Bars of iron are put into each of the three beakers as shown here. In which beaker—A, B, or C—would you expect the iron to show the most corrosion ? [Section 20.8]

Video transcript

Welcome back everyone in which of the four beakers illustrated is the iron bar expected to exhibit the greatest corrosion. Let's begin by recalling that corrosion is going to be a redox process in which our medal will transfer electrons to another species being are corroding agent. And in this case our coding agent can be oxygen or an electrolyte solution. And so we can imagine that we have iron reacting with oxygen. So iron would be a solid oxygen as a gas and we would form Iron Oxide Feo two solid where recognized that iron because it's a metal in its standard state will have on the react inside an oxidation number of zero, whereas the oxygen di atomic molecules on the reaction side is also in its standard state and will also have an oxidation state of zero. But on the product side recall that when oxygen is is in a compound, it has an oxidation state of negative two. And for iron, we would need to solve for its oxidation number and so we would form an equation where we have x minus two is equal to the overall charge of our product, which is neutral. So it's set equal to zero and solving for X. We would add two to both sides to see that on the product side, Our oxidation number, or X is equal to plus two. And so our oxidation number of iron is equal to plus two and so recognized that oxygen went from having an oxidation state of 02 plus two. And so iron is oxidized because its oxidation number increases and so we can write out its oxidation reaction in which we have solid iron which forms are iron two plus catty on as a product. And that cary in charge means that we have a release or loss of two electrons. Now focusing on what happens to oxygen. We see that it went from an oxidation state of zero to a negative oxidation state on the product side. And so oxygen is reduced. And to write out its reaction or its reduction reaction, we have oxygen gas which forms or rather we should say, which is balanced with water on the product side, in which we would need two moles of water as a product since we need two moles of oxygen on both sides of our equation. And recall that reduction will require acidic protons. And so we're going to need to balance out our hydrogen because we have four moles of hydrogen on the product side and none on our reactant side of this reduction. So we're going to expand it and add four moles of protons as a second product in this reduction of oxygen. The last thing to remember is we need a bounced charge overall. And so because we have a net charge of plus four on the reactant side, we can cancel that out by adding four electrons as our third reactant. And so we would have a neutral charge on both sides of the reaction. So this is what represents our, sorry, our reduction of oxygen. And we need to combine these two equations together. And so we're going to need to have the same amount of electrons, we only have two electrons in our oxidation of iron. So we're going to take the entire reaction and multiply it by a factor of two. And this is going to give us now two moles of iron solid Yields two moles of our iron catalon And four electrons. And so canceling out the four electrons in both reactions. We would be able to take our reduction and oxidation reaction to come up with a net ionic equation. But what we want to focus on is that as we stated, we need H plus or protons for our reduction. So we want a more acidic solution. So we automatically will observe that beaker A with a ph of four is going to have a much higher concentration of protons due to that low ph which will promote reduction and therefore be useful in the corrosion of iron. Since we understand that it's redox, let's go ahead and also consider beaker B, which has a ph of seven, recall the auto ionization of water at a ph of seven, meaning that we won't have any available age plus which we need for our redox. And so we would rule out bigger B because it would exhibit low corrosion. Moving on to bigger. See we have a ph of nine and bigger D, which has a ph of 11, Recall that when we have a ph that is greater than nine iron will not corrode. And so we can also rule out bigger C. And D. Meaning that the only correct choice with a bigger that exhibits the greatest amount of corrosion would be bigger A due to its low ph and availability of high concentration of protons to promote the reduction of oxygen for the corrosion of iron. And so A would be our final answer to complete this example. I hope this was helpful and let us know if you have any questions.