redox reaction?
• We leave out the spectator ions from the reaction.
How do I identify whether the reaction is in acid or base?
• Look at the basic laws of the chemical reaction is it losing or gaining a hydrogen, if an atom is losing a hydrogen it is going to be an acid in which it becomes a conjugate base and if it is gaining a hydrogen atom it is a base in which it becomes a conjugate acid.
• Where are they getting the water (or the H+), where do the electrons go when something is oxidized, and what is the difference between oxidation states and charges?
• The water is from the reaction taking place in an acidic solution - which basically (no pun intended) requires the reaction to occur in water. The H+ ions come from the acidic solution as well, as they are present in all acidic solutions.
• How do we balance a reaction in which the same element disproportionates and gets reduced and oxidised ?
(1 vote)
• You write the same element in both half-reactions. If your reaction in basic solution is
Cl₂ → Cl⁻ + ClO₃⁻
you write the half-reactions as

5×[Cl₂ +2e⁻ → 2Cl⁻]
1×[Cl₂ + 6H₂O → 2ClO₃⁻ + 12 H⁺ + 10e⁻]

6Cl₂ + 6H₂O→ 10Cl⁻ + 2ClO₃⁻ + 12H⁺
12 H⁺ + 12OH⁻ →12 H₂O

6Cl₂ + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 6H₂O
• At why is it that the charge of the oxygen totals to 2-? I thought that because there are 7 that it would be -14? Could someone explain this to me please?
• I had the same question when I first watched this video. However, I think it is because the molecule has a 2- charge, not necessarily each individual oxygen atom within that molecule. I do believe, however, that each individual oxygen atom does have a 2- oxidation state, which would total 14-, but then each chromium atom has a 6+ oxidation state, cancelling out 12 of those negative charges (because there are two chromium atoms), thus resulting in the 2- overall charge of the molecule.
Someone please correct me if I'm wrong!

Still confusing the way it was explained though, I agree.
• At why are we taking hole molecul of chromate for half reaction? Why not individually Cr2 and O7.
Thanks.
• Cr2O7 (dichromate) is a polyatomic ion. It was bonded to something else but dissolved in a solution. It is easier to think of polyatomics as single units. Since it is already 1 ion, it cannot ionize, or split, anymore.
• Why we have to balance redox reaction this way why we don't just get it balanced like we balance other equations?
• It is to remove the redundancies in an equation. In normal algebraic equations, redundancies like your typical leftover "x" or "y" are easy to remove because there are no electrons that change the quantifiable identity of each variable, if that makes sense. In order to keep track of electrons, we must be extra careful and account for charge of an atom. For example, Na+ is not the same as Na (the neutral one has an extra electron, which must be accounted for).
• I thought that the redox half of the reaction was where the substance gained electrons and oxidation reaction was where the substance lost charge?
• I normally remember it like " Eat electron and get reduce " It always work when you are confused :-)
• Thank you very much for this clear, excellent explanation!

I still have a question about a specific reaction. It is MnO4 - + H2C2O4 --> Mn2+ + CO2

I can see that MnO4- --> Mn2+ is the oxidation, since electrons are lost. But what happens with the other reaction? I'm confused, since every oxidation requires a reduction, but there are no charges on the other parts.

For the first reaction, I'd write the following: 8 H+ + MnO4- + 5e- --> Mn2+ + 4H2O

But I have no idea how to continue. How can I proceed in such a case as it seems to differ from the equation you show in the video? Thanks for your help!
• You are wrong from the very beginning!
Mn04- ----> Mn2+ is a reduction reaction (** Since oxidation number of Mn in MnO4- is +7 & in Mn2+ is +2). It can be seen that it is a reduction reaction since it has gained electron (+7 ---> +2)

And H2C2O4 ----> CO2 is oxidation reaction (** Since oxidation number of C in H2C2O4 is +3 & in CO2 it is +4). So its oxidation as electron is lost. (+3 ---> +4)

Basicallly the charges doesnt matter in the beginning...
but as you go on you have to balance the charges.

you determine whether it is oxidation or reduction by change in oxidation number not the charges
• At -
Why is 12 added to the chromium? Where do you find the 12 being added?