Missing Scribe For Friday March 2 (sorry)

Well, on Friday, we received worksheets. We had an option to either hand it in on Friday or Monday, but now it would be too late. Well the sheet that was handed in was a review, to check if you know what oxidation and reduction was. The other was about Balancing Equations using a different method, if a redox reaction occured. Here is a guide to interpreting that sheet:

Firstly, find the oxidation numbers, but remembering to use the "Rules for Determining Oxidation Numbers."

Lets use Ag₂S + Al -> Al₂S₃ + Ag

Ag: is reduced because it is gaining one electron. +1 to 0 (more negative number)
S: remains the same, so it isn't oxidized.
Al: is oxidized because it is losing three electrons. 0 to +3 (more positive number)

Multiply the subscripts with the number of electrons lost or gained.

Therefore:

Ag₂S: 2(because you have two Ag atoms) X 1 electron (number of electrons that was gained)

Al: 1(because you only have one Al atom) X 3 electrons (number of electrons that was lost)

Now we have to balance. So that we have the same number of electorns gained and lost. We find their lowest common multiple...Hm...2 and 3, 6 goes into both, so lets multiply both to get 6:

Ag₂S: 2 electrons X 3 = 6
S: 3 electrons X 2 = 6

Use what you multipled there as your coeffcients: so 3 for Ag₂S and 2 for S:

3Ag₂S + 2Al -> Al₂S₃ + Ag

Note: Check both sides are balanced, you may need to balance the rest of the equation. For this case, since you have 6Ag on the reactant side, then you need a 6 in front of the Ag in the end product side. That is all.


Therefore, your answer should be:

3Ag₂S + 2Al -> Al₂S₃ + 6Ag