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 { }
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
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
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.
Yes, that final post looks good. Good work.
To balance the oxidation half-reaction using the method of half-reactions, follow these steps:
Step 1: Write the unbalanced half-reactions.
The given equation is:
Cr2O7 {2-} (aq) + F{-} (aq) = Cr {3+} (aq) + F2(aq)
For the oxidation half-reaction, identify the species that is being oxidized. In this case, it is F{-}, which is converted into F2. Therefore, the oxidation half-reaction is:
F{-} (aq) → F2(aq)
Step 2: Balance the element being oxidized.
To balance the number of atoms of the element being oxidized, add water (H2O) to the side that is deficient in oxygen. In this case, the left side is deficient in oxygen, so add water to that side:
F{-} (aq) → F2(aq) + H2O
Step 3: Balance the oxygen atoms.
To balance the oxygen atoms, add hydrogen ions (H+) to the side that is deficient in hydrogen. In this case, the right side is deficient in hydrogen, so add hydrogen ions to that side:
F{-} (aq) → F2(aq) + H2O + 2H+
Step 4: Balance the charge.
To balance the charge, add electrons (e-) to the side that is deficient in negative charge. In this case, the left side is deficient in negative charge, so add the appropriate number of electrons:
F{-} (aq) + e- → F2(aq) + H2O + 2H+
Therefore, the balanced oxidation half-reaction in acid solution is:
2F{-} (aq) → F2(aq) + 2H2O + 2H+
The oxidized reactant is F{-} (aq), and it gets oxidized to F2(aq).