welcome to another episode of Terry's notes and today we are going to be looking at oxidation and reduction now oxidation can be defined as the gain of oxygen by a substance or the loss of electrons and reduction can be defined as the loss of oxygen by a substance or the gain of electrons now some people use the term oil rig in order to help them remember how to explain what is oxidation and reduction so oxidation is the loss of electrons and reduction is the gain of electrons so that's how we have oil rig and a reaction in which oxidation and reduction occurs at the same time is called a Redux reaction right as in reduction and oxidation now what you need to be able to do is to determine what is called um the oxidation number oxidation number indicates the oxidation state of an element in its free state or in a compound so there are some rules that you need to remember the oxidation state of atoms of an element in its free state is zero so sodium the oxidation state of sodium is zero and the C of chlorine and chlorine gas the oxidation number is zero next rule is the oxidation number of elements that exist as simple ions in ionic compounds is the same as the charge on the ion so if we have something like sodium chloride we know that in sodium chloride we have na+ and cl minus the oxidation state of sodium is+ one which is simply the charge on the ion and the oxidation number of chlorine is minus one which is just the charge on the ion as well another example is magnesium oxide um we know we have mg2+ ions present and we also have O2 minus ions present so the oxidation number of magnesium is plus two and the oxidation number of oxygen is minus 2 another rule that we have is the sum of all the oxidation numbers of elements in a compound is zero so if you have a compound like magnesium chloride the oxidation number of magnesium is plus2 and the oxidation number of chlorine is minus1 so if we have so it's plus two for magnesium and minus one for chlorine so since we have two chlorine present it'll be 2 * by minus1 so this will give us min-2 so minus so plus 2 - 2 equal 0 so the sum of all the oxidation numbers of elements in a compound is zero the next rule is that the sum of all the oxidation numbers of the elements in a radical is equal to the charge on the ion so if we have an something if you have something like M4 minus which is a manganate ion the and we were asked to find the oxidation state or the oxidation number of Manganese we let the oxidation number be the oxidation state be X and oxygen now we know that oxygen is typically min-2 so it's 4 * -2 and this will be equal to the charge on the radical here is minus1 so all we need to do is to solve for x and we will get the oxidation state of Manganese so x - 8 = -1 so therefore x = -1 + 8 so X is equal to + 7 so the oxidation state of manganese in this case is+ 7 then we have substances that have variable oxidation States so hydrogen is usually plus one except in hydrides where it is minus one so for example sodium hydride sodium is + one and hydrogen in this case is minus one right typically hydrogen is+ one um in the case of oxygen it is usually minus 2 except in peroxides where it is minus one all right now let's do some examples some typical examples that would come in an examination question we want to determine the oxidation number of the element on the line okay so we have the solutions across here so the first example is K2 cr207 and we want to find the oxidation number of chromium so we start off by writing out our equation um the overall charge is zero so and potassium is in group one so it is the oxidation number is plus one so it'll be 2 multiplied by + 1 we got this two from here and we are letting the oxidation number of chromium be C be X so it'll be 2 multiplied X for the second term in this expression and then we have a seven the oxidation number of oxygen is minus 2 so it'll be 7 * - 2 so this will give us 2 + 2x x - 14 = 0 and when we solve for x we end up with X = to + 6 so that is the oxidation number of chromium in this example in this other example we have km4 and we want to find the oxidation number of Manganese or the MN potassium is in group one so the oxidation number is+ one plus X which is what we need to find and we have four multip by the oxidation number of oxygen is minus 2 and that is equal to zero so we solve for x and we end up with X is equal to+ 7 notice that we don't just write s we have to put Plus in front of it the next example is the sulfate ion s so42 minus so we let the oxidation number of the sulfur be X so X plus we have four oxygen so it's 4 by -2 now this min-2 is the oxidation number of oxygen right and we're going to put that equal to the overall charge of this ion is minus 2 so we have X Plus Open brackets 4 * min-2 equal -2 and if we solve for x we get + 6 our next example is the cl3 minus ion and we want to find for chlorine in this case we have we let the oxidation number of the chlorine be X and we have three for the oxygen but the oxidation state of oxygen is min-2 so it's 3 * by-2 that's how I got this and the overall charge is minus one so we need to put our expression equaling minus one we solve for x and we get + five so the oxidation number of chlorine in this case is + 5 the next example is the ammonium ion and we want to find the oxidation number for nitrogen so we let the oxidation number be X so it'll be X plus we have a four here so it's four multip by we know hydrogen is always plus one except in in peroxides so it'll be 4 multip plus one and the overall charge of the ammonium ion is we we just seen all we seeing here is a plus so that is really + one and when we solve for x we end up with x = -3 so the oxidation number of the nitrogen is minus 3 now this is where students get confused as well when we look at oxidizing and reducing agents now when a substance is acting as an oxidizing agent it is reduced in the process this is very important now the trick with this is not to Lo learn both definitions you learn one and you can remember the other one so when a substance is acting as an oxidizing agent we need to know that it is reduced in the process when a substance is acting as a reducing agent it is oxidized in the process right so don't get confused so what we are saying is if we have a substance a reacting with a substance B and this substance is an oxidizing agent the oxidizing agent will oxidize substance B but in the process this substance which is substance a will be reduced okay do you get that if a is an oxidizing agent and it reacts with substance B it will oxidize substance B and substance a will be reduced in the process let's look at the reverse now if substance p is reacting with substance Q q and substance p is a reducing agent the reducing agent will reduce substance q but in the process substance P will be oxidized all right so it's don't get confused with the statements just understand what is happening now you need to be able to give examples of oxidizing agents so the first example is acidified potassium