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The following has an E? jed value of -1.2 V [Cr(OH)4]" + 3e Cr + 4 OH" What is the EPox value for this equation ?1/3Cr + 4/3OH"I/3 [Cr(OH)4] " +...

Question

The following has an E? jed value of -1.2 V [Cr(OH)4]" + 3e Cr + 4 OH" What is the EPox value for this equation ?1/3Cr + 4/3OH"I/3 [Cr(OH)4] " + e

The following has an E? jed value of -1.2 V [Cr(OH)4]" + 3e Cr + 4 OH" What is the EPox value for this equation ? 1/3Cr + 4/3OH" I/3 [Cr(OH)4] " + e



Answers

Solve the equation $E=I Z$ for the remaining value.
$I=10+4 i, \quad E=88+128 i$

Hello. So today we're going to be looking at thes reduction potentials and we want to figure out this reduction potential right here. So how can we do that? Well, let's see what thes reduction potentials are telling us while we have this becoming this and then this becoming this. So how about we take a look at our oxidation states? So the oxidation state of this here is so oxygen is negative two and there are four of them. So that's negative. Eight and then we have ah, too high traditions that are positive one. So that's negative. Six. So this has to be a positive six oxidation state. And here we have two oxygen's each negative too. So we've got negative four. So this has to be positive for So we're being reduced from positive 16 positive for in oxidation state between these two. So it's gaining two electrons and now we take a look here. This is the same as above. It's positive six. But here we see that it's positive. Three. And so we consort of write it like this. We start off with this compound right here, which is the most oxidized. Then we reduce it once here and then we reduce it a second time. So this production potential right here is referring to this part. And then this reduction potential right here is referring to this and we want to figure out this part. So it seems like if we took this energy change when we subtracted this energy change, we would get this energy change. However, sell potentials. You can't add and subtract sell potentials, but you can add and subtract free energies. So let's figure out the free energy of this right here, so still touchy is equal to negative Z fair days, constant and standard self potential or reduction potential. So free energy of this compound right here, coming this compound that would be gaining three electrons SOS Z is three fair days constant and then 0.4 to 8. And then self potential of this becoming this that would be gaining two electrons. So Z is too. And then the self, the reduction potential a 0.646 So now we can just take this and subtract this from it. So 0.4 to 8 times three would be negative 1.284 Saturday subtracting. Okay, two times. 20.646 Yeah, is negative. One point tune nine to Fair day. So that would become 0.0 eight. Fair day. And now let's make that into a reduction potential. So Negative z Fair day reduction potential. And so we know that c Well, we're going from a plus four oxidation state to a plus three. So Z would just be one. And so we get that the reduction potential off this compound right here. Coming. This compound right here is negative. 0.0 eight boots. So there you go.

To answer this question. The first thing that we need to do is calculate the silver concentration found in the saturated solution of silver phosphate. To do this, we need to use Caspi. Here's the K SP expression where K S P is going to be equal to the silver concentration cubed multiplied by the phosphate concentration. That product will then be the case P value that is provided of 1.1 times 10 to the negative 12. Phosphate concentration will be equal to the Mueller scalability and the silver concentration. Because of this, three will be three times to Mueller sell ability. So plugging these expressions into this expression will get K sp at 1.1 times 10 to the negative 12 is equal to three x cubed times s rearrangement of this equation gives us s equal to 1.1 times 10 to the negative 12 divided by three cubed which is 27 raised to the 1/4 power. This gives us 4.49 times 10 to the negative three Moller now knowing the silver concentration in a side saturated with silver Fox State and the silver concentration on the other side at 0.140 We can now use the nerds equation. Well, sorry. One step back. This is s but silver concentration is three. Yes, get 1.3 times 10 the negative three. We can use these two silver concentrations and the nearest equation Because this is a concentration cell. When the concentrations of silver are equal under standard conditions one Mueller, the self potential standard is zero. So s a will be equal to the cell standard minus 0.5916 divided by the number of moles of electrons being transferred which is just one divided by And this is sometimes the tricky part. What do we put up top? Do we put this up top? Or do we put the 0.140 up top? Well in line notation They tell us that the saturated solution is serving as the an ode because it is serving as the an ode. A node is where oxidation occurs. Oxidation is going to be a silver medal going to Silver Ion. So this silver ion concentration is involved in the a node and that is the saturated solution. So we put the 1.35 times 10 the negative three up top. And then the cathode is the one that contains the 10.140 molar concentration of silver. This then gives us a cell potential of 0.119 volts.

So with this problem, we are given a reaction, and we are given that our standard potential this cell is equal 2.5 volts on. We're gonna calculate the delta G of the cell. So we're gonna use that equation that is there in red. So our delta g of the cell that in stands for moles in this reaction there is just one s are in his one r f is our Faraday's constant, which is 9.65 times 10 to the fourth. And then our yeast cells, What was given to us A 0.5. And then when you multiply all three of those out, uh, you get about 48,000 we're gonna put that in terms of killer jewels. So negative 48.3 village.

So here it is given three is equals two e to the power off three. So long since it is a cube root so we can write he is equals to que brute off. So therefore, our final answer is easy cause to cube root off the


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