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/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.
Sure, I can help you with that! Balancing redox reactions can sometimes be confusing, but I'll guide you step-by-step through the process.
First, let's start by identifying the oxidation states of the elements in the given reaction:
PdCl4^2-(aq) + Cd(s) → Pd(s) + 4Cl^-(aq) + Cd^2+
To find the oxidation state of Pd in PdCl4^2-, you can use the information your teacher provided. Each chloride ion has a charge of -1, so 4 chloride ions would contribute a total charge of -4. Since the overall charge of PdCl4^2- is -2, the Pd ion must have a charge of +2 to balance out the negative charge from the chloride ions.
So the oxidation state of Pd in PdCl4^2- is +2.
Now, let's write the half-reaction for PdCl4^2-:
PdCl4^2- → Pd
Since the oxidation state of Pd changes from +2 on the left side of the equation to 0 on the right side (since Pd is in its elemental form), we can write the half-reaction as:
PdCl4^2- → Pd + 2e-
This half-reaction represents the oxidation half-reaction, which means it shows the loss of electrons.
I hope this helps! If you have any more questions, feel free to ask.
Of course, I can help you! Balancing redox reactions can sometimes be tricky, but I'll guide you through the process step by step.
First, let's determine the oxidation states of the elements involved in the PdCl4^2- ion. In this case, the chlorine atom has a fixed oxidation state of -1, so the total charge contributed by the chloride ions is -4 (4 x -1).
To figure out the oxidation state of the palladium (Pd) atom, we can use the fact that the overall charge on the PdCl4^2- ion is -2. Since there are four chloride ions contributing -4, the Pd atom must contribute a +2 charge to make the total -2. Therefore, the oxidation state of palladium in PdCl4^2- is +2.
Now, let's write the half-reaction for the PdCl4^2- ion. Since palladium is being reduced (from an oxidation state of +2 to 0), the half-reaction will involve the gain of electrons (e^-).
Palladium half-reaction:
PdCl4^2- (aq) + 2e^- -> Pd (s) + 4Cl^- (aq)
By adding 2 electrons to the left side, the charge on palladium is decreased from +2 to 0, balanced by the 4 chloride ions with a total charge of -4.
Now, if you're entering this reaction into a program, make sure you include the stoichiometric coefficients (numbers in front of each species) to balance the atoms as well. For example:
2(PdCl4^2-) (aq) + 4e^- -> 2Pd (s) + 8Cl^- (aq)
Remember, this half-reaction represents the reduction of PdCl4^2- in the overall reaction you provided. To balance the full redox equation, you'll also need the half-reaction for the Cd(s) oxidation.