A voltaic cell that uses the reaction PdCl42-1aq2 + Cd1s2 ¡ Pd1s2 + 4 Cl-1aq2 + Cd2+1aq2 has a measured standard cell potential of +1.03 V. (b) By using data from Appendix E, determine E°red for the reaction involving Pd.

Video transcript

Hello. Everyone in this video, we're given this overall reaction right over here were also given the resell value as well as the standard reduction potential of aluminum. And in this program we're trying to do is to determine the standard reduction potential of our gold, our au. So first we go ahead and break down this overall reaction by looking at each oxidation number of the materials and the reactions and see what's going on exactly. So let's first take a look at the first material that is A. U. B. R. Four negative. So the oxidation state of B. R. Is always negative one And the oxidation number of a. U. is plus three. Now, for our 2nd story material of A.L.. That's its natural elemental state. So the oxidation number for any adam is natural. Mental state is zero. Okay, now for our product side. So this is the starting materials and then we have our products. So the first product we're gonna go ahead and evaluate is gold in a solid state. So again, it's in this natural and mental state. So the oxidation number is zero. Then we're dealing with B. R minus. So we have a minus one charge. The oxidation number is negative one for a. L. Three plus, same as browning. We have a plus three charge. So the oxidation number is plus three. Now we can see here then if we have broken this down, we see that we have a change in oxidation state for gold, it's going from positive 3-0. Let's actually put this in writing. So for gold we're going from Plus 3-0. So the oxidation number decreases, meaning that we have a reduction And this occurs at the cathode, not for our aluminum. You can see here that the oxidation state goes from zero to positive three. So the oxidation number has increased. Mean that it's been oxidized and this occurs the anodes. So then our half cell reactions will be that the A. L. Three plus. Since it's being oxidized, we're acting it with three electrons to give us a loom no solid. And the standard reduction potential of this is said to be 1.6 -1.662V. Again this at the node. Now for our gold we have a U. B. R four minus Reacting with three electrons to give us gold in a solid state As well as four moles of the bombing and ion. And the center reduction potential for this is unknown. We're trying to determine this and of course this occurs at the cathode. So now we're going ahead to use the calculation or the equation for R. S. L. Let's recall that that equation is that the E cell is equal to the standard reduction potential of our cathode minus that of our a note. So we have the value given to us as 2.516 volts equal to the center reduction potential of our cathode, which we do not know minus a negative 1. volts for the center production potential at the anote, we can get that the standard reduction potential of our catholic, then Physical two 0.854V. So the so therefore, then the standard reduction potential Of gold is equal to 0.854V. Then this is going to be my final answer for this problem. Thank you all so much for watching.