Posted by Christie on Tuesday, November 17, 2009 at 6:44pm.
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!

electrochemistry  DrBob222, Tuesday, November 17, 2009 at 7:15pm
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.
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