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Redox reactions

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1. I've been given some redox reactions and I've got to write the balanced overall symbol equation for homework! I did most of the questions a couple of weeks ago, and I've just returned and feel like I've forgotten what to do :P The equations are...

ClO (-) ----> ClO3 (-) AND ClO (-) -------> Cl (-)

2ClO (-) + H20 -------> ClO3 (-) + Cl (-) + 2H (+) + 2e(-)

Can you write an answer like this? I feel like having the two H+ ions and 2 electrons in the products is pointless and I've done something wrong in my working out!

2. Your two half equations are:

ClO- + 2H+ + 2e- -> Cl- + H2O
AND
ClO- + 2H2O -> ClO3- + 4H+ + 4e-

#1 x2 gets rid of the e- problem
3. (Original post by Pigster)

ClO- + 2H+ + 2e- -> Cl- + H2O
AND
ClO- + 2H2O -> ClO3- + 4H+ + 4e-

#1 x2 gets rid of the e- problem
Ahh that makes sense thank you! If I were to do that could I also cancel out the water - as there will be 2H2O on both the reactants' and products' side?
4. (Original post by VetStudentOf2017)
Ahh that makes sense thank you! If I were to do that could I also cancel out the water - as there will be 2H2O on both the reactants' and products' side?
Redox is a mechanical process (working it out, I mean). I.e. follow the steps, follow them in succession and apply them, and you will get the right result

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS
5. (Original post by apronedsamurai)
Redox is a mechanical process (working it out, I mean). I.e. follow the steps, follow them in succession and apply them, and you will get the right result

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS
I see, thanks so much for your help
6. (Original post by VetStudentOf2017)
I see, thanks so much for your help
http://www.sciencegeek.net/APchemist...dox/redox5.htm
7. (Original post by apronedsamurai)
Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS
That looks like you're saying that if you add two half equations: one with water on the LHS and one with water on the RHS, then you'd add the waters...

Which is nonsense. You cancel the waters.

In the OP's case, there are two waters on both side (once you've multiplied to cancel the e-). The overall redox equation has no water in it. NOT four.
8. (Original post by Pigster)
That looks like you're saying that if you add two half equations: one with water on the LHS and one with water on the RHS, then you'd add the waters...

Which is nonsense. You cancel the waters.

In the OP's case, there are two waters on both side (once you've multiplied to cancel the e-). The overall redox equation has no water in it. NOT four.
I didn't say that, and I'd recommend you moderate your tone.

With regards to my previous comment:

"Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS"

That was intended as a hypothetical figure, to illustrate my point, not in reference to the original post or the values contained therein.
9. (Original post by apronedsamurai)
Redox is a mechanical process (working it out, I mean). I.e. follow the steps, follow them in succession and apply them, and you will get the right result

Water will be aggregated, and then placed on ONE side of the equation. So if with the two half equations you have 4 H2O on LHS, and 4 H20 on RHS, there will be 8 H2O total, on either LHS or RHS
No it wouldn't. You simplify the balanced equations, so if your combined equation has an equal amount of water on the reactants and the products, you simplify it down and remove the waters, not just add the products to the reactants or vice versa.

Take the oxidation of ethanol with dichromate:

two half equations:

CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-
Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O

Multiplication is three of the first, two of the second so the combined equation is:

3CH3CH2OH + 3H2O + 2Cr2O72- + 28H+ -> 3CH3COOH + 12H+ + 4Cr3+ + 14H2O

You don't then add the 3 H2O of the left to the 14 of the right, you take them away to give

3CH3CH2OH + 2Cr2O72- + 16H+ -> 3CH3COOH + 4Cr3+ + 11H2O

(Original post by VetStudentOf2017)
Ahh that makes sense thank you! If I were to do that could I also cancel out the water - as there will be 2H2O on both the reactants' and products' side?
Yes you could. You simplify the reaction as much as possible, so if there's 2H2O on both sides you remove them from both sides, if there's 3H2O in the reactants and only 2H2O in the products then you have just the one H2O in the reactants.
10. (Original post by Stiff Little Fingers)
No it wouldn't. You simplify the balanced equations, so if your combined equation has an equal amount of water on the reactants and the products, you simplify it down and remove the waters, not just add the products to the reactants or vice versa.

Take the oxidation of ethanol with dichromate:

two half equations:

CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-
Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O

Multiplication is three of the first, two of the second so the combined equation is:

3CH3CH2OH + 3H2O + 2Cr2O72- + 28H+ -> 3CH3COOH + 12H+ + 4Cr3+ + 14H2O

You don't then add the 3 H2O of the left to the 14 of the right, you take them away to give

3CH3CH2OH + 2Cr2O72- + 16H+ -> 3CH3COOH + 4Cr3+ + 11H2O

Yes you could. You simplify the reaction as much as possible, so if there's 2H2O on both sides you remove them from both sides, if there's 3H2O in the reactants and only 2H2O in the products then you have just the one H2O in the reactants.
Was going with what I was shown here :\

http://www.sciencegeek.net/APchemist...dox/redox5.htm

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