DrBob222 you were so helpful earlier in assuring me I was on the right track with the ionic equations. Your very gifted in chemistry. I have spent from a short time after that post until now banging my head against the wall trying to figure out how to balance an oxidation reduction reaction then show half reactions. I have looked at many examples and when I calculate one of my problems, I get the same charges on both sides. I realize I don't know how to determine if the charge changes. In the powerpoint I have, it shows tie lines and different charges on the products - how does one know this? Below is one example I have been working on: Original problem reads Fluorine reacts with water to yield hydrofluoric acid and oxygen. So my initial set up looks like this:
F(s)+H2O(l)=HF(aq)+O2(g) Note-the state of fluorine was not provided so I assumed it's a solid but really don't know how it was intended in this reaction. So in my notes I have step 1 is write the unbalanced equation, which is what I wrote above. Step 2, determine oxidation state of all. This, I believe is where I'm going wrong initially because I have F(-1), H(+1)and I'm not sure if I multiply that by the 2 subscript or not, then I have O(-2) on the left I have the same H(+1), F(-1) O(-2) and again unsure if I multiply by the subscript 2. So without understanding how to tell a charge has changed, of course they are going to look the same on both sides - unless I do multiply by the subscripts, then it looks like the products would be overall more negative. In the event you can help get this through to me, I will go ahead and list the other steps because I have questions about them also. Step 3, show electrons using tie lines. Great, but I don't know which ones I tie together - is it just any that have changed their oxidation state number? Step 4, use coefficients to equalize - not sure about this step either then step 5 is to balance. Phew. That's just a lot of confusion. Hope you can help by explaining and using my example then I will try to figure out my others.
Let me forget the steps you go through and go through your problem step by step; then you ask questions if needed. But I think all will be clear. First, F is a gas. The periodic table consists of 5 gases (+ the noble gases but let's not worry about them right now), 2 liquids, all others are solid. That makes it easy.
H, N, O, F, Cl are gases. Look at the periodic table and see where they are. Move the H over mentally to the C spot and you see it looks like this.
H N O F
with Cl under the F. And ALL of these are written as H2, N2, O2, F2, Cl2 (you also have Br2 and I2 below them) but those are not gases. Br2 and Hg are the only two liquids at room T. Everything else is a solid.
F2(g) + H2O(l)= HF(aq) + O2(g)
Here are the rules for determining oxidation state.
http://www.chemteam.info/Redox/Redox-Rules.html
F2 is zero (EACH F is zero), H is +1 EACH and O in H2O is -2. On the right side, H is +1 for EACH H and F is -1 for EACH F. On the right oxygen is zero for each. So F has changed from zero to -1 for EACH F and O has changed from -2 to zero for EACH O. The next step, and most students leave it out and wonder why the equation won't balance, is to make sure you are talking about the same number of atoms that have changed; i.e., F2 goes to 2HF (you add that 2 as a preliminary number (it may change later) and 2H2O goes to O2 (add that 2 to the H2O and that is temporary also. It may change later).
So 2F in F2 is zero going to 2HF at -2 for BOTH F ions. That is a gain of 2 electrons. Draw a line from F2 to 2HF and write in gain of 2e. Then 2O atoms in H2O is -4 (O is -2 EACH) and it goes to zero. Draw a line connecting those and write in loss 4e.
Next step. Make the electron loss = electron gain. You must multiply the 2e gain x 2 to make 4e gained and multiply the 4e lost by 1 to make 4e lost.
Next step. Where you have those connecting lines, multiply those coefficients by the multiplier factor you used so its 1 x 1 = 1 for the O2 and 1 x 2 = 2 for the H2O. That 2 we placed there for the 2H2O as a temporary number is now a solid number. Then 2 x 1F2 = 2 for F2 and 2 x 2HF = 4HF (those coefficients DID change. Those now are solid numbers).
Next step. Now you check everything to make sure things are balanced. There are three things to check.
1. The atoms must balance. I have 4 F on both sides; 4 H on both sides; and 2 O on each side. Atoms balance.
2. Charge must balance. I have zero charge on left and zero on right. OK.
3. The electron loss must equal electron gain. That's ok.
Equation is done.
And I'm not gifted; I've been doing this for awhile and practice makes perfect although I'm not perfect yet but I'm working on it. :-)
I'm glad to hear that I was able to help you earlier with the ionic equations. I understand that balancing oxidation-reduction (redox) reactions can be quite confusing, but I'll do my best to explain the steps to you using your example of the reaction between fluorine and water.
Step 1: Write the unbalanced equation.
You have correctly written the unbalanced equation as:
F(s) + H2O(l) = HF(aq) + O2(g)
Step 2: Determine the oxidation states of all elements.
The oxidation state (also known as the oxidation number) is an indication of the charge that an atom would have if all of the shared electrons in a compound were assigned to the more electronegative atom in each bond. In this case, the oxidation state of F (fluorine) is -1, and the oxidation state of H is +1. Oxygen (O) usually has an oxidation state of -2 when in compounds, but in this case, it's bonded to a more electronegative fluorine atom, so it has an oxidation state of -1.
However, you make a good point about the subscripts. The subscripts in a chemical formula indicate the number of atoms of each element in the compound, but they do not affect the oxidation state. So you should not multiply the oxidation state of an element by the subscript. In your case, F has an oxidation state of -1, H has an oxidation state of +1, and O has an oxidation state of -2, regardless of the subscripts.
So, both sides of the equation have F(-1), H(+1), and O(-2), and this indicates that the charges are balanced. However, charges alone might not give you the complete picture in a redox reaction.
Step 3: Show electrons using tie lines.
In a redox reaction, electrons are transferred from the reducing agent (the species that gets oxidized) to the oxidizing agent (the species that gets reduced). To represent this, you can use tie lines, which connect species that are involved in an electron transfer.
In this case, fluorine is being reduced from an oxidation state of 0 to -1, meaning it is gaining an electron. On the other hand, oxygen is being oxidized from an oxidation state of -1 to 0, meaning it is losing an electron. So you can represent this transfer of electrons with a tie line connecting F and O.
F(s) + H2O(l) = HF(aq) + O2(g)
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Electron transfer
Step 4: Use coefficients to equalize the number of electrons transferred.
The number of electrons transferred in the redox reaction should be the same for the reducing agent and the oxidizing agent. So, in this case, you need to balance the number of electrons transferred between F and O.
Since F is gaining one electron and O is losing two electrons, you can multiply the equation by 2 to equalize the number of electrons transferred:
2F(s) + 2H2O(l) = 2HF(aq) + O2(g)
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Electron transfer
Now, the number of electrons transferred on both sides of the equation is balanced.
Step 5: Balance the remaining atoms.
Now, you can balance the atoms (other than the electrons) in the equation. In this case, you have two H atoms on the left side and two H atoms on the right side, so hydrogen is balanced.
However, you have four O atoms on the right side (two in HF and two in O2), but only two O atoms on the left side (in H2O). To balance the O, you can add a coefficient of 2 in front of H2O to give:
2F(s) + 2H2O(l) = 2HF(aq) + O2(g)
Now, all the atoms are balanced, and you have a balanced redox equation.
I hope this explanation helps clarify the steps involved in balancing redox reactions using oxidation states, tie lines, and coefficients. If you have any further questions or need help with other examples, please let me know!