So within a given redox titration, we have the analyte and we have our titrant. We're going to say here that it is common for the analyte to exist in multiple oxidation states while in solution. So, it'll possess multiple charges. And we're gonna say when we do any type of quantitative analysis, it has to be converted to just one oxidative state. Now, we're going to say, for example, iron itself can exist as 2 different forms. It can exist as iron 2 or iron 3. Now, if we have a solution that contains both of these ions, that won't help us in terms of our redox titration. What we would do is we could basically add either a reducing agent or an oxidizing agent to this solution of these 2 ions and in that way get rid of 1 of them. Once we've gotten down to 1 oxidation state for our iron, we can utilize that ion and either do an oxidation with it or a reduction. Now, a common type of oxidizing agent and it tends to be used with these ions is your Cerium 4 ion. Cerium 4 ion is one of the strongest oxidizing agents that exists. So, as an oxidizing iron, that iron, that means iron is going to become more positive. Since these are the two forms that it can exist in, it has to react with Fe2+ removing an electron from it means that iron now becomes Fe3+ This here, it oxidizes iron 2 by removing its electrons so it accepts that electron so now it becomes Ce3+. And that's how these oxidizing agents and reducing agents behave. They either help to oxidize one of the forms of a particular ion so at the end, our solution possesses only one particular oxidation state for that given ion. From there, we then continue with our redox titration. Now, here if we're talking about reducing agents, we call them auxiliary reducing agents. We're gonna say an auxiliary reducing agent represents an easily oxidized metal. Typically, we have Zinc and Silver, but also less commonly used, Tin and Cadmium. Here, there are reducing agents, so they're going to help reduce one of the ions for a given solution and it's part of what we call the pre-reduction step. We're going to reduce one of these ions so that at the end, we have only one of the ions remaining that can then undergo some type of redox titration. Now, we're going to say that the 2 types of reduction columns that exist, in the first one, we have what's called the Jones Reductor and here this is just your reduction column is filled with a zinc amalgam. That's basically zinc bonded to mercury. We're going to say in the process, the zinc metal reduces the analyte while it is oxidized. Here are zinc amalgam. It's going to help to reduce the analyte. In the process, we have our zinc here. That's 2+ and remember, if it's helping to reduce the other species, that means it itself is being oxidized. That's how it loses these 2 electrons. Then here, we have our mercury liquid that has basically decoupled itself from the zinc amalgam. Now, our other common reduction column is called the Walden reductor. Here, this one is filled with basically little pebbles or granules of solid silver metal. And we're going to say while the analyte solution is infused with acid. The solution itself is acidified. Here we have our silver solid. It's going to react with the chloride ion. It gets oxidized and becomes plus 1 and that's how it's able to combine with the chloride ion that's minus 1. Here, this represents the electron that it's lost. That electron that it's lost will then go to the analyte and reduce it so that we have only one oxidation state for that particular analyte. Remember, the whole point of an auxiliary reducing agent is to donate an electron or more to an analyte in order to have only one oxidation state at the end. Once we get to that one ion, we can do a redox titration with that given analyte. Now that we've talked about auxiliary reducing agents, come back and see what we say in terms of auxiliary oxidizing agents.
Analyte Oxidation State - Video Tutorials & Practice Problems
Auxiliary Oxidizing and Reducing Agents
Analyte Oxidation State
Video transcript
Analyte Oxidation State
Video transcript
Alright guys. So now we're going to take a look at our auxiliary oxidizing agents. We're going to say the auxiliary oxidizing agents basically just oxidize the analyte and in the process, they themselves are reduced. Because they are oxidizing the analyte to just one oxidative state, we say that this is part of our pre-oxidation stage. So, we're going to create or keep the analyte that has the more positive charge which therefore corresponds to the greater oxidation number. Now, we're going to say here that peroxydisulfate which is this represents a powerful oxidizing agent that works in conjunction with the silver catalyst. So here this reaction, we have peroxydisulfate reacting with silver ion. Here it's going to be reduced which is why we have the electrons here as reactants. In the process, we create our sulfate ion and then here this represents our new oxidizing agent that can then react with another species in order to oxidize it. Now, we're going to say here that this oxidizing mixture can oxidize, for example, manganese II ion to permanganate ion. It could oxidize cerium III ion to cerium IV ion. That shows you the power of this oxidizing agent mixture because cerium represents one of the strongest oxidizing agents out there. But this auxiliary oxidizing agent is so strong that it’s even able to oxidize cerium. Now from there, we also have our dichromate. Well, here we're going to have chromium III ion being oxidized to dichromate ion. We have here, this is our vanadium cation. And then, we have V O2+. And this is just our vanadium IV oxide cation. Now, beside peroxidisulfate, we also have hydrogen peroxide which is H2O2. It represents another powerful oxidizing agent. This one though can work within an acidic or basic environment. Here in this general equation, we have our peroxy acid, hydrogen peroxide operating within an acidic environment. Again, it's being reduced so that it can oxidize something else. In