You need to learn to do these by yourself and me doing them will not help you. Here is a site that will help or if you have trouble show what you can do and explain fully what you don't understand about the next step.
Cr goes from +6 on the left(for each) to 3+ on the right. F^- goes from -1 on the left to 0 on the right. The place where students get stuck on this particular type is at the beginning. The first thing you do is Put a 2 for Cr^3+ on the right and a 2 for F^- on the left. Then balance it by the redox method. The reason for that is that you MUST compare equal quantities; i.e., 2 Cr and 2 F.
The site is here.http://www.chemteam.info/Redox/Redox.html
Use the method of half reactions to balance the oxidation half reaction in acid solution as well as provide the oxidized and reduced reactants.
Cr2O7 {2-} (aq) + F{-} (aq) = Cr {3+} (aq) + F2(aq)
Charges are in { }
4 answers
Well thank you very much for the site, it helps! I've worked out the problem myself, please tell me if this looks okay?
Cr2O7{2-} +F-=Cr{3+}+F2
6e^- + Cr2O7^3- + 14H^+ = 2Cr^3+ + 7H2O
2F^- = F2 +4e^-
Then I multiplied the top by 2 and the bottom by 3 to get a GCF for the electron coefficients.
My final product was 6F^- + 2Cr2O7^2- + 28H^+ = 4Cr^3+3F2+14H2O
Cr2O7{2-} +F-=Cr{3+}+F2
6e^- + Cr2O7^3- + 14H^+ = 2Cr^3+ + 7H2O
2F^- = F2 +4e^-
Then I multiplied the top by 2 and the bottom by 3 to get a GCF for the electron coefficients.
My final product was 6F^- + 2Cr2O7^2- + 28H^+ = 4Cr^3+3F2+14H2O
Correction!!
Cr2O7^2-(aq)+ 6F^-(aq)+14H+ = 2Cr^3+(aq)+3F2(aq)+7H2O(l)
Is what I get, I put electrons on the wrong side for F to F2.
Cr2O7^2-(aq)+ 6F^-(aq)+14H+ = 2Cr^3+(aq)+3F2(aq)+7H2O(l)
Is what I get, I put electrons on the wrong side for F to F2.
Yes, that final post looks good. Good work.
2 answers