# electrochemistry

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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 -

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/Redox-Rules.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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