[Baydon Jefferies]

Hi could someone please point me in the right direction on this redox question?

Question: Write half equations, and a balanced redox equation, for the reaction between an acidic solution of hypochlorite ion and ferrous iron
             
OCl- / Cl-           Fe²⁺ / Fe³⁺

So far I think I have the half equation for the ferrous ion ( Fe²⁺ +e^- ---- Fe³⁺) but I do not understand the hypochlorite side or how or what the final product would be (Fe(ClO)₂??)
Any suggestions would be greatly appreciated.

[Borek]

You have chloride (Cl^-) listed as your product.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Scroll down to see what to do with the second halfreaction.

[Baydon Jefferies]

This forum is excellent for correspondence students, thanks a bunch for that tip. Do you think I'm on the right track with this?


OCl? + H₂O  +2e--------- Cl^- + 2OH-
Fe²⁺ --------------------------Fe³⁺ +1e-   x 2
---------------------------------------------------------------------------------
OCl^- +H₂O + 2Fe²⁺ ---------CL^- +2OH- + 2Fe³⁺

Redox equation: OCL^- + H₂O+ 2Fe²⁺ ---------CL^- + 2OH^- + 2Fe³⁺

[Borek]

Looks OK, just watch this capital L in chlorine.

[UnintentionalChaos]

You said it is occuring in acidic solution, so what you have needs a slight adjustment.
The basic protocol for these equations is to balance non oxygen and non hydrogen atoms first.

OCl- ------> Cl-

Chlorine is balanced on both sides. Next, you balance oxygen by adding water.

OCl- -----> Cl- + H2O

Then, hydrogen is balanced with H+ ions

2H+ + OCl- ------> Cl- + H2O

Lastly, charge is balanced with e-. The left hand side of the equation has a charge of +1 and the right has -1. Add two electrons to the left hand side to make it -1 as well.

2H+ + OCl- + 2e- ----> Cl- +H2O

Next, the iron ion. The non-oxygen and non-hydrogen elements are balanced already.

Fe(2+) -------> Fe(3+)

Since there are no loose H or O, proceed directly to electrons.

Fe(2+) -------> Fe(3+) + e-

The equations are then multiplied to balance electrons on both sides so they cancel out when added. Add the equations together.

2Fe(2+) + 2H+ + OCl- -----> 2Fe(3+) + Cl- +H2O

Since this is occuring in acid solution, the H+ are left alone on the reactants side of the equation and you are done.

If this was taking place in basic solution, add a hydroxide to both sides to eliminate all of the H+ as water (In this case, 2OH- to both sides).

2Fe(2+) + 2H2O + OCl- ----> 2Fe(3+) + Cl- +H2O +2OH-

Eliminate the repeated waters to finish up the equation for basic solution.

2Fe(2+) + H2O + OCl- ----> 2Fe(3+) + Cl- +2OH-

This is the basic method that I was taught. The individual steps help a lot with understanding what to do.