Determine whether or not each metal, if coated onto iron, would prevent the corrosion of iron. b. Cr

Video transcript

hey everyone in this example, we're asked whether coating iron with cadmium metal will prevent the corrosion of iron. So what we should recall is that to prevent the corrosion of an iron, the metal must be more easily oxidized. So we should recall that in order to determine whether our metal is more easily oxidized, that would correspond to a lower value for our self potential. So a lower value for our self potential means that our metal is oxidized first or more easily oxidized. So what we should recognize is we have the prompt which is proposing coding our iron metal with cadmium, meaning that we would have a reaction with cadmium and iron. So we're going to write out the half reactions of that reaction. So for our full reaction we have cadmium metal reacting with iron And it's three plus Catalan form where we would produce our product cadmium two plus as a caddy on and the solid iron metal. Now writing out our half reactions which come from our reactant. We would begin with our first reactant which is our solid cadmium Which produces as a product, the Cadmium two plus Catalan. Now we should recognize that we have a net charge of zero on the reactant side here and a net charge of plus two on the product side for this half reaction. So we want to go ahead and cancel out this net charge of plus two by adding two electrons to our product side. And so this would give us a net charge of zero on both sides of our equation. Now because we recognize that we added electrons to our product side, we would recognize that this half reaction occurs as an oxidation half reaction. Now moving on to our second half reaction, which comes from our second reactant, which would be our iron three plus cat ion we would produce as a product the solid iron metal. Now we should recognize that on our react inside we have a net charge of plus three, whereas on our product side we have a net charge of zero. So we're going to expand our react inside and sorry about that, we're going to expand this react inside to cancel out that net plus three charge by adding three electrons to our reactant side. And because we recall that when we add electrons to a reactant side, as we did here, this half reaction is going to occur as a reduction. And so now that we've outlined which of our half reactions is the oxidation and which is the reduction. We're going to be able to calculate our cell potential. Now we should recall that our oxidation half reaction occurs at the anodes of our voltaic cell. Whereas our reduction half reaction occurs at the cathode of our voltaic cell. And so our next step is to refer to our standard reduction potential table. We would find this table either online or in our textbooks and we're going to look up the for the oxidation of the iron three plus Catalan. In this table, we would see that it has a self potential value equal to negative 0.36 volts. Whereas for our oxidation of the cadmium two plus catalon. According to this table, that will have a self potential equal to a value of negative 0. volts. And so now that we've calculated our self potential values for both our oxidation half reaction in our reduction potential or our reduction half reaction, we can compare these two values and see that our value negative 20.40 volts is a lower value. We can say negative 0.40 volts is less than negative 0.36 volts. And so therefore we can say solid cadmium will be oxidized first because it has the lower self potential value. And because we know that cadmium as a medal will be oxidized first. We can say therefore cadmium will prevent the corrosion of iron. So that's a yes. And so this statement here will be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video