Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (c) reduction of Ce4+ to Ce3+ by H2O2,

Video transcript

Welcome back everyone. In this example, we have a true or false statement. The reduction of Cobalt three plus to Cobalt two plus by peroxide can occur spontaneously in an acidic solution under standard conditions. So let's recall some facts. The first being that when we have a reduction potential that is more positive in value, this will correspond to a spontaneous reaction. But when we have a reduction potential that is more negative in value, this will correspond to a non spontaneous reaction. We want to write out our half reactions first for the formation of hydrogen peroxide where we have oxygen gas which gains or rather reacts with two moles of our hydrogen proton and releases two electrons to form our peroxide product. Recall that this has a reduction potential value from our textbooks of 0.68 volts. Now writing out our reaction for the formation of our Cobalt two plus ion, we begin with our Cobalt three plus ion. So that's co and three plus as the exponent there, this would be gaining an electron to produce our Cobalt two plus ion as a product where we should recognize that we went from an oxidation number of our ion charge being plus three. So the oxidation state is plus three here where it went to plus two here. So we have a decrease in oxidation number, meaning that this is going to be our reduction half reaction which occurs at our cathode. We recall moving on, we want to show how our hydrogen peroxide h2o 2 is again formed from our oxygen gas plus our two moles of our hydrogen proton plus two electrons. Where for our molecule, hydrogen peroxide, we would recall that we have an oxidation state of minus one for each of our two oxygen atoms. So that would be minus one times two, which would be negative two. But because we recall that the oxidation state for hydrogen is plus one, these or this minus two oxidation number would be canceled out, which is why we actually have an overall neutral oxidation state here for hydrogen peroxide. And on our product side, we go to having a oxidation state for hydrogen of plus two where we gain two electrons so that it's canceled out and we now have an oxidation state of minus two or correction. Sorry. So the electrons are not going to cancel out our oxidation state here. We would just have the oxidation state of plus two coming from the two moles of our protons where we multiply it by that plus one oxidation state to get a oxidation state of plus two. So we have an increase in our oxidation state, meaning that this will occur as an oxidation at the a node half cell. We want to come up with an overall equation. But before we can do so, we need to bounce out our electrons. And so we're going to multiply this first equation by a factor of two so that we can have two electrons as we have in this second equation. And what we would have when we add these together is a total where we have two moles of our Cobalt ion plus two moles of our or rather we did not multiply the second equation. So we would have one mole of our hydrogen peroxide. This is then going to produce on our product side, our two moles of our Cobalt two plus ion plus are one mole of oxygen gas plus are two moles of protons. Where now that we have this overall reaction, we want to go back to our cell potential charge of our reduction reaction here for Cobalt. And recall that this value is a value of 1.8 to 1.82 volts here. And so now that we have this overall reaction, we can calculate our standard cell potential e degree where we would take the difference between our cell potential of our cathode reaction minus the cell potential of our a node reaction. So plugging in what we know we determined above that our cell potential of our cathode reaction is our reaction with our Cobalt ions which had a cell potential value of 1.82 volts from our textbooks subtracting from the cell potential of our anote half reaction, which was for our hydrogen peroxide reaction. We noted as 0.68 volts. So plugging that in we have 0.68 volts from our textbooks, this difference gives us a result for our standard cell potential equal to 1.14 volts. And because we see that this is a positive cell potential value, we understand that the reduction of Cobalt three plus to Cobalt two plus by hydrogen peroxide will occur spontaneously. And so going back to our statement given in our prompts, it says that this reduction will occur spontaneously in an aesthetic solution under standard conditions. And we would consider this statement as true as we've proved below from our work. So our final answer is going to be that the statement given is a true statement. I hope everything I reviewed was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.