hello everyone my name is Iman welcome back to my YouTube channel today we're going to be continuing our lecture on oxidation reduction reactions the first thing we want to talk about today is how to balance Redux reactions this is going to involve us recalling our general rules for balancing Chemical Equations I have them listed here the first thing that you want to do is determine what reaction is occurring what are the reactants what are the products and what are the physical States involved once you figured that out you can go ahead and write the unbalanced equation looking at the unbalanced equation you're going to balance one atom at a time starting with the substance that has the most elements present if necessary in this step you can use fractional coefficients to balance the simplest substance but in the fourth step you're going to want to get rid of those fractional coefficients so if there are any multiply each coefficient by a factor that's going to result in the smallest whole number coefficient then last but certainly not least double check that each atom is indeed balanced now we have an added consideration here for Redux reactions to balance them both the net charge and and the number of atoms must be equal on both sides of the equation Now by assigning oxidation numbers to the atoms in both the reactants and products you can identify which atoms are oxidized and which atoms are reduced now this step is essential because it helps to determine how many electrons are lost in oxidation and gained in reduction now the most common method for balancing Redux equations is the half reaction method it's also known as the ion electron method in this method the equation is separated into two half reactions the oxidation part and the reduction part now each half reaction is then balanced separately and then they're added together to give a balanced overall reaction so what we want to do here is go over some of the detailed steps on how to balance Redux reactions and then we're going to jump into a practice problem in that practice problem these steps will make more sense the first thing is you're going to want to identify the oxidation and reduction half reactions so you're going to separate the overall Redux reaction into two half reactions one that represents oxidation and one that represents reduction then you're going to go ahead and you're going to balance each half reaction you're going to balance atoms other than oxygen and hydrogen first then you can balance oxygen atoms and hydrogen atoms a side note here for balancing the oxygen and hydrogen atoms in an acidic solution you can balance oxygen atoms by adding water to the side that is oxygen deficient and you can balance the hydrogen atoms by adding hydrogen ions in basic Solutions you can instead use hydroxide and water to balance also you want to balance charge so you're going to add electrons to one side of the reaction to balance the charge the number of electrons added should equal the change in oxidation state so that it ensures that the total charge on both sides of the half reaction are equal the next step is to equalize electron transfer multiply each half reaction by a coefficient so that the number of electrons that is lost in the oxidation half reaction equals the numbers of electrons gained in the reduction half reaction and again this is going to ensure that electrons are conserved in the overall reaction then you're going to combine those half reactions you're going to add the two half reactions together and you're going to ensure that electrons cancel out appropriately you're going to combine the reactants and products to obtain the overall balanced Redux reaction and then of course last but not least you want to verify that everything is indeed balanced with these rules in mind let's go ahead and tackle a practice problem this example wants us to balance the following Redux reaction using the half reaction method what we have here is the perangan ion reacting with the iodide ion to form molecular iodine and the manganese 2 ion the first thing that we want to do is assign oxidation numbers so that we can determine the oxidation half reaction and the reduction half reaction let's start with our per manganate ion the most electronegative element here is going to be oxygen it's going to have an oxidation number of Min -2 and there are a total of four oxygen atoms so we have a total of minus8 now something that's important to keep in mind is that the sum of the oxidation numbers of the atoms that are present in a polyatomic ion like permanganate is going to be equal to the charge of the ion here that's going to be minus1 in that case the manganese is going to have an oxidation number of + 7 because + 7 - 8 gives us a total of minus1 and that matches the charge of the ion the next one we want to work on is our iodide ion what is the oxidation number here well what we have is a monatomic ion the oxidation number for a monatomic ion is going to be equal to the charge of the ion so the oxidation number here is going to be equal to minus one just like the charge of this iodide ion next up we have molecular iodine the oxidation number here is going to be zero remember the oxidation number of a free element is zero always last we have our manganese 2 ion again this is a monatomic ion so the oxidation number is equal to the charge of the ion which in this case is going to be+ 2 now the next thing we can do is figure out what our oxidation half reaction is and what our reduction half reaction is let's go ahead and take a look at our iodide ion and compare it to molecular iodine here the iodine atom