electrochemistry
posted by Christie on .
I am soooo confused! I was given this PdCl4^2(aq)+Cd(s)Pd(s)+4Cl^(aq)+Cd^2+ I am supposed to give the 1/2 reaction for The PdCl4^2 My teacher said you could figure out the charges on the atoms by, in PdCl4^2, taking the subscript and mult. by the superscript to get the toal charge on all O atoms(there are 4 so 2*4=8 so to have the molecule be balanced the Pd would have a +8 charge. A friend told me to say O=2 so you take 4(2)+Pd=2(thecharge on the molecule) in this case Pd would =+6. In eaither case when I put the equation into the program wiht e to balance out the charges it is wrong. Can you help me please! Thank you!

Part of the reason for confusion may be that there is no oxygen in any of your work except in the explanation. Here is a site that will help you determine the oxidation state of almost anything.
http://www.chemteam.info/Redox/RedoxRules.html
For PdCl4^2:
Cl is 1. There are 4 of them for a total of 4. So what must Pd be in order to leave a 2 charge on the ion. Obviously, Pd must be +2.
Let's check it. +2 for Pd, 4 for Cl, total is +2 + (4) = 2.
So PdCl4^2 + 2e ==> Pd + 4Cl^
You can balance this another way without knowing the oxidation state for Pd but I don't recommend it because it doesn't teach oxidation numbers and you need that for later. However, here is how you do it.
Write the half reaction from what is given.
PdCl4^2 ==> Pd + 4Cl^
Now count the charge on the left and right. On the left I see 2. On the right I see 4. So we add electrons to the left to make it balance, like this.
PdCl4^2 + ?e ==> Pd + 4Cl^
How many electrons must we add so we have 4 on both sides. Of course, that is 2e. So we have
PdCl4^2 + 2e ==> Pd + 4Cl^
4 = 4
Works every time.
For the Cd.
Cd ==> Cd^+2 + 2e
I hope this helps you get unconfused.