COMEDK 2024 Chemistry | Top 30 Electrochemistry Questions for #COMEDK

hello students welcome back to our Channel dsha Karnataka so as you all know your Comet K exam is approaching so we have started our Series where we are discussing the top 30 questions of all the important chapters from chemistry so we have already completed it for Solutions chapter so do watch it so now in this video I'll be discussing the top 30 question from one of the most high weightage topic that is electrochemistry so you can easily expect three to five questions from this chapter so revise all formulas and then please watch this video and Sol try solving these questions so these questions most of them are your pyq so you can understand what type of questions are asked from this chapter okay so if you able to solve this questions you can assume that this chapter is ready for your exam okay so question levels are moderate to difficult okay so let us see these questions so starting with your question number one so calculate the standard cell potential in BT of the cell in which the following reaction takes place so we have fe2+ to fe3+ ag+ to AG okay so now e of AG + 2 a is XV e of fe2 + 2 Fe is yv and E not of Fe 3 + 2 Fe is zv now we have to calculate the E not of the cell so e not of the cell is equal to e not of cathode so which is the cathode here which is undergoing reduction that is ag+ 2 AG minus E of anode that is your Fe + 3 to Fe +2 right but if you see the question carefully Fe + 3 to Fe + 2 value is not given rather we are given with fe2 + 2 Fe and fe3 + 2 Fe value so using these two datas we have to derive that value first so that simple way of doing that is your latima diagram we have Fe fe2 fe2+ and fe2+ to fe3+ okay now let this value be E1 okay this value be E2 and this value be E3 okay now according to this one what we can write is N3 E3 is equal to N2 E2 + N1 E1 okay now I'll mention the values here what is E3 value here E3 is fe3 fe2 fe3+ okay then E2 value is what E2 value is fe2+ to fe3+ okay and E1 value is fe2 fe2+ okay these three values I have mentioned now let us see the first one N3 E3 from here to here how many electrons are involved three electrons are involved so N3 will be three now if you see in the question fe2 fe3+ value is z right so if I want to calculate this is the reduction potential so oxidation potential will be negative of it so this one will be 3 * of minus Z that is equal to N2 E2 N2 is 1 and E2 value this we have to calculate plus N1 here number of elect involved is two and E1 again you can see here the reduction potential is given but I want the oxidation potential that is fe2 fe2 plus so it will be reverse of this one that is minus of Y then what is my E2 here E2 value will be 2 y - 3 Z okay but this is what E2 is your Fe 2 + 2 Fe 3 + but if you see the question in the formula I need Fe + 3 to Fe +2 so I can take the reverse of this one so I'll get 3 Z - 2 y okay that is your e of Fe + 3 2 F + 2 right so now we can substitute this value in the formula so what is the E not of silver so for silver it is X x minus Fe + 3 to F2 that is 3 Z - 2 y so if I solve this I'll get x + 2 y - 3 Z okay so the correct answer for this question is your option number c so little lengthy but a good question so please practice this one this one is from your uh Nur equation and latima diagram okay next question okay so here is a big announcement for all of you since your Comet K exams are approaching and many requests have come to us to start a test Series so we have started a absolutely free Comet K mock test Series where there will be three mock test you'll be getting all the video Solutions and the PDF of the solutions in detail so the first test is starting from 20 9th April so you can download the question paper from our WhatsApp channel so the link of the WhatsApp channel will be there in the description section of the video okay and you can also download it from the community section of this channel so just go there and download it and attempt the test in the exam conditions and come back to see the solutions okay so let us go to our second question find the value of uh equivalent conductance of potas Alum so first this first you should know the formula of potas Alum so in potas Alum we have K2 al2 s so4 whole 4 okay this is the formula so just we have to add for this one for k+ it is 73.5 so 73.5 into 2 plus for aluminium it is 189 so 189 into 2 and then for sufer it is 160 so 160 into 4 so if we add this we'll be getting around 1165 okay so the correct answer for this question is option number B okay next question question number three when a lead storage battery is discharged so what happens so sulfuric acid is consumed so here h204 okay gets converted to pbso4 so the correct answer for this question is option number D next during electrolysis of a solution of ag3 9650 kums of charge is passed through the solution