JEE MAIN & ADVANCE 11th PCM Chemistry -Redox Reaction Demo Videos

Buy Now

Hello students let us continue discussion on redox reaction. Now in a last module we will discussing about calculation of  equivalent rates of oxidizing agent  and reducing agent and we also defined 3 important term Z factor today we will continue to discuss some example on equivalent rate calculation and then we will see to particular special cases Fec2o4 and Cu2S. let us start some example Find equivalent weight of Kbro3and Br2 in the following reaction . now you can see in this reaction 10 electrons being exchange , Aur hume yaad hai Z factor kis tarah se calculate karna hai so let us begin with KBro3, so we can say that 2 moles of KBro3number of electrons exchange or a gained in this case equal to 10, so there for we can say that for 1 mol of KBro3 number of electrons gained   so we can say that 1 mole of kbro3number of electrons gained will be equal to 10/2 that is five now Z factor KBro3 will become 5 and there for Equivalent weight Equal to molecular weight of kbro3 divided by 5 abhi me exact weight nahi nikal raha, me just symbolically represent kar raha hu and that is important for Br2 we see that number of electrons exchanged are 10 and since one mole of Br2 we need to see that nearly 10 electrons we have to use so Z factor of Br2 becomes 10 And equilent weight equals o molecular weight of bromine divided by 10. Let us see some example when Hno3 is oxidize into NH3the equilent weight of Hno3 will be how much? HNO3 changed In ammonia , sabse pehle hum inke   oxidation no. likh lete hai Hno3 will be equal to 5 and you know the algebirc method  to calculation of oxidation number , now in ammonia the oxidation number is -3 so we can see that we can calculate the change in oxidation number since there is 1 atom of nitrogen it will be equal to +5 –(-3) that is equal to 8, yaad rakhna change in oxidation is initial oxidation number  minus final oxidation number so ye jo 8 hai so this is become the Z factor so the equivalent weight equal to molecular weight divided by 8, let us see the next example . The Equivalent weight of H2So4 in the following reaction is humme ek redox reaction given hai aur usme H2SO4 ka equivalent weight pucha jar aha hai now can I say the z factor for H2SO4 will be equal to 2  , it should be wrong because it is not acid base reaction . it should be analyze like a redox reaction , then we focus on  loss and gained electrons , now what should be strategy , first we will find out the overall loss and gain of electrons in the redox reaction matlab jaise balanced karte hai uss process se loss and gained electrons nikalenge to do this we can focus on SO2 or na2Cr2O7, now we will use this information calculate Z factor for H2So4. Just remember there is 1 mole of H2SO4 involved in the reaction so let us focus on solution the oxidation no. of chromium K2cr2O7 is plus 6 and Cr2(So4)3 is plus 3 , since there are 3 atom of chromium we can see that number of electrons exchanged overall will be equal to six and there for we can say that for 1 mole of H2SO4 number of electrons exchanged  will be also equal to six .Z factor   for H2SO4  is 6 and equivalent weight will be equal to molecular weight divided by 6 so I remembered redox reaction me acid and base ka equivalent weight jo hota hai  wo loss ya gained of electrons se nikalte hai . now this calculation could  have be done if you just analyzed So2 and So4 two negative, so lets us take a another example the equivalent of Hno3 molecualr weight is just 63 in the following reaction is so we have given the reaction between copper and Hno3 to give copper nitrate .aur hamare pass 4 option hai . we analyzed this reaction understand the Hno3 is here just not acting on a agent but it is also act oxidizing  agent    so Hno3 ke dual  role hai, so it is act as a oxidizing agent and it is also provide acidic medium now find out the loss or gain of electrons to do this better copper because copper ka sirf oxidation ho raha hai Hno3 reduced bhi ho raha hai , Hno3 ke ander no ko jara dhayn se dekhenge  so using this information we will be calculate the Z factor then we will gate the equivalent weight. Lets us see the solution so we can see that the oxidation number in Cu is zero and that in copper nitrate is +2. Since there is 3 copper involved we can see the change in oxidation number is -6. If we have to do NO remember  we have only two NO corresponding to change in oxidation number . so we will find that is HNo3 +5 and that in NO is +2 . so the change in oxidation number is +6. To chahe me copper se karu , chahe Hno3 se karu  dono sahi loss and gain in oxidation number is 6aur hona bhi chahiye so number of electrons overall changed is 6 so there for 8 moles of HNO3number of electrons is equal to six  for 1 mole of Hno3it is 6/8 that is ¾  so equivalent weight will be equal to molecular weight divided by ¾  so we can say it will be 63*4/3 . now let us focus some special cases we focus on FeC2O4. FeC2O4 is a good reducing agent now fe^2 positive is oxidize  to Fe^3 positive . ab iss reaction dekh sakte hai reactant side pe charges 2+, and product side pe Charges 3+ now to balanced the charge since atoms  are already to balanced a charged we simply need to charged an electron to the product side now C2O4negative it is oxidize co2 and we can see that yah ape atoms pehel balanced karne padenge  so just multiply the product side by 2 now to balanced the charged reactant side got two negative charged and product side is zero so to balanced the charge we simply add two electrons to the product side ab me dono reaction ke liye dekh raha hu overall fec2o4 pe dono component oxidize ho rahe hai so FeC2O4 overall losses the total of 3electrons and hence equivalent weight of FeC2O4 will be equal to molecular weight  divided by 3 let us see another special case Cu2S. it is also good reducing agent now in this case Cu+ is oxidize to Cu^2+ again charge balanced karte hai reactant side pe charges  1 postive and product side pe  charges 2+. So to balanced the charged we simply add an electron to the product side . now since there are two copper we can say overall 2 electrons lost for copper . now sulphide ions is oxidizes to SO2first we balanced the atoms we to do add 2 H2o on reactant side oxygen ko   balanced karne ke liye and Add 4H+ to the product side yah ape assume kar liya medium in acidic in nature . so we add this and balanced  the charge we see that  reactant side has total  charge 2 negative  and product side 4+. To do this we will now add 6 electrons to product side and balances the charge  so 2 electrons copper ke and six sulphide ions ke  so CU2s losses 8 electrons and hence its equivalent  weight Molecular weight divide by 8, Thank you

Increase your scores by Studying with the BEST TEACHERS – Anytime and anywhere you want

Open chat
Can we help you?

Download App