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Two redox reactions are shown below along with their measured voltages.2 AgZn2 AgZn2+156V2 Au3+3 Zn2 Au3 Zn?t2.26 VArbitrarily set the 2Agt/2Ag half equal to 0 and ...

Question

Two redox reactions are shown below along with their measured voltages.2 AgZn2 AgZn2+156V2 Au3+3 Zn2 Au3 Zn?t2.26 VArbitrarily set the 2Agt/2Ag half equal to 0 and then determine the Zn/Znz+ half-reactior voltage.Use your Zn/Zn2+ voltage to calculate your 2Au3+/2Au half-reaction voltage

Two redox reactions are shown below along with their measured voltages. 2 Ag Zn 2 Ag Zn2+ 156V 2 Au3+ 3 Zn 2 Au 3 Zn?t 2.26 V Arbitrarily set the 2Agt/2Ag half equal to 0 and then determine the Zn/Znz+ half-reactior voltage. Use your Zn/Zn2+ voltage to calculate your 2Au3+/2Au half-reaction voltage



Answers

Calculate the voltage and identify the cathode for a cell in which the natural reaction between the following electrodes happens:
$$\mathrm{Zn}^{2+}(a q)+2 e^{-} \rightleftarrows \mathrm{Zn}(s)$$
$$2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \rightleftarrows \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q)$$

Let's first look up the reduction potential for the 2/2 reactions that are given to us. The 1st 1 is a reduction half reaction, and it has a reduction potential of zero. By definition, it is the standard hydrogen electric. The second half reaction is an oxidation half reaction on the corresponding reduction. Half reaction Is this one here, shown with a reduction potential a positive 0.771 volts. If this half reaction is reversed, as shown in the question than oxidation is occurring. So this is the an ode. So it's going to be the self potential equal to the Catholic potential, minus theano potential or zero minus 0.771 giving us the negative 0.771 volts.

Schedule. Take sales of this equation. So I'm ever gonna have beakers collected by assault bridge and then some electrodes that is likely to each other. The sides always an ode to Catholic electrons, always pulled that ways will, when we have to write out the half reactions. So, at my catheter, my analyst, I have oxidation. So oxidation is before I lose electrons. Production is where gave electron. So right I have over here to or PBS goes to And that has a potential for the next one in. And no, I mean for the cathode. Uh huh. To a g waas. You look no potential is your point really same thing on this one? So my oxidation is to say hello to feel Omar lose. Why production is this one bridge mood. So at the at, the family have to CEO too. Do you feel to modernise polls to elect her arm? It has a voltage. Is your report 95 or so Potentials Castle, huh? To, uh, Marchioness, which had herself pretend toe no point before floats safety here My oxidation is sink to using two plus then both of these are involved in the production edition So am I. So and I have the electrons flow this way. This was my okay. Half reaction for the oxidation. It was a self made for the oxidation and that's it.

Okay, so we have to determine the half reactions, um, for this Redox reaction, and then figure out the overall, um, the overall cell potential here. Um, So the way that you do that is first for at the half reactions. Right. So, um, the creations already balanced in terms of, um, in terms of mass strike geometry. So there's two irons and one cabbie. Um, excuse me. One cabin on both sides. So we're going to balance the half reactions. Um, so two irons, three are going to iron two plus. So that means that each, um, each iron is accepting an electron. So and there were two of them. So we have to add two electrons to that. And for cadmium, cadmium is going from cadmium zero to cadmium two plus, So that is a change of two electrons as well. Um, So, um, we're gonna make two electrons be the product, so it's charged balanced as well, Which is great. Um, so which is which, Right, So this reaction is, um, iron is being reduced here, so this is gonna be your cathode and, um, cadmium is being oxidized. This is gonna be your an ode. So the last thing we need is to find these hat for actions on our table, and then we're going to, um we're gonna do some subtraction. So there's two ways that you can go about doing this. So, um right, so the reduction potentials that in your book are reduction. Pretend the potential, the equations that in your book are reductions in the table, and they're given the potential for that reduction. So for cadmium, which is actually being reduced, um, that's so value is minus for 0.403 volts. Um, and for the reduction of, um oh, I mess it up. Sorry. That's the value for the reduction of cadmium, right? But, um, you're it's actually being oxidized in this reaction, but I'll there too, I'm getting at. Is there two ways you could do this math on gonna walk through both. So the iron three to iron to reduction cell is positive 0.771 So you can see that because Iran has the more positive value. That half reaction's gonna be a cathode, which we determine. They told us already by giving us the equation. So there's two ways that you could do this? Um, uh, The first way to do it is the way that it's in the book, which is the cathode minus the an ode Pretend reduction. So you're so you're attracting the cathode reduction potential from the an ode reduction potential. So it's confusing because reduction is not happening at the an ode. But you if you're not gonna flip the sign like you're supposed to, um, then you could do subtraction, and I would take whichever way you're comfortable with and stick with it the whole time. So, um, the textbook doesn't have you flip. So the textbook just says, take the values that are in the table and always subtract reaction it cathode minus reaction it. And so if we do that, um, you're going to gets you're going to get, um, you sell is equal to cathode, which is 0.771 volts minus negative zero points for 03 volts. This is your an ode, and that's going to give you 1.174 volts. Right? So the Converse way to do that is the right answer. The the other way to do it is, um, if you, um what you're supposed to do so if there's a question that says, What is the potential of this oxidation reaction? Um, that you have to find that half reaction in your table and flipped the sign, right? So all of the all I said before all of the Rio, all the reactions that are in the table are reduction reactions and the e values that are given or for the reduction. So if you're going to, um, if that metal is being is actually at the an ode, it's being oxidized. So it's losing electrons. It's not gaining two electrons, like in the like. The table says it is. You have to flip the sign. So if you're gonna flip the sign for cadmium here, the actual, um, the actual e not of the oxidation is going to be positive 0.403 volts at the an ode. So if you're going to do that, if you're going to account for the fact that you're gonna flip the charge, then you have to add the potentials together. So it's not what the book says, the book says. Keep the values the same. It's attract if you're gonna flip the value of the an mode. So this is a more positive values. So that means that reduction is, um, more positive value. This is You can call this the oxidation potential with That's not really what people say. If you flip the reaction, it means that this is the spontaneous reaction to happen. Um, without any outside energy being put it in. So that shows you that the reduction off cadmium is not spontaneous, which is why it's the and out here. Um, so if you're going to do that, then you would add the values together, right? So then you would take e cell. Um, and you would go 0.774 and then you would add the an ode potential, which is positive 0.403 and you're gonna get the same answer. So depending on which way you're teacher tells you to do, it might be different from the way the book tells you to do it. So I showed you both ways, and I'll go over this again in other problems that I'm doing that, um, that

This question deals with electro chemistry and with voltaic cells, those are spontaneous electrochemical cells. And the first example were asked to look at is a reaction between lead and silver. And what we're told according to what forms in the process here is that led undergoes this half reaction gives up two electrons to form the lead, two plus cat eye on and those electrons get transferred to silver to farm solid silver. This reaction is what takes place at the an ode an old as ours where oxidation takes place. This reaction takes place at the cathode. That's always a reduction takes place. Reduction is a gaining of electrons an oxidation, a loss of electrons. Little pneumonic that can help you remember that by the way, is an ox. An ode oxidation. And red cat reduction takes place at the catholic. And so basically what we have going on here is that lead is going to be giving up electrons. Electrons are going to flow from always the nano towards the cathode. Lead is going to be giving up electrons supplying them to the silver plus, can I on two producer played out silver medal while lead becomes oxidized in the an ode compartment over here. So again, this is my an ode where the oxidation is taking place. This is my cathode and that's what's happening to the electrons throughout the course of this cell. Well, the next example that asks us to draw a sketch of a voltaic cell involves at the anodes, iodide ions being oxidized to iodine with this half reaction and claro um dioxide compound at the other end of the spectrum here, at the cathode that's being reduced to some um chloride ions. So the electrons are being furnished here at the an odd where oxidation is taking place. The catholic over here is where the reduction is taking place. And the electrons are going to flow from this iodide solution through an inert or usually kurban electrode because we don't have a metal available here to help transfer the electrons. The same place is going to be same case is going to be true at the cathode compartment here we're going to have in a nurture usually kurban electrode to help transfer the electrons and the electrons as always will flow from the an ode their source towards the cathode and salt bridge of course, will connect the two components to um in essence be able to drive this process to happen. The last example of voltaic cell for us to look at is probably most complex one because it involves oxygen as a gas and usually some sort of a vessel that can contain a gas, like a glass vial here with the electrode. Um like again in a nerd electrode that can run through that gas is vile to supply electrons And so the half reaction that's occurring in the anodes or oxidation component side over here is zinc being oxidized to zinc two plus cat eye and and supplying some electrons as always, the electrons will flow from an ode two cathode in this case, at the catholic, the reduction process that's taking place is oxygen gas is in the presence of an acidic solution here, picking up the electrons that are furnished from zinc and producing water as a product of that reaction. And um again, the overall half reaction looks something like this. As the oxygen becomes oxidized to its for water form, it's going to react with those hydrogen ions in the solution here to produce water as an ultimate product.


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