what will be the mass of silver deposited at the cathode so this question is from your Fades first law right so in Fades first law we have m is equ Al to Z into I into T this it is nothing but charge so I can write z into Q now what is z m mass divided by n into 965 so silver M mass is 1 108 for silver the reaction is AG + 1 electron gives AG so number of electrons is 1 and then 9 6 5 0 0 and charge is 9650 okay so if we solve this we'll be getting around 10.8 right so the correct answer for this question will be option number B that is 10.8 the next question limiting Mar conductivity of nabr so we need nabr right so first of all there should be na so all the options are having na okay next option should have BR so here it is BR here also it is BR but here you see NaOH and nabr there's nothing to be cancelled but you see second option NaCl and KBR is there so what extra is formed here KCl extra is formed and then that KCl has to be subtracted so the correct answer for this question will be option number B easy question from your cra's law so many questions have been found in K rla Cas It generally not many questions were there from K but in Comet K you have more questions also from this one col res then the E not of the cell for this reaction is given the calculate the value of equilibrium constant so this formula this question is little change here it is min - 0.827 7 okay so now if you solve here e not of the cell is equal to minus sorry plus 2.303 RT by NF so we can put it by 0.059 by n log of K right this value is there so I'll erase this one so now we can solve it n value is your 1 right so we can have log K is equal to e not of the cell is - 0. 8277 7 n is = 1 by 0.059 okay so just we can for example we can take it like 827 by 59 okay for easy calculation 8 27.7 by this one so if you can see this value will be coming around 14 okay minus 14 so K value will be equal to 10 the^ of -40 so the correct answer for this question will be option number a okay so like you know 14 siga is somewhere around 84 right so somewhere around 800 is there so the answer you can guess is 10 ^ of-4 okay next is question number seven predict the pH of the following half cell if oxidation potential is 3 uh 3.5 okay so now if you see this question oxidation potential how to calculate e of the cell or E of the electrode e of the electrode is equal to e not of the electrode minus 059 by n log of H+ right so now if you calculate what is e not e of the cell is given to you 0.305 and for hydrogen electrode we know for hydrogen electrode the E not value is assumed to be zero so minus I'll take it as 06 for EAS calculation 1 log of H+ right now if I solve this one this is35 / 06 is equal to log ofus log of H+ so minus log of H+ is nothing but pH value so this one we can take it like3 by 06 right so we'll get around five right so the correct answer for this question will be option number c that is 5.85 okay so now let us see the next question calculate the value of uh infinite mol conductance using the appropriate mol conductance of the electrolytes given in the table so first one is your so the question is given we have to find for h o a c okay so first you find where this OAC part okay the acetate ion where it is present it is present in na o so I'll take Lambda of n a o a c right now I need H value so from where where I'll get i'll get from Lambda of HCL now if you see these two if I add these two I'll getting extra is NAC so I have to subtract it from nccl right so what are the values I'm getting this is 91 plus HCL is 426.7 point5 so if you solve this one the correct answer for this question is 39.7 so the correct answer is option number c very easy question so you can see cold R every year so these are some of the previous year papers so col's law questions are asked again and again okay so that is one point what you have to revise properly then is reduction potential for a cell reaction MN plus plus n electron M the Nur equation for the electrode potential for any concentration measured with respect to standard hydrogen electrode is so we know the equation is MN Plus+ give you m so e not of the cell minus RT by NF log of product by reactant so 1 by m n plus concentration so the correct answer for this question is option number a here what is the wrong this m concentration is not taken here if you are taking log it has to be 2.303 that is why this is wrong and here what is the problem here it is uh e of the cell is equal to I think both the options here okay this is 1 by m right the product by reactant so this concentration is taken to be 1 divided 1 by MN plus so this equation is also incorrect next which which one of the following has a potential more than zero okay so if we see that will be the Formula E of the H I'll write like H + 2 H2 is equal to e of H + 2 H2 minus 059 by 1 log of 1 by H+ concentration right so now if I take here 1 