I'm gonna cry in a minute so please help!

[Fallacy202]

Okay, I'm trying to balance redox equations, but the half-equation method DOES NOT WORK!!!!

Say, for this equation:

[math]

\ce{IO_3^-+I^- ->[\text{acid}]I_3^-}

[/math]

 

Okay, so for the oxidation half:

[math]

\ce{I^- -> I_3^-}

[/math]

 

How does one balance this??

 

And the reduction?

 

[math]

\ce{IO_3^- -> I_3^-}

[/math]

 

The I in IO3 is reduced, but the oxygen is oxidized. How do I get around that?? Anyhow it just doesn't work and I've stared at this question for about 6 hours still can't figure it out I'm about to eat my own hw plz help!!!

[jdurg]

The Half-Reaction method doesn't always work. What is the initial equation that has to be balanced?

[Fallacy202]

The equation my textbook gave me was to balance this redox reaction in acidic environment:

[math]

\ce{IO_3^- + I^- ->[\text{Acid}]I_3^-}

[/math]

[jdurg]

This is a bit of a tricky one because the reaction of iodate and iodide in an acidic environment does NOT make the triiodide ion. Instead it makes elemental iodine which THEN combines with excess iodide to form the triiodide ion.

 

So the first thing you have to do is alter your reaction to:

 

IO3(-) + I(-) => I2(s)

 

This is done in an acidic environment, so the H+ ions from the acid will take care of the oxygen atoms from the iodate ion and form some water. This leaves you with a reaction of:

 

IO3(-) + 5I(-) + 6H(+) => 3I2(s) + 3H2O

 

With an excess of iodide (I-) in solution, you'll get some triiodide forming. Therefore, if we want all of that I2 to become I3-, just add three more I(-) to the equation to give you an overall equation of:

 

IO3(-) + 8I(-) + 6H(+) => 3I3(-) + 3H2O