[Bat3996]

I am having some serious trouble with a redox equation in my chemistry homework and would greatly appreciate some help.

The equation is as follows:

 H2O2(aq)   +   I¯ (aq)  -->  IO3¯ (aq)

The redox equation is taking place under basic condition. I really don't know what to do with the IO3-. In all of my teachers examples there were 2 products that formed the oxidation reduction, and not just one. How do I split the equation into 2 parts with only one product. Please help.

[AWK]

Write down OH- ions on the left side. Water is a usual product of oxidation on the right side.

[Bat3996]

Something like this then:

H2O2(aq)   +   I¯ (aq)  -->  IO3¯ (aq) goes to 2OH¯ + H2O2 + I¯ --> IO3¯ + 2H20

?

Thanks for your help.

[Nobby]

Somebody erased my further post.

Solving of Redox equations:

Reduction: H₂O₂ + 2 e^- => 2 OH^-


Oxidation: I^- + 6 OH^- => IO₃^- + 3 H₂O + 6 e^-

First equation x 3

3 H₂O₂ + 6 e^- => 6 OH^-

Addition of both equations and elimination of electrons and OH^-

3 H₂O₂ + I^- => IO₃^- + 3 H₂O

[Bat3996]

Thanks, but can you explain where you got the 2 OH- from in the reduction step. What was on the right when you started that.

Also what was the original equation for the oxidation step. If it was I- -> IO3- why did you add 3 h20s to the right?

[Nobby]

Simple receipe for solving redox equations:

Alkalaine condition:

Reduction left side add water right side add OH^-

Oxidation left side add OH^- and right side add water

in our example the Peroxide creates 2 OH^-

acidic condition

Reduction: left side add H⁺ and right side add water

Oxidation: left side add water and right side add H⁺

this example:

H₂O₂ + 2 H⁺ + 2 e^- => 2 H₂O


I^- + 3 H₂O => IO₃^- + 6 H⁺ + 6 e^-

again first equation x3


3 H₂O₂ + 6 H⁺ + 6 e^- => 6 H₂O

Addition and elimination

3 H₂O₂ + I ^- => IO₃^- + 3 H₂O

[Bat3996]

I still don't understand. Why does Hydrogen Peroxide create 2 OH-s?

In my next problem  NO3¯ (aq)  +   NH3(aq)   =>   NO2¯ (aq)

Nitrogen's oxidation number changes 3 times and N is the only thing that changes in the entire equation o.O. How do I know how many OHs to add. Can I have the N from NO3 being the reduction part of the equation and the N from the NH3 be the oxidation part of the equation both going to NO2-?

Also for the synthesis of the half equations if there is only 1 product how do you determine what to do with th eproducts in the half equations,

[Nobby]

H₂O₂  equal HO-OH , this decompose to 2 OH^-

Reduction

NO₃^- +  H₂O + 2 e^-=> NO₂^- + 2 OH^-

Oxidation

NH₃ + 7 OH^- => NO₂^- + 5 H₂O + 6 e^-

first equation x 3

3 NO₃^- +  3 H₂O + 6 e^-=> 3 NO₂^- + 6 OH^-

Addition

3 NO₃^- + NH₃ + OH^- => 4 NO₂^- + 2 H₂O

[Bat3996]

In the second problem both of your equation had the product of NO2-, but in the first problem only one of them involved IO3-. Why is that?

Also I want to understand why I am adding H2O to the right and OH to the left in the oxidation step.

[Nobby]

In the first problem

we have the reduction of Peroxide to Water and the Oxidation of Iodine to Iodate.
4 different compounds!!

In the second problem we have the Reduction of Nitrate to Nitrite and the Oxidation of Ammonia to Nitrite.
 3 Different compounds!!!

It is called also Syn-Proportion

The addition of Water OH^- or H⁺ I explained above. It is a kind of law to do so.

[Bat3996]

How did you know Peroxide was going to water. It didn't say that it was going into water in the equation.

[Nobby]

That is practical expierence. Peroxide can be reduced to water or oxidised to Oxygen.

So if Iodine get oxidised, then peroxide can only be reduced.

[Bat3996]

It's ok, theres no way I can figure this out anyway, but I figured it was against the rules.

[Borek]

theres no way I can figure this out anyway

It may look hard at first, but once you get used to it, it is not that difficult.

In redox reactions water, OH^- and H⁺ are always used to balance charge, hydrogen and oxygen. When you are told reaction takes place in acidic solution, you use water and H⁺, when it takes place in basic solution, you use water and OH^-. It requires some experience and training, but the basic idea is simple.

For example if you need oxygen atom on the left hand side (LHS) of the equation and reaction takes place in acidic solution, you can add water molecule on the left and two H⁺ on the right - net difference in atoms is just one oxygen on the left (charge is not balanced, but that's OK - you will balance it using electrons later).

If you need oxygen atom on the LHS and solution is basic, you can add two OH^- on the left and water molecule on the right - again, net effect in atoms is that there is one oxygen atom left on the left.

Reactions with hydrogen peroxide are not much different - this is the same juggling with H₂O, H⁺ and OH^-.

You can try to polish your skills here: http://www.chemistry-quizzes.info/quizz.php?m=n&s=equation-balancing-redox

[Bat3996]

Well my teacher told us he meant to put that on the acid worksheet and not the base worksheet which is why I probably still have no idea what's going on