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 ______.
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.
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.
To balance the equation using the oxidation number method, we need to assign oxidation numbers to each element in the equation.
Let's start by assigning oxidation numbers:
H2O2 + I- → I2 + H2O
Let's assume the oxidation number of hydrogen (H) is +1 and oxygen (O) is -2.
For H2O2, the overall charge is 0, so the sum of the oxidation numbers must be 0. Since there are two hydrogen atoms, each with an oxidation number of +1, the oxidation number of oxygen must be -1 in H2O2.
For I-, the overall charge is -1, so the oxidation number of iodine (I) must be -1.
Next, let's look at the products:
In I2, since it is an element, the oxidation number is 0.
In H2O, the overall charge is 0, so the sum of the oxidation numbers must be 0. Since there are two hydrogen atoms, each with an oxidation number of +1, the oxidation number of oxygen must be -2 in H2O.
Now that we have assigned oxidation numbers, we can identify the element being oxidized and the element being reduced:
The element being oxidized is iodine (I). Its oxidation number changes from -1 in I- to 0 in I2. Thus, iodine is being oxidized.
The element being reduced is oxygen (O). Its oxidation number changes from -1 in H2O2 to -2 in H2O. Thus, oxygen is being reduced.
Now, let's determine the oxidizing and reducing agents:
The oxidizing agent is the species that causes the other species to be oxidized. In this case, H2O2 is causing iodine (I) to be oxidized. So, the oxidizing agent is H2O2.
The reducing agent is the species that causes the other species to be reduced. In this case, I- is causing oxygen (O) to be reduced. So, the reducing agent is I-.
Finally, let's determine the total number of electrons transferred:
To do this, we need to look at the change in oxidation number.
Iodine (I) changes its oxidation number from -1 to 0, which means the oxidation number increases by 1. This means 2 electrons are lost per iodine atom.
Oxygen (O) changes its oxidation number from -1 to -2, which means the oxidation number decreases by 1. This means 2 electrons are gained per oxygen atom.
Since there is one iodine atom and two oxygen atoms involved in the reaction, a total of 2 + 2(2) = 6 electrons are transferred during the reaction.