mol log 1 is equal to 0 so we will get this value as 0 this is already 0 so - 0 59 by 1 log of 1 by H+ concentration right but now if you see here log of 2 so if I substitute here 2 so -59 by 1 log of 0.5 right so now if you solve this this value will be negative so negative negative we'll get a positive value positive value will be more than zero so the correct answer for this question will be option number B okay now next question is when 9.6 9655 colums of electricity so again this is from your Fades law similar question we have solved m is equal to Z into it so8 / 1 into 965 here charge is 965 so that is equal to 1.08 the correct answer for this question is option number c okay next corosion of iron is an essentially an electrochemical phenomenon where the cell reactions are so we know Fe gets oxidized to fe2+ and the dissolved oxygen in water is reduced to O minus okay so this is direct for fact based question correct answer for this question is option number a next question number 13 the EMF of a cell containing sodium and copper electrod is 3.05 volt so we know e not of the cell is equal to e not of cathode so which is the cathode here copper okay so you should always remember left side is for oxidation right side is for reduction okay so cathode minus anode so e not of sodium okay now the question is asked that if the electrode potential of copper is 0 so this value is 3.05 and this value is 0.34 and I need to find for sodium then what is the reduction or electrod potential of sodium so if I solve this we'll get 0.34 minus 3.05 that is equal to - 2.71 volt so the correct answer for this question is option number a okay so we know sodium is a metal its reduction potential will be negative so that is why we are getting negative reduction potential of 2.71 volt okay very easy question against from Nur equation next the space in the dry cell is filled with so dry cell is filled with M2 zncl2 and nh4cl so the correct answer for this question is your option number c so this question is from your commercial batteries okay next question the standard electrode potential of Daniel cell is 1.1 volt what is the standard gives free energy for the reaction so we have this formula Delta G is equal to minus NF e of the cell right now at equilibrium for this one Daniel cell is your copper Zin cell where n value is 2 so I can substitute 2 into 965 into e not of the cell is 1.1 volt so if I solve this the correct answer will be -200 12.3 K per mole okay so the correct answer for this question is option number B okay very easy direct based questions now coming to next question the cathode reaction in the dry cell so if you know dry cell again just now we discussed M2 nh4 will be there reduction will take place so the correct answer for this question is option number B zinc is the anode so we can eliminate these two options directly okay next question Lambda not M for nh4 CL NaOH and NaCl are given calculate for nh4 o so let us first note down the values for nh4 cl the value is 130 then Lambda M for NaOH is 248 and then Lambda KN of NAC is 126.5 we need to calculate for nh4 o right so first of all you see in which compound we will get nh4 that is from nh4 CL so we'll take first that one next we need o o is present in NaOH so just add that one Lambda not of NaOH now you see we will take nh4 and we'll take o what is left extra NAC that has to be subtracted so Lambda KN of NAC so if I add this one 248 sorry first is your 130 We'll add it serially so NH um first is nh4cl that is 130 plus NaOH is 248 and then na is 126.5 so if we solve this we'll be getting around 21.5 so the correct answer for this question is option number a okay direct easy based question next in the electrochemical cell the reaction will be fible when so reaction for reaction to be fible Delta G has to be negative right that is the first condition so we can eliminate all these options from here now if Delta G is negative we have this formula right so if this one has to be negative this value has to be positive because if this value also comes out to be negative then negative negative will become positive and Delta G will become positive which is not required so e not of the cell must be positive so correct answer for this question will be option number a okay next the standard EMF of the cell ZN zn2+ is 1.56 volt if the standard reduction potential of AG is 0.8 Vol the standard oxidation potential of zinc is okay so this question is again from your uh calculation of EMF of the cell very simple question see e not of any cell is equal to e not of cathode so which is my cathode here AG + 2 AG okay so please remember left side is always anode and right side is always cathode there's one easy way to remember this you can remember loan Lo stand L stands for left o stands for oxidation a stands for anode and the charge of the anode is negative right so there is always left