goes from minus1 Ox oxidation number to zero it is losing electrons so this is our oxidation half reaction I'm going to go ahead and write it right here now what about our reduction half reaction let's take a look at our permanganate ion and compare it to our manganese 2 ion here manganese goes from + 7 oxidation number to +2 it is gaining electrons so this is our reduction half reaction now that we have that figured out our next step is to go ahead and balance these half reactions we're going to start with our oxidation half reaction now this oxidation half reaction involves iodide ions transforming into molecular iodide here here we have two iodine atoms here we have one to balance that we're going to add a coefficient of two here now that we have these atoms balanced the only thing we're concerned with is also balancing the charges by adding electrons where they're needed the charge on the reactant side is min-2 because we have two iodide ions and the charge on the product side is zero to to balance this we're going to add two electrons to the product side now our oxidation half reaction is all good and balanced we can go ahead and move into our reduction half reaction now here we're going to first balance the manganese atoms since one manganese atom is present on both sides it's already balanced so then we move on to balancing the oxygen atoms all right and we're going to do this by adding water on the reactant side we have four oxygen atoms so I'm going to go ahead and add four water molecules to the product side so that we now have four oxygen atoms on the reactant side four oxygen atoms on the product side then we're going to balance the hydrogen atoms on the product side we have a total of eight hydrogen atoms so we're going to want to add eight hydrogen ions to the reactant side to make sure that that now our hydrogen atoms are balanced then finally we want to balance the charges by adding electrons let's look at our reactant side here we have a minus1 charge and then we have eight plus charge from our hydrogen ions this gives us a total charge for our reactant side of plus 7 on the product side we only have a charge of + two from that manganese 2 ion the only way that we can balance these charges is if we add five electrons to our reactant side when we do that we now have a total charge of plus two in the reactant side that perfectly matches our plus two charge on the product side now that we have balanced these reactions we want to combine them back together to combine the half reaction we have to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction so here for our oxidation half reaction we have a total of two electrons on the product side and in our reduction half reaction we have a total of five electrons on the reactant side this is good they're on opposite sides so that when we add this together they're going to cancel each other out but the problem is we have two and five those aren't going to properly cancel each other out what we're going to do is multiply each half reaction by the appropriate coefficient that's going to get us to have an equal number of electrons on both sides that means I'm going to multiply the oxidation half reaction by five and I'm going to multiply the reduction half reaction by two this is going to give us the following oxidation half reaction now that we've multiplied it by a coefficient of five and we're going to get the following reduction half reaction now that we've multiplied it by a coefficient of two and now that we have matching number of electrons one on the product side one on the reactant side when we add these together they're going to cancel each other out and that is exactly what you want now we can go ahead and write the overall reaction adding these balanced half reactions together gives us 10 iodide ions plus two permanganate ions plus 16 hydrogen ions and that gives us on the product side five molecular iodine molecules plus two manganese 2 ions plus eight water molecules and there you have it we have now solved this problem let's go ahead and move into our final conversation before we move into our last and final objective to conclude this section it's important to revisit and clarify the terminology that we used for oxidation reduction and their corresponding ing agents the phrasing it may vary slightly across different classes or textbooks but the underlying Concepts remain the same for oxidation oxidation refers to the loss of electrons that is the allpurpose definition in general chemistry oxidation is characterized by an increase in the oxidation state of an element in organic chemistry oxidation is often associated with the gain of oxygen or the loss of hydrogen for instance when a molecule loses hydrogen atoms or gains oxygen atoms it is said to be oxidized for reduction the all-purpose definition is that you gain electrons it is the gain of electrons in general chemistry reduction is the process that is marked by a decrease in the oxidation state in organic chemistry reduction us usually involves the loss of oxygen or the gain of hydrogen now the substances that facilitate these processes are known as oxidizing and reducing agents an oxidizing agent causes another substance to be oxidized chemically it is the substance that is reduced because it gains the electrons lost by the oxidized substance for an oxidizing agent to function function it must contain an element in one of its higher oxidation States and ready to accept electrons on the other hand reducing agent causes another substance to be reduced it is the substance that is oxidized in the reaction as it donates electrons to the