Use the data in Appendix D to predict whether the following reactions can occur under standard-state conditions. (d) Reduction of I₂(s) by H₂SO₃(aq)

Video transcript

Hello everyone in this video. We want to use the given reduction potential blow to identify whether the reduction of cl two gas by HN 02 in its greatest form will occur spontaneously at standard state conditions. And these are the values here. Alright, so given these reduction potentials, We have the reaction cl two Which is a quiz that reacts with the two electrons and this yields the two moles of RCL -, which is the reduction potential here is equal to 1.358V. Alright, so for our N 02 or N 03 minus rather, it's a good state to give us a church end to a quiz. We have to balance the oxygen's by adding H 20. And then we can go ahead and balance our hydrogen is by adding R H plus and lastly balancing out the electrons. So if we do all this, that we get that the final equation is going to be where we have N + minus again, which is a quiz, reacting with three moles of R H plus, which is a quiz. As well as two electrons. This will yield are products of H and +02, which is a quiz and H +20, which is liquid. And the reduction potential here then is 0.934 volts. Alright, so let's go ahead and rewrite our reactions. So for my reduction, which happens at the cathode, we have our cl two which is Aquarius reacting with two electrons and this yields two moles of cl minus. Just a curious the reduction potential here is already said 1.358 volts. And for our oxidation reaction and this occurs then at the anodes. This reaction is that we have a church N 02 which is a quiz, reacting with H 20. And its liquid state. This yields N 03 minus which is a quiz and three moles over H plus, which is also Aquarius and the two electrons. The standard reduction potential here is equal to negative 0. volts as we just flip the product and reactant. Alright, so combine these two reactions, we can get the overall reaction. So what we can catch out we can see is going to be the two electrons, but everything else will remain the same. So the overall reaction then is equal to, let's see from my starting materials, we still have our cl two, we have our H N 02, which is a quiz. And we have our H 20. And its liquid state. This will all yield two moles of r C l minus, one mole of n 03 minus And three moles of my age plus. Alright, now, for the cell value, this is equal to the E. Of the reduction. So at the cathode minus that of The oxidation, of course that is at the anodes, we would have these values and that is 1.358V plus are negative 0.934 volts. We put this into the calculator, we get the sl value is equal to 0.4 to four volts. So this value determines if it's spontaneous or non spontaneous. Since here we see that we have a positive value. This is of course greater than zero, and if we have A. S. L. Value that's greater than zero. This means that the reaction is spontaneous in the four direction. So we can conclude then that the reduction of cl to aggress by R H N 02 degrees well occur spontaneously. And this right here is going to be my final answer for this problem. Thank you all so much for watching.