How do you balance oxidation-reduction reactions. I have all the steps, but I don't understand.
If you have all the steps it should be just a matter of applying them. If you don't understand how to follow those directions I don't know that it will do much good for me to give you another set, or an extended set, of directions which might be just a difficult to follow. So let's do this.
You write the equation to be balanced, then write the first step and tell me what you don't understand about it. I will get you through it step by step. That may be the easiest way since we are not face to face with a black board. By the way, let me know if these are molecular equations to be balanced or if you are to split them up into the reduction part and the oxidation part, balance them, then add them to obtain the final equation.
Oxidation reduction reactions. Assume all reactions occur in aqueous solutions.
H2O2 + Ce(SO4)3 ---> Ce2(SO4)3 + H2SO4 + O2
The first thing I did was write out all compounds.
O2 -1 ---> O2 + 2e-
Ce +3 ---> 2Ce + ?
2SO4 -2 ---> 3SO4 + ?
I'm not sure what to do now and exactly where the numbers come from
Are you sure Ce(SO4)3 is correct on the left. Could that be Ce(SO4)2?
You have a good start. In small numbers, write the oxidation state of each element above it, from left to right.
For example, for H2O2, above H put +1 each and above that put +2 (since there are two of them). Above oxygen, the top number is -2 and the bottom number is -1 each.
For Ce2(SO4)3 on the right. Since sulfate is -2, then -2 goes just above SO4 and -6 goes above that since 3 x -2 = -6. That makes the total for Ce, the top number, +6 and each is +3. I would not break SO4 down to give number for S and for O since you can see it is SO4 on the left and SO4 on the right; therefore, it isn't involved in the redox part of the equation. We can make those polyatomic ions balance later.
Here is a site that gives instructions on how to determine oxidation states if you need any help.
(Broken Link Removed)
I can't go further until I know the formula for Ce sulfate on the left side. I think it must be Ce(SO4)2 in which you made a typo.
Balancing oxidation-reduction (redox) reactions can be a bit challenging at first, but with a systematic approach, it becomes easier to understand. Here is a step-by-step guide to help you balance redox reactions:
Step 1: Identify the species that is being oxidized and the species being reduced. Oxidation involves losing electrons, while reduction involves gaining electrons.
Step 2: Write down the half-reactions for oxidation and reduction separately. In a balanced redox reaction, both half-reactions should have the same number of atoms of each element on both sides of the equation.
Step 3: Balance the atoms in each half-reaction by adding coefficients where necessary. Start with the elements other than Oxygen and Hydrogen and then balance Oxygen atoms by adding water (H2O) molecules and Hydrogen atoms by adding protons (H+).
Step 4: Balance the charge in each half-reaction by adding or subtracting electrons (e-). The number of electrons gained in the reduction half-reaction should be equal to the number of electrons lost in the oxidation half-reaction.
Step 5: Multiply each half-reaction by the appropriate factor to equalize the number of electrons transferred. This is done to ensure that the total change in electrons is equal in both half-reactions.
Step 6: Add the two balanced half-reactions together, canceling out any common terms. Make sure that the atom balances, charge balances, and electron transfer are consistent for both half-reactions.
Step 7: Check if the final equation is balanced by counting the number of atoms of each element and confirming that the charge is balanced.
Remember, practice is key when it comes to balancing redox reactions. The more you practice, the better you will become at recognizing patterns and applying the steps effectively. There are also online resources and software available that can help you practice and solve more complex redox reactions.