H2O2 + I- → I2 + H2O (basic conditions)
I will do one in detail and leave the rest for you.
1. First make sure the elements being oxidized/reduced are made equal; i.e., for H2O2 ==> H2O that will be H2O2 ==> 2H2O and for I^- to I2 that will be 2I^- ==> I2.
2. Next determine which elements are oxidized/reduced. For H2O2 that is oxygen. Write the half equation and show electrons lost or gained this way.
H2O2 ==> 2H2O. 2 O changes from 2- on the left to 4- on the right so the equation is H2O2 + 2e ==> 2H2O
3. Add the charge on the left and right and add OH^- (basic solution) to the appropriate side along with H2O to balanced like this.
4. The charge on the left is 2- from the electrons. On right is is zero so add 2 OH^- on the right to balance the charge
4a. H2O2 + 2e ==> 2H2O + 2OH^- for the OH^-, then add water
4b. H2O2 + 2e + 2H2O ==> 2H2O + 2OH^-
4c, I like to check this to make sure it's balanced. Atoms balance, electron change balances, charge balanced.
4d. So this half equation is done.
You can do the I. I left the easy one for you. Just follow the rules I laid out above. After you've finished the I^- --> I2, the multiply the I^==> I2 half equation and the H2O2 half equation by whatever number you need to make the electrons in the two half equations equal, then add everything together.
To answer the others.
The element oxidized: oxidation is the loss of electrons.
The elements reduced: reduction is the gain of electrons.
Those two are all you ever need to worry about BECAUSE you ALWAYS know the element oxidized is the reducing agent.
The element reduced is the oxidizing agent.
Post your work if you run into trouble. Balancing redox equations is one of the easiest things you will ever do in chemistry BECAUSE the rules and steps leaves all of the guess work out of it. No thinking to do after you get the hang of it.
Can someone help?
For the reaction below, use the oxidation number method to balance the equation under basic conditions. SHOW ALL YOUR WORK. Complete the statements at the end.
H2O2 + I- → I2 + H2O (basic conditions)
The element oxidized is ______ and the oxidizing agent is __________.
The element reduced is ______ and the reducing agent is __________.
The total number of electrons transferred during the reaction is ______.
3 answers
Should I find the oxidizing agent and reducing agent before or after I balance the equation? Also thank you!!
It doesn't matter. In a redox equation the loss of e is oxidation and gain of electrons is reduction. Similarly, the element losing electrons (oxidation) is the reducing agent and the element gaining electrons (reduction) is the oxidizing agent. In fact from what I did you already know that oxygen changed from 1- on the left for each O to 2- on the right for each O. so O had to gain electrons. That makes it reduced and the oxidizing agent. So you know I^- must be oxidized before you even write the equation. So I^- goes from 1- for each I on the left to zero on the right So you know that must be oxidation and the reducing agent. By the way if you go far enough in redox equations there are some where you have two materials being oxidized or two materials being reduced in the same equation. Those are a little more challenging but a lot of fun to do.
1 answer