posted by Marissa .
Balance this equations for redox reactions in basic solution.
Fe(OH)3(s) + OCl- ¨ FeO42- + Cl-
If you don't know the system for balancing redox equations you need to learn it. This is what I use and I will give you one of the equation.
Step 1. Writ the half equation.
OCl^- ==> Cl^-
Step 2. Write oxidations states of elements and add electrons to account for change in oxidation state.
Cl is +1 on left and -1 on right; therefore, add 2electrons to the left side like so.
2e + OCl^- ==> Cl^-
Step 3. Count charge and add OH (in this case but add H^+ for acidic solutions) to the appropriate side.
charge on left is -3; on right is -1 which means add 2OH to right side.
2e + OCl^- ==> Cl^- + 2OH^-
Step 4. Add H2O (usually to the opposite side where OH or H were added).
H2O + 2e + OCl^- ==> Cl^- + 2OH^-
Step 5. Check to make sure it is balanced. ALWAYS CHECK IT.
You do the Fe the same way. It will be easier to do, I think, if you separate the half equation like this.
Fe^+3 ==> FeO4^-2
I got the following, but I'm not sure if I did it correctly:
2Fe(OH)3 + 3OCl- ->2FeO4-2 + 3Cl- + 4H+ + H2O
If someone could let me know whether I solved it correctly or not, that'd be nice. :)
no its not right, its a basic solution so no H+ but OH-