Balancing the redox rxn in an acidic solution

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205192823
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Joined: Wed Sep 18, 2019 12:19 am

Balancing the redox rxn in an acidic solution

Postby 205192823 » Thu Mar 12, 2020 6:24 pm

How does this solution get balanced: Cl2O7 (g) =H2O2(aq) ---> ClO2(aq) + O2(g)?

MingdaH 3B
Posts: 133
Joined: Thu Jul 11, 2019 12:17 am

Re: Balancing the redox rxn in an acidic solution

Postby MingdaH 3B » Thu Mar 12, 2020 6:49 pm

Divide it into the partial reactions, where Cl is reduced and O2 is oxidized

Jacob Motawakel
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Re: Balancing the redox rxn in an acidic solution

Postby Jacob Motawakel » Thu Mar 12, 2020 6:59 pm

After separating the redox reactions into two half reactions, you add H2O to balance oxygens in each half reaction, and add H+ to balance hydrogen in each half reaction.

Jared_Yuge
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

Re: Balancing the redox rxn in an acidic solution

Postby Jared_Yuge » Thu Mar 12, 2020 7:00 pm

From their oxidation numbers determine the species being oxidized and reduced and then from their unbalanced equations balance the half reactions, then have the electrons in both equations cancel by multiplying both equations to make them cancel, then you can combine the equations, cancel stuff out, and voila u got the balanced redox equation.

Jared_Yuge
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

Re: Balancing the redox rxn in an acidic solution

Postby Jared_Yuge » Thu Mar 12, 2020 7:02 pm

The difference between balancing in an acidic or a basic solution is when you are balancing H. If you have an acidic solution you just add H+ to the side that needs H+. But with a basic solution you need to add H2O to the side that needs the H+ and OH- to the other side, so that the net difference is just H+

205192823
Posts: 100
Joined: Wed Sep 18, 2019 12:19 am

Re: Balancing the redox rxn in an acidic solution

Postby 205192823 » Thu Mar 12, 2020 7:12 pm

Ohh so is it only H+ that you add in acidic solutions to balance it out and use no H20? Because when I solved this problem I got 6H+ + Cl2O7(g) ---> 2ClO2- (aq) + H2O + 5e-...I got this problem partially wrong and I was wondering why I did.


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