Use the data in Appendix D to predict whether the following reactions can occur under standard-state conditions. (a) Oxidation of Sn²⁺(aq) by Br₂(aq)

Video transcript

Hello everyone. So in this video we're trying to predict whether the oxidation of copper plus by our iodine will go ahead and take place under standard state conditions. So if we go ahead and refer to our standard reduction potential table. So if we do that, I gather this information here that will go ahead and write in green. So let's I to a solid Reacting with two electrons to give us two moles of our iodine and ions And has a value of 0.54V. Our second reaction, we see a C U two plus our copper two plus carryin reacting with one electron here to give us our copper plus one carry on. And that has a value of 0.154. So we can see here that it's an oxidation of our copper plus with our by our iodine solid. So we'll just go ahead and actually switch this reaction. Our second reaction here, we'll go ahead and switch those two. And we do so we'll get our copper plus being our starting material and then the C two plus acquis and our one electron will be our products. So now we go ahead and evaluate these reactions. So we see from the first reaction here without iodine, we have gained two electrons because of this two plus here because we have gained electrons. So we've gained electrons since we gain electrons, we associate this with a reduction and this of course occurs at the cathode. Now for the reaction with copper we see here that we're losing elections and we're losing last chance, like we said, an oxidation occurs, this occurs at the node. Okay, so let's go ahead and actually recall some information. Let's go ahead and recall that if we have a positive E cell value, so greater than zero, this is spontaneous in the forward direction. And then if we have a negative sl value, so less than zero, there's going to be non spontaneous in the forward direction. Let's go ahead and calculate our value. Then the equation for sl is E cell is equal to the E catheter minus the E A node. So then plugging the values, we have 0.54V - R 0.15V. We get the value then of RS. L. is equal to positive 0.39V. So because this value is positive, which is greater than zero, this will then be spontaneous In the four direction. So we can then conclude that the oxidation reaction will occur under standard state conditions. So then this right here is going to be my final answer for this problem. Thank you all so much for watching