manganate 7 sorry this is a typo here should be seven so we typically write it as km4 H+ and it is an oxidizing agent when it acts as an oxidizing agent it changes from purple to colorless because oxidizing agents are reduced now you need to understand the explanation as well the M4 minus ion changes from oxidation state + 7 to mn2+ so it goes from oxidation state plus 7 to plus 2 the plus 7 oxidation state is purple and the plus to oxidation state is colorless so that is why we are getting a purple to colorless color change another example of an oxidizing agent is acidified potassium D chromate 6 which is written as K2 CR2 207 H+ all right when it acts as an oxidizing agent we know that it will be reduced in the process it changes color from Orange to green and the explanation is the chromium or the chromate ion is changing from oxidation state plus 6 to+ three the plus 6 oxidation state is orange and the plus three oxidation state is Green ion tree salts can act as an oxidizing agent when it acts as an oxidizing agent it is reduced it changes color from yellow to pale green and what is happening is that the plus three oxidation state of the ion is changing to plus two plus three is where we have the yellow color and plus two is the greenish color in the case of concentrated sulfuric acid which can also act as an oxidizing agent what happens is that we get a pungent gas being produced in this case sulfur dioxide is being produced when concentrated sulfuric acid behaves as an oxidizing agent another oxidizing agent is nitric acid and the change is that a brown gas is produced in this case nitrogen dioxide is being produced and oxygen gas and chlorine gas are also oxidized in agents now we need to be able to give examples of reducing agents potassium iodide is a reducing agent and the color change is colorless to Brown and the explanation for this is that iodine is being produced all right Ion 2 salts also acts as reducing agents and they change from pale green to yellow in which case we are moving from oxidation state plus2 to+ three if it's if the Ion 2 salts are acting as a reducing agent it will be oxidized so that is why it is going from oxidation state plus2 to plus three hydrogen sulfide is also a reducing agent in this case a yellow precipitate is formed which is sulfur concentrated hydrochloric acid is also a reducing agent in this case a yellow green gas is evolved in which is chlorine gas other reducing agents are carbon carbon monoxide gas and hydrogen gas now there are substances that can behave as an oxidizing agent and a reducing agent two examples are sulfur dioxide and acidified hydrogen peroxide so they can act as both an oxidizing and a reducing agent now how do we test for an oxidizing agent if we want to test for an oxidizing agent we can use we can just add potassium iodide in which case the color changes from colorer Brown because iodine is being produced or we can add an AKs Ion 2 salt and the change in this case is a pale pale green to yellow or we can use hydrogen sulfide so what you need to do try to remember at least two of them and note the change that is taking place because once we observe the change then it gives us a positive indication that an oxidizing agent is present if we want to test for a reducing agent we can add potassium manganate 7 and the color change we get is purple to colorless or we can add acidified potassium D chromate 6 and the color change is orange to green or we can add an iron tree salt in which case the color change is from yellow to pale green so you need to remember a few of these and note the color change all right these are typical examination questions so for each of the following equation identify the substance that is being oxidized and give a reason for your answer now when you get a question like this the first thing you need to do the question is asking for which substance is being oxidized now if a substance is being oxidized if I were to just draw a vertical line like this and we have oxidation number and this is zero if we have an increase in oxidation number it is called oxidation right so the first thing you should look for is look for substances that have an increase in oxidation number now in this case zinc the oxidation number is zero and look on the right hand side the oxidation number has changed to plus two so the question is asking which substance is being oxidized in this case we can easily tell that zinc is the oxidation number is increasing from 0 to plus two so the answer is zinc for the second one if we look at the ion the oxidation number is plus two and the oxidation number after is plus three if we are moving from plus2 to plus three that is an increase in oxidation number so therefore it is the fe2+ that is being oxidized right not the chlorine chlorine the oxidation number is zero and the chloride ion is minus one therefore we have a decrease in oxidation number which a decrease in oxidation number is really a reduction all right um now this one is a little trickier now let's start with the easy ones nitrogen is zero CU it exists in its um standard state in this case we have copper solid so this is also zero so if we look at the Copper in this case here we know that oxygen is min-2 therefore the copper is +2 now copper is going from plus two to zero so therefore it's a decrease so that is reduction so it can't be the copper in the case of nitrogen the oxidation number of the nitrogen is -3 because you know that hydrogen is + one so therefore in nitrogen must be plus three and nitrogen is going from minus 3 to zero so on a number line or a vertical axis if we have zero line here and we have minus 3 and we are going to zero we are increasing oxidation number therefore it is the ammonia that is being oxidized okay in this next question you have to identify the oxidizing and reducing agents in the equation below so since we have ions present we can we know the oxidation number of ion here is Plus 2 ion is + three manganese is +2 I'm just looking at the charge on the ion here it's nothing complicated and in the case of um the manganate ion well we did this in one of the previous examples the manganese is plus 7 all right so which is the oxidizing agent if something behaves as an oxidizing agent it will be reduced so therefore we should see a decrease in oxidation number now the manganese or the um manganate ion is moving from + 7 to+ 2 therefore it is decreasing oxidation number therefore the manganate ion is being reduced if it is being reduced it is behaving as an oxidizing agent I needed to understand this logic all right manganese is moving from plus 7 to plus2 all right therefore it is being reduced and if it is being reduced it is behaving as an oxidizing agent right in the case of the fe2+ we are moving from plus2 to plus three if we are moving from plus2 to plus three it means that the fe2+ is being oxidized and if it is being oxidized it is behaving as a reducing AG agent all right so I just want you to understand the logic how we came up with this