side the anode one where oxidation takes place so e not of left side is anode and right side is your cathode so e not of cathode that is your right side minus E not of anode that is ZN 2 + 2 Z we are taking the reduction potentials so e not of the cell is how much given to is 1.56 volt okay and E not of silver is 0.8 VT and minus E not of ZN 2 + 2 ZN so what will be the value of e not of ZN 2 + 2 ZN so that is your we'll get around 0.76 minus okay now the question has asked you to calculate the oxidation potential so oxidation potential will be the reverse of this one that is + 0.764 so the correct answer for this question is option number B that is plus 0.76 Vol the next question is on the m conductivities the m conductivity of NaOH NaCl and bac2 at infinite dilution are given so let us note down their values first so for Lambda M of NaOH is 2481 into 10 ^ of -2 s m squ mole inverse next one is given for NaCl so for NaCl the value is 1. 265 into 10 ^ of minus 2 so it is very important that you note down the values correctly then we need for uh B2 the B2 value given is um 2.8 into 10 ^ ofus 2 now question is asked for barium hydroxide okay now if you see this properly we need first first is barium in which of the following we'll get barium that is the third one so I can take Lambda of barium chloride now the next one is required Hydro o ions o ions is present in NaOH but you see we require two times of O ions so I'll take them take this one two times of O ions right now if you see this one what is the extra that we are forming here so we are getting two chlorines here and 2 O here right and to na here so we can subtract we need berium hydroxide the extra is Na and Cl so I will subtract min-2 * of Lambda not of NAC so just substitute this values for barium chloride it is 2.8 into 10^ of -2 Plus for sodium it is 2 * of 2.48 1 into 10 ^ of -2 then minus 2 * of NaCl is 1.26 5 into 10^ ofus 2 so once you solve this the value you'll be getting around is 5232 into 10 ^ of minus 5 the correct answer for this question is option number a okay okay now let us go to the next question the consider the half cell reactions e not cell for the given reaction is so what we have to do is two reactions are given we have to identify how can we get the third reaction from these two reactions so now you see in any reaction the electron number should be same here the number of electrons is 2 here the number of electrons is one so let us multiply this equation with two so what is the equation first I'm having mn2+ + 2 electron is MN and the second one is 2 mn2+ is giving me 2 MN 3+ okay plus 2 electrons now if I add these two reactions okay so here see in like enthalpies and other questions we used to multiply the two to the value also but in this type of in electrode potential we need not add okay so do not multiply this one with two clear now if we add these two reactions what I will get so 2 mn2+ + 1 mn2+ I'll get 3 mn2+ right then this two electrons this two electrons will get cancelled and here we have mn+ 2 MN 3 + right so now what we have to do is just we have to Simply add these two values so I'll get - 1.18 and - 1.51 so if I add this one I'll get - 2.69 so now if you see Delta G value will be -2 into F into - 2.69 so we will get plus value so the reaction will not be spontaneous or the reaction will not be forward so - 2.69 and no so the correct answer for this question is option number D okay now coming to next question how many gam of cobalt metal will be deposited when when a solution of cobalt chloride is electrolyte is electrolyzed with a current of 10 ampere for 109 minutes okay so this is again the question is from farad law so we have to use the formula m = z into I into T So Zed is what marol mass by n by 965 so Cobalt Mar mass is 59 and if you see the charge is here Cobalt + 2 so n will be 2 into 965 now current given is your 10 amp and then time given is 109 minutes which must be converted to seconds okay so now if you solve this um you'll be getting the value around 20 okay so the correct answer for this question is option number B that is 20 okay now coming to next question to okay so the e e not of the cell is given to increase the EMF of the cell what we should do so first let us write the equation E of the cell or is equal to e not of the cell - 059 by 2 log of now if you see in the log we have chlorine minus ion okay Square zn2+ ion and then we have in this set pressure of chlorine now they are asking question that the E EMF of the cell can be increased right now this term whatever value comes in the this term this term has to be subtracted from E not of the cell to get e of the cell now if this term is small okay this value is a small value then we will be subtracting a small number and as a result e of the cell will increase so we have to see how we can do that this value comes out to be a small number now let us see