substance being reduced a reducing agent must contain an element in one of its lower oxidation States and it's supposed to be prepared to lose electrons now we can move into our final objective here we're going to revisit the topic of net ionic equations we have discussed net ionic equations plenty of times in this playlist and we've discussed the three main steps to writing our net ionic equation the first step you want to write the molecular equation you want to write the reactants and the products in their undissociated form then in the Second Step the total ionic equation you want to write your aquous reactants and products in their dissociated form keeping in mind that our solids our insoluble salts are going to stay as is then you're going to get rid of your spectator ions these are the ions that appear in both the reactant side and the product side you're going to get rid of those and what you will be left with is your n ionic equation now a common example we have gone over several times is the reaction between silver nitrate and sodium chloride so here is the reaction in its undissociated form here is the total ionic equation where we write our aquous Solutions in their dissociated form and once we get rid of The Spectator ions which are our sodium ions and our nitrate ions we are left with the net ionic equation now in chapter 4 we discussed reaction types and when we did we left out the rationale for why certain elements come together and others do not now that we have discussed oxidation reduction reactions it should be more clear that the gain and loss of electrons drives the formation of many compounds especially ionic ones so what we're going to do now is we're going to re revisit many important reaction types and understand their basis in oxidation reduction reactions let's go ahead and get started with combination reactions combination reactions are when two or more species come together to form a product now in this reaction we have hydrogen and Florine gas combined to form hydrogen fluoride to understand this reaction we can go ahead and break it into two half reactions that separately show the oxidation and reduction processes here is our oxidation half reaction in this half reaction we have molecular hydrogen it loses electrons indicating that it is oxidized each hydrogen atom here goes from an oxidation state of zero to an oxidation state of + one notice that here we've added two hydrogen ions so that we can balance our hydrogen gas and to balance the charge we've also added two electrons here we see our reduction half reaction in this half reaction molecular Florine gains electrons indicating that it is produced each Florine atom in F2 goes from an oxidation state of 0o to min-1 and to the reactant side we've added two electrons to balance the minus 2 charge on the product side now we can combine the two half reactions to get an overall balanced Redux reaction this is our overall balanced Redux reaction and it shows the transfer of electrons from hydrogen to Florine now here hydrogen acts as a reducing agent because it donates electrons and it is oxidized from an oxidation state of zero to + one Florine here acts as an oxidizing agent because it accepts electrons and it is reduced from an oxidation state of zero to minus1 now in this action there are no spectator ions because all of the species are actively participating in the Redux process and the entire process involves the direct transfer of electrons from hydrogen to Florine resulting in the formation of hydrogen fluoride now that we've talked about combination reactions let's talk about decomposition reactions this is where one product breaks down into two or more species here we're going to look at the decomposition of ammonium D chromate let's go ahead and write the half reactions what you see here is the oxidation half reaction in this half reaction the ammonium ion loses electrons indicating oxidation the nitrogen atoms in the ammonium ion they have an oxidation state of Min -3 which changes to zero in nitrogen gas and here we've added eight hydrogen ions to the product side to make sure that the hydrogen atoms are balanced and that the charge is also balanced here you see the reduction half reaction so in this half reaction the DI chromate ion gains electrons indicating reduction the chromium atoms in dichromate ion have an oxid xation state of + 6 which changes to+ three in chromium 3 oxide now we can go ahead and combine the two half reactions and we're going to get the following overall balanced Redux reaction now the nitrogen atom in the ammonium atom or in the ammonium ion acts as a reducing agent because it donates electrons and it's oxidized from an oxidation state of-3 to0 now the chromium atom in the D chromate ion acts as an oxidizing agent because it accepts electrons and it is reduced from an oxidation state of + 6 to +3 now in this reaction there are no spectator ions because all of the ions here are actively participating in the Redux process the nitrogen from the ammonium ion is oxidized and the chromium from the DI chromate ion is reduced and the result is the formation of nitrogen gas chromium 3 oxide and water let's go ahead and talk about combustion reactions next combustion reactions these are described as a fuel usually a hydrocarbon that's mixed with oxygen and it forms carbon dioxide and water so our example reaction here involves methane and oxygen gas they react together to give us carbon dioxide and water and we can go ahead and write the half reactions here you see the oxidation half reaction in this half reaction carbon in methane all right the carbon in