Hello, students, let us continue our discussion on redox reactions. Now, in the last modules we will be discussing about calculation of equivalent weights of oxidising agents and reducing agents and we also defined a very important term z factor. Today we will continue to discuss some examples on equivalent weight calculation and then we will see two particular special cases FeC2O4 and Cu2S. Let us start with some examples.
Find the equivalent weight of KBrO3 and Br2 in the following reaction. Now, you can see that in this reaction 10 electrons are being exchanged. And we remember how we have to calculate z factor. So, let us begin with KBrO3. So, we can say that for 2 moles of KBrO3 number of electrons exchanged or rather gained in this case will be equal to 10. So, therefore we can say that for 1 mole of KBrO3, number of electrons gained. So, we can say that, so, we can say that for 1 mole of KBrO3, number of electrons gained will be equal to 10 by 2 that is 5. Now, the z factor of KBrO3 will be 5 and therefore equivalent weight will be equal to molecular weight of KBrO3 divided by 5. I am not getting the exact weight, I am just representing it symbolically and that is what is important. Now, for Br2 we see that the number of electrons exchanged are 10 and since there is only 1 mole of Br2 we need to see that there is only 10 electrons that we have to use. So, the z factor of bromine becomes 10 and equivalent weight will become molecular weight of bromine divided by 10.
Let us see the next example. When HNO3 is oxidised into ammonia means NH3, the equivalent weight of HNO3 will be how much? Now, HNO3 is being changed into ammonia. First of all we will write their oxidation number, nitrogen in HNO3 will be equal to plus 5, you know the algebraic method to calculate the oxidation numbers. Now, in ammonia the oxidation number of nitrogen is minus 3. So, we can see that, we can calculate the change in oxidation number, since there is only one atom of nitrogen it will be equal to plus 5 minus, minus minus 3 that is equal to plus 8. Keep in mind, change in oxidation number is initial oxidation number minus the final oxidation number. So, this 8 is there, this is what it will become the z factor and therefore equivalent weight is molecular weight divided by 8.
Let us see the next example. The equivalent weight of H2SO4 in the following reaction is. They have given us a redox reaction and in that they are asking for weight of H2SO4. Now, can I say that the z factor for H2SO4 will be equal to 2? Actually it would be wrong, why, because it is not an acid base reaction. It should be actually analysed like a redox reaction, which means we have to focus on the loss and gain of the electron. So, what should be your strategy? First we will find out the overall loss or gain of electrons in the redox reaction. It is like balancing, and with that process we have to get the loss or gain of electrons. To do this, we can either focus on SO2 or sodium dichromate that is Na2Cr2O7. Now, we will use this information to calculate the z factor of H2SO4. Just remember that there is only 1 mole of H2SO4 involved in the reaction. So, let us focus on the solution. The oxidation number of chromium in K2Cr2O7 is plus 6 and that in Cr2SO4 thrice is plus 3. Since there are two atoms of chromium we can see that the number of electrons exchanged overall will be equal to 6 and therefore we can say that for 1 mole of H2SO4 number of electrons exchanged will be also equal to 6. So, z factor of H2SO4 is 6 and the equivalent weight will be equal to molecular weight divided by 6. So, remember that in redox reaction acid and base equivalent weight will be got from loss or gain of electrons. Now, this calculation would also have been done, if you have just analysed SO2 to SO4 2 negative.
So, let us take a look at the next example. The equivalent weight of HNO3, molecular weight is 63, in the following reaction is. So, we have given a reaction between copper and HNO3 to give you copper nitrate and NO and H2. And we are having four options. Now, to analyse this reaction understand that HNO3 here is not just acting as an acid but it is also acting as an oxidising agent. So, HNO3 has dual roles, it is going to act as an oxidising agent and it will also provide the acidic medium. Now, we will find out the overall loss or gain of electrons. And to do this better would be copper because only oxidisation of copper is done. HNO3 is reducing, so, in HNO3 we have to observe NO very carefully. So, using this information we will calculate z factor and then we will get to the equivalent weight part. Let us see the solution. So, we can see that the oxidation number of Cu in copper is zero and that in copper nitrate is plus 2. Since there are 3 copper involved, we can see that the change in oxidation number is minus 6. Now, if we had to do NO, remember we have got only 2 NO corresponding to the change in oxidation number. So, we will find that nitrogen in HNO3 is plus 5 and that in NO is plus 2. So, the change in oxidation number is plus 6. Even if I do through copper or from HNO3 that means NO, from both the loss and gain of oxidation number will be 6 and it should be like that. So, number of electrons exchanged overall is 6 and therefore we can say for 8 moles of HNO3 number of electrons is equal to 6. For 1 mole of HNO3 it will be 6 by 8 that is 3 by 4. So, equivalent weight will be equal to molecular weight divided 3 by 4 or we can say it will be 63 times 4 by 3.
Now, let us focus on some special cases. We will focus on ferrous oxalate, FeC2O4. FeC2O4 is a good reducing agent. Now Fe2 positive is oxidised to Fe3 positive. Now, in this reaction we can see on the reactant side the charge is 2 positive and on the product side the charge is 3 positive. Now, to balance the charge since atoms are already balanced, to balance the charge we simply need to add an electron to the product side. Now, C2O4 negative 2 is oxidised to CO2 and we can see that here we will have to balance the atoms first. So, we just multiple the product side by 2. Now, to balance the charge, reactant side has got 2 negative charge and product side has 0. So, to balance the charge we simply add 2 electrons to the product side. Now, I am seeing both the reactions here that overall FeC2O4 both the components are oxidised. So, FeC2O4 overall loses a total of 3 electrons and hence equivalent weight of FeC2O4 will be equal to molecular weight divided by 3.
Now, let us see another special case, Cu2S, cupra sulphide. It is also a good reducing agent. Now, in this case Cu positive is oxidised to Cu 2positive. Again we will balance the charge. On the reactant side charge is 1 positive and on the product side charge is 2 positive. So, to balance the charge, we simply add an electron to the product side. Now, since there are 2 copper, we can say overall 2 electrons are lost for copper. Now, sulphide iron is oxidised to SO2. Now, first we will balance the atoms. So to do that, we have to add 2 H2O on the reactant side to balance the oxygen and 4H positive on the product side. Here we have assumed that medium is acidic in nature. So, we add these and then to balance the charge we see that the reactant side has a total charge of 2 negative and product side has 4 positive. To do this, we will now add 6 electrons to the product side and this balances the charge. So, 2 electrons of copper and 6 sulphide iron. So, Cu2S loses the total of 8 electrons and hence its equivalent weight will be equal to molecular weight divided by 8.
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