if we increase the Z and 2 plus value then this log value will increase then overall value will increase so we have to subtract a bigger number right so then the value will not the E of the cell will decrease in that case now Z and 2 plus should be decreased if we take Z and 2 plus and we decrease it so this log value will be a small number overall will be a small number and we will be subtracting a small number from this one so we can say e of the cell can be increased in this case CL minus should be increased again if CL minus is increasing right the whole value will increase log value will be a bigger number we are again going to subtract a bigger number so e of the cell will again decrease in that case now partial pressure of the cl2 is decreased so if P of the cl2 will decrease that denominator is decreasing so the fraction value will increase log value will increase the whole value this one will increase and then e not of the cell we have to subtract a bigger number e of the cell will decrease in that case so the correct answer for this question will be option number B okay so this is very conceptual question next question the reduction potential of an electrode can be increased by so if you see again this one reduction potential of any electrode e is equal to e minus 2.303 RT by NF to log of 1 by m n+ right so this is the reduction potential of any electrode formula increasing the area of the electrode so there's no relation in of area in the formula so this one is is not relevant now decreasing the temperature again the same thing see if we decrease the temperature this value will decrease this value will be a small number and we are going to subtract a small number from this one so the reduction potential will increase so this can be an answer now if you increase the temperature if we increase the temperature this overall value will increase we are subtracting a bigger number from E not value so e value will become less so this one is definitely not the answer now decreasing the concentration of metal ion so if you decrease the concentration here if the value is decreasing overall it will be a bigger value right so we are subtracting a bigger value from E not of the cell then e of the cell will again decrease so to increase then we have to decrease the temperature which is directly related the correct answer for this question will be option number B next a uh an electric current is passed through silver water silver and water voltmeter connected in series the cathode of silver voltmeter weighs 0.54 G more after the electrolysis the volume of o2 liberated is so this question is very very important okay so here you have to first understand that some amount of silver is deposited so number so on one side silver is deposited on another side oxygen is deposited so I can say number of equivalents of silver deposited is equal to number of equivalents of o2 deposited okay these both are equal now what is number of equivalents of silver deposited how to calculate that we have this formula m is equal to Z into it so Z is what M Mass okay divided by n into 965 into I into T now what is your equivalent Mass if you know equivalent mass is nothing but given mass by Mar Mass into n Factor so this is my given so I can write mass divided by m Mass okay this value into n that is equal to I into T / 9650 this is nothing but number of equivalence right so now we can calculate for magnesium what is the number of equivalents that is deposited okay so now the question is what we have to calculate the number of equivalents of magnesium sorry the number of equivalents of silver so Mass deposited is 100 uh 0.054 right marar mass is 108 and into n so n Factor here is your 1 so if I solve this I'll get 0 0 and 5 right so now the number of equivalent deposited of oxygen will also be same okay so what is the number of equivalents of oxygen that is 0.5 right now we have to calculate the volume so here something you should know one equivalent of oxygen occupies 5.6 L now from where this number has come so if you know in oxygen we have this formula O2 minus 22 minus gives O2 + 4 electron right equivalent volume if I want to calculate that is equal to molar volume by n Factor so that is 22.4 is the m volume n factor is 4 that is 5.6 l so that is why one equivalent of oxygen occupies 5 .6 L now how many equivalents of oxygen we have 005 equivalent so one equivalent occupies 5.6 l so this many equivalents will occup occupy how much so volume occupied will be number of the volume occupied by one equivalent is 5.6 into .5 okay so now if you solve this one we will be getting around um the answer we'll be getting can the multiply here so 2 80 we are getting then we have this many decimal places so we'll be getting around 2.8 CM Cub okay just multiply so this is liter so we can tell into 10 ^ of 3 for Mill right so that is your 5.6 into 5 so we'll get 2.8 