the process of its reduction, it's going to become 2 moles of water as a liquid. Now, in basic solutions, it oxidizes. Okay. So here in basic solutions, it could oxidize cobalt II ion to cobalt III ion. It could oxidize iron II to iron III. It could also oxidize manganese II ion to manganese IV oxide. Now, if we switch it from a basic solution to an acidic solution, it now acts like an auxiliary reducing agent. So now it's going to reduce. Here, it could reduce dichromate ion into chromium III ion or it could reduce permanganate ion into manganese II ion. It's more versatile than peroxydisulfate because depending on the solution that it's in, it can act either as an auxiliary oxidizing agent and oxidize, or it can act as an auxiliary reducing agent and reduce. Other commonly used oxidizing agents also include silver oxide. In organic, we tend to talk about silver compounds that act as oxidizing agents as toluene’s reagent. For those of you who haven't taken organic, don’t worry about it. For those of you who have taken organic, you might have forgotten. Just a little quick bit of outside information. And we also have sodium bismuthate which is NaBiO3 as another special type of oxidizing agent. But out of these 4, remember when it comes to redox titrations, it's not uncommon to have your analyte exist in multiple oxidation states. So it's your responsibility to use either an auxiliary reducing agent or an auxiliary oxidizing agent to get that analyte down to just one single charge. Once we do that, we can then carry on with our redox titrations. Titrations. So just keep in mind the jobs of your auxiliary oxidizing agents versus your auxiliary reducing agents.
Analyte Oxidation State Calculations
Video transcript
So continuing with this idea of oxidizing agents and reducing agents, let's answer each of the following questions based on the following half reactions. From the half reactions that are given, we can see that they're all written as reductions, where our electrons are written as reactants. Remember, in a reduction, your electrons are written as reactants, and what we have here are all these cell potentials for each one of these half reactions. Now, for the first one, it's asking which is the strongest oxidizing agent. So, let's think about what an oxidizing agent is telling us. Remember, if you are the oxidizing agent, that means that you have been reduced. If you've been reduced, that means you represent the cathode. And remember, with the cathode, we have the gaining of an electron. And if you're gaining an electron, that means you must be on the same side as the electrons, and if you are the cathode, you should have the largest cell potential present. Since all of them are written as reductions, we can compare them all. If some are written as reductions and some as oxidations, you'd have to reverse all of the reactions so they all are either written as reductions or as oxidations. It's more customary to have them all written as reductions where your electrons are reactants. Now we can see here that the largest cell potential is with the first equation where the cell potential is 1.36 volts. So we know that our answer will lie in the first half-cell reaction. We're saying here you're gaining electrons here on the same side with the electrons. If we look at our two choices, we have chloride ion or chlorine gas. It's the chlorine gas that's on the same side with the electrons, so that would be our answer. The strongest oxidizing agent presented here would be the chlorine gas.
Now, it's asking us basically for the opposite. We're looking for the strongest reducing agent. So if you're the reducing agent, that means you've been oxidized. If you've been oxidized, you represent the anode. Now remember, the anode should have the smallest cell potential. And remember, if you're the anode, you're undergoing oxidation, that means you're losing electrons, which means you won't be on the same side with the electrons because you've lost them. So if we take a look, the smallest one is the bottom reaction, or talking about vanadium, and our choices are either vanadium solid or vanadium 2 ion. Here again, we want to be away from the electrons. So the strongest reducing agent here would have to be the vanadium solid as my answer. So, V solid would be our answer for this question.
Now finally, it says, Will iodide reduce chlorine gas to chloride ion? Alright. Let's think about what it's saying. Here, it's saying that for chlorine to be oxidized, it would have to be the anode. It would have to have the smaller cell potential. So here, the equation with iodine does have a smaller cell potential than the equation with chlorine, and if it's being oxidized, it would have to be on the side away from the electrons. Here, the iodide ion, yes. It is on the side that is not next to the electrons. So, my iodide ion would reduce my chlorine gas and as a result become oxidized itself? So here we'd say yes. So remember, for questions like this, you really have to go back and take a look and remember the principles that we've talked about in the past. Leo, the lion goes: If you're losing electrons, you're undergoing reduction. So you're the oxidizing agent. If you've been reduced, you represent the cathode. The cathode typically has the larger cell potential. Remember, this is true when we're dealing with a spontaneous electrochemical cell, when we're dealing with a galvanic or a voltaic electrochemical cell. These are spontaneous. So the cathode should have the higher cell potential and the anode should have the smaller cell potential. When we're talking about non-spontaneous electrochemical cells, like the electrolytic cell, everything is reversed. The cathode would actually have the smaller cell potential value, and the anode would be the larger cell potential value. You'd be indicated which type of cell you're dealing with. Here, we're assuming that we're dealing with a spontaneous electrochemical cell. So we're dealing with a galvanic or a voltaic cell.