Given the following standard reduction potentials at 25 °C, (a) b...

Question

Given the following standard reduction potentials at 25 °C, (a) balance the equation for the reaction of H2MoO4 with elemental arsenic in acidic solution to give Mo3+ and H3AsO4 and (b) calculate E° for this reaction.

Accepted Answer

Hello. Everyone in this video essentially balancing out a chemical reaction. So we're given a chemical reaction sort of written in just words, we're going to translate this to purely just chemical form our chemical terms. So we have on our star material side, H C. L. O reacting with sulfur and its solid state to give us a church to S. 03 and cl two in this case a state. Alright, so the first thing to do is to break this big equation down into half reactions and then we'll balance it out from there. So let's start off with the first half reaction being H C L. O. Acquis. And that gives us cl two gas. So the first thing I wanna do is at our auction source, so we can see on the start material side we have one adam of oxygen but on the right we have none. So go ahead and add oxygen source by adding H 20. I'll start off with one mole of H 20. Well see now that on the right side we have to auction atoms on the left side, we have one, so we'll add a coefficient on the left side of two. However, we can see now that because we add this coefficient of two, we have two oxygen atoms but this side only has one. So again, Barabbas is out by adding a coefficient of two to Rh- 20. And another problem occurs because we have now four atoms of hydrogen but this has only two. So you kind of just bounce back and forth. Then we can throw off the ratio. So we need a source of hydrogen somehow, purely just hydrogen. How we can do this is adding in two moles of our hydrogen ion. Of course that is going to be Aquarius. So you do. So now you can see that this side and this side has all bounced out items of hydrogen, oxygen and chlorine. So this is now considered balance in terms of just our elements. Now let's kind of focus on our charges. So if we balance out our charges here we can see that this side has a plus two charge and this side is going to be on neutral. How we can balance our charges to make this side neutral is by adding electrons we just need here two electrons because these two negative energies and these two positive, we'll go ahead and cancel to give us neutral. Alright, so now moving on to the other half reaction. So for this one we have silver earth sulfur in a solid state to give us a church to S. 03, which is of course a curious so again we're gonna go ahead and start off by balancing out our oxygen atoms. We see on the right side there we have three moles of or three atoms of oxygen on the left side, we have none. So adding our oxygen at a source by adding H 20. So at three most of H 20. Alright, so now that I've done, so we can see here that we have six atoms of hydrogen on the right side, we only have two, So go in and add in just purely hydrogen by adding in hydrogen ion. So we'll add in four H plus cat ions on the right side if we do so we can actually see that everything is actually balanced now. So that was quite easy compared to the other one. As for our bouncing out of our reaction for our charges, we can see that we need to have let's see it's on the left side, everything is neutral on the right side. We do have this four H plus. So to balance out the charges, we can add four electrons to this side over here. Or actually we'll add four height or electrons over here to give the neutral ST So have four electrons. So now that we have done so we can see that this will all be neutral because these four negative energies will balance out with these four positive energies. All right. So one thing that we have to balance out for both of these have reactions will be our electrons. So we see here on the bottom we have four but on top we only have two. So going to actually multiply this whole reaction by two and once we do so I just scroll down a little bit here to give us more space. So once we do that and combine these two, we will get four electrons plus four h pluses Plus four HCLO, which is a Chris To give us two cl and four waters. And then for the other half reaction, which is right as is. So we have three H 20 reacting with sulfur and assault state. That gives us a church to S. 03, which is a chris reacting with four H plus is again, that's a chris and four electrons. Okay, so now that we can cancel some things out. We can see here foremost, our two electrons or four electrons will cancel out. And we see that our four high regions will cancel out. And lastly our water. What kind of cancel out? All right. And our neck reaction then will be so far in a solid state With four moles of HCLO, which is a Chris. And that's it for the starting material side. For the product side we will have two moles of C guests reacting with one mole of H two S 03. All right. And this is going to be our balance next direction for this problem. Thank you all so much for watching

Given the following standard reduction potentials at 25 °C, (a) balance the equation for the reaction of H2MoO4 with elemental arsenic in acidic solution to give Mo3+ and H3AsO4 and (b) calculate E° for this reaction.

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