methane loses electrons indicating oxidation so the carbon atom in methane it has an oxidation state of Min -4 which changes to + 4 in carbon dioxide now we can look at the reduction half reaction as well in this half reaction oxygen molecules gain electrons indicating reduction each oxygen atom in oxygen gas has an oxidation state of zero which changes to min-2 in water and here we use water to balance the oxygen atoms and we also have eight electron here in the reactant side to balance charge we balance charge with these eight hydrogen ions that we've added to the reactant side to be balanced with the eight hydrogens we have from water on the product side and similarly going back to our oxidation half reaction here we have eight electrons to balance the charge as well now we can go ahead and write our overall balanced equation all right we can go ahead and write our balanced overall equation and we get the following equation there are no spectator ions here now the carbon atom in methane again acts as the reducing agent because it donates electrons and it is oxidized from an oxidation state of Min -4 to+ 4 the oxygen molecule acts as the oxidizing agent because it accept electrons and it's reduced from an oxidation state of 0 to min-2 now combustion reactions they often have complex half reactions depending on the type of fuel used you can see it here as well it's pretty complex and it's a lot more species involved than some of the other reaction types we've looked at in this instance the carbon in methane is oxidized and the oxygen is reduced wonderful let's move on to the next reaction type double displacement reactions double displacement reactions involve the switching of counter ions now because all ions generally retain their oxidation state these are not usually Redux reactions so for example in this reaction between silver nitrate and hydrochloric acid we form nitric acid and silver chloride again in this double displacement reaction there's no change in oxidation States so the net ionic equation really just highlights the formation of the precipitate without altering the charges of the ions and so in this net ionic equation we have silver ions and chloride ions combined to form solid silver chloride demonstrating that exchange of ions now the remaining ions hydrogen ion nitrate ion they stay in solution as spectator ions and they're not involved in the formation of the precipitate now next up we want to talk about disproportionation reactions this is a specific type of Redux reaction where a single element under goes both oxidation and reduction within the same reaction and it results in two different products this type of reaction is actually quite common in biological systems and it's utilized by many enzymes a a example is cataly cataly helps protect cells from oxidative damage by breaking down hydrogen peroxide which is a potentially harmful byproduct of metabolic processes so the reaction for that is two hydrogen peroxide molecules break down to two water molecules and oxygen gas now in hydrogen peroxide the oxy oxidation state of each oxygen atom is minus one and the oxidation state of the hydrogen is plus one there's two of each and that gives us um a perfect neutral compound when we sum up the oxidation numbers then in water the oxidation state of Oxygen's actually min-2 all right and in molecular oxygen the the oxidation number for oxygen is zero so one of the oxygen atoms in hydrogen peroxide is oxidized from an o oxidation state of minus1 to0 forming oxygen gas this means it loses electrons and it's been oxidized and then the other oxygen atom in hydrogen peroxide it's reduced from an oxidation state of minus1 to min-2 forming water so this means it gains electrons so in this single reaction oxygen undergo both oxidation and reduction and this is the definition of a disproportionate ation reaction lastly let's briefly cover oxidation reduction titrations these are a type of titration that's used to determine the concentration of an unknown solution through a Redux reaction so very similar in setup to acidbase titrations Redux titrations just involve the transfer of electrons instead of protons the goal is to reach the equivalence point where the amount of titrant added is stochiometric L equivalent to the quantity of the analyte in the solution indicators that change color at specific voltage values can then be used to Signal the equivalence point in redo titrations Now quickly there are two types of Redux titrations I just want to quickly Define first type is ideom metric titrations these involve the use of iodine as the titrant iodine can act as both an oxidizing and reducing agent and this method is particularly useful for determining the concentration of substances that can oxidize iodide ions to iodine or reduce iodine to iodide ions and the end point is usually detected by a starch indicator which forms a blue black complex with iodine then we have potentiometric titration this is a method where no chemical indicator is used instead the progress of the titration is monitored by measuring the electrical potential difference the voltage of the solution with a voltmeter and as the titration progresses the voltage changes and this change is used to determine the equivalence point so this is analogous to following an acid base titration with a pH meter instead of a color indicator and they are highly accur accurate and they're often used when precise determination of an end point is required so with that we've completed all our topics for this chapter I hope that this was helpful please let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors