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How to get the balanced redox equation in this sample problem?

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Summary: Our chemistry teacher gave a sample problem regarding balancing a redox equation. I managed to proceed to a considerable distance, but I can't seem to understand how to get the final balanced equation. Please help me out.

The problem is; reaction of K2Cr2O7 with SO2 in the presence of H2SO4(aq) giving Cr3+ ions and SO4 ions as main products.

I could complete the following steps :

Oxidation half reaction : SO2 ------------------ > SO4
Reduction half reaction : Cr2O72- ------------------- > Cr3+

2H2O + SO2 ------------------ > SO4 + 4H+ + 2e-

6e- + 14H+ + Cr2O72- ------------------- > 2Cr3+ + 7H2O


Then I multiplied the "2H2O + SO2 ------------------ > SO4 + 4H+ + 2e-" equation by three so that I can add the both equations by equalizing the electron amount.

Then I added the both like this....

6H2O + 3SO2 + 14H+ + Cr2O72- ------ > 3SO4 + 12H+ + 2Cr3+ + 7H2O

Then simplified it into this :

2H+ + Cr2O72- + 3SO2 -----> 2Cr3+ + 3SO4 + H2O

But the ultimate balanced equation that I should get is,

H2SO4(aq) + K2Cr2O7(aq) + 3SO2(g) ------ > Cr2(SO4)3(aq) + K2SO4(aq) + H2O(aq)

right?

How can I proceed to get the ultimate balanced equation? I can't understand how to do it. Please help me on this. Thank you very much in advance.
 

Borek

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You need to add a bunch of spectators. Start by writing chromium sulfate on the right in the molecular form, then add potassium cations and sulfate anion on the left to convert ions to other molecules, and the same amount on the right to keep the reaction balanced.
 
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You need to add a bunch of spectators. Start by writing chromium sulfate on the right in the molecular form, then add potassium cations and sulfate anion on the left to convert ions to other molecules, and the same amount on the right to keep the reaction balanced.
What are spectators? And why should we add them?
 

Borek

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Do you know the difference between net ionic reactions and complete reactions?
 
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Do you know the difference between net ionic reactions and complete reactions?
Yes. Just after you mentioned, I did a small research and found out about the both types of reaction equations. I have a small idea about spectators now. They are just like normal conventional spectators in a game show, they are watching at what happens in a reaction, and also with them only, the ones that react to form a complete product come with. Isn't it?

So why do we need to add a couple of spectators here?
 

symbolipoint

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Oxidation half reaction : SO2 ------------------ > SO42ˉ
Reduction half reaction : Cr2O72- ------------------- > Cr3+
Having not yet checked to see what you took care of yet, my suggestion with these two half-reactions is this:

  1. Balance each of the two for at least the atom which changes in charge; and then include the necessary number of electrons in each half-reaction.
  2. Now balance the two half reactions together for the number of electrons. What this is supposed to mean, is to make sure to adjust one or both half reactions so they each use the same number of electrons.

Once that is done, you can take care of the rest of the balancing of atoms or ions.
 

Borek

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27,963
2,470
So why do we need to add a couple of spectators here?
When you compare the reaction you are asked to give as a final solution with the reaction you already have, it is apparent spectators are missing. For example - you should have K on both sides, as a cation in a salt. There are no K+ in the net ionic reaction that you have balanced - so you need to add them. As long as you add the same number of them on both sides, reaction will be still balanced.

This is just the opposite of the process of preparing the net ionic reaction - typically you start with a full reaction and cancel out ions/moles that didn't change. Here you have to bring them back.
 
50
10
When you compare the reaction you are asked to give as a final solution with the reaction you already have, it is apparent spectators are missing. For example - you should have K on both sides, as a cation in a salt. There are no K+ in the net ionic reaction that you have balanced - so you need to add them. As long as you add the same number of them on both sides, reaction will be still balanced.

This is just the opposite of the process of preparing the net ionic reaction - typically you start with a full reaction and cancel out ions/moles that didn't change. Here you have to bring them back.
Thank you for showing me the way. Really appreciate it @Borek ....
 
50
10
Having not yet checked to see what you took care of yet, my suggestion with these two half-reactions is this:

  1. Balance each of the two for at least the atom which changes in charge; and then include the necessary number of electrons in each half-reaction.
  2. Now balance the two half reactions together for the number of electrons. What this is supposed to mean, is to make sure to adjust one or both half reactions so they each use the same number of electrons.

Once that is done, you can take care of the rest of the balancing of atoms or ions.
Thank you @symbolipoint
 

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