okay so the correct answer for this question is option number D please check this question once again I'll repeat once again what we have done number number so AG is getting deposited on one side and oxygen is getting deposited on another side number of equivalents of AG deposited will be equal to the number of equivalence of o2 deposited so from the given data .54 g 0.54 g of AG is getting deposited so from that we can calculate how many equivalents of silver got deposited and same will be the number of equivalents of oxygen got deposited now the point is to calculate the volume so one equivalent volume of oxygen occupies 5.6 l so so these many equivalence will occupy how much volume so 05 into 5.6 okay so the value we'll be getting is 2.8 CM Cub okay so now let us see the next question the volume of H2 Li obtained at STP when magnesium obtained by passing a current of 05 ampere through the molten mgcl2 for 32.2 minutes is treated with uh excess of dilute HCL is approximately equal so this question is very similar to the previous one so the question is we have to calculate uh first again on one side H2 is getting deposited on another side magnesium is getting deposited so according to this we'll get number of equivalents of magnesium deposited will be equal to number of equivalents of H2 deposited right so first let us calculate the number of equivalents of magnesium deposited so if we have this formula that is uh m is equal to Z into I into T right so I can calculate here m / the molar mass of magnesium into n into 965 so that is equal to into I into T right so here we have mass deposited is um we have to calculate here so I can write number of equivalents of mg will be I into t so current is your. 5 aamp and time is 32.2 minutes then convert it into seconds divided by n factor is 2 because magnesium is getting deposited so we need not consider this n Factor here because we have taken this n factor for number of equivalents so here this value is M ided by m Mass into n this is nothing but number of equivalents right the one which we have solved in this one this formula okay so this one is will be equal to divided by 965 0 so number of equivalence we got5 into 32.2 into 60 ided 965 clear now this one is also equal to the number of equivalence of hydrogen deposited now like how in the previous question we multiplied the volume occupied by one equivalent of hydrogen so the volume occupied by 1 equivalent of H2 so that is your 11.2 L now how this value is coming if you have seen uh H2 reaction is h++ 2 electrons is H2 right so now if I I want to calculate the equivalent volume it will be molar volume by n Factor so what is the mol volume for hydrogen it is always 22.4 L the N Factor here is 2 so we'll get 11.2 l so this value into 11.2 you have to do okay then the units are in cm Cub so just multiply with 10 ^ of 3 to get in terms of ml so the correct answer will be here 112 that is option number D okay so this calculation also is the previous question and this question is very very similar you have to first calculate the number of equivalents once you get number of equivalents multiply it with the equivalent volume okay so for hydrogen the equivalent volume is 11.2 L and for uh the previous one was your oxygen for oxygen the equivalent volume is 5.6 so mol volume by n Factor you have to do okay so this question answer will be 11 112 CM Cube okay now let us see the next question this question is your from farad Second Law right so when there is cells are connected in series M1 by M2 will be equal to E1 by E2 so same quantity of current is passed through copper sulfate and ag3 the solution 2.7 G of silver is deposited okay okay then the amount of copper deposited is okay so we have this formula mass of silver mass of copper deposited by mass of silver deposited will be equal to the equivalent weight of Copper by equivalent weight of silver okay so now if you see here mass of copper deposited we have to calculate and mass of silver deposited is already given 2.7 G now what is the formula for equivalent weight so equivalent weight is nothing but Mar mass by n Factor okay molar mass by n Factor so now if you see for copper the mar mass is 65 uh 63.5 and the N factor is 2 right because copper 2+ gets reduced to give you with two electrons it get reduced to give you copper so we have this one 63.5 / 2 and the equivalent weight of silver will be 108 by 1 because silver reduction potential how does it take place ag+ + 1 electron is giving you a so n Factor here will be 1 so we will have 108 by 1 okay so now just solve this one M of Cu or mass deposited of copper will be 63.5 into 1 ided 2 into 108 into 2.7 okay so now if you solve this the value will'll be getting around is 0.8 so the mass of copper deposited in this series will be your 0.8 G so correct answer for this question is option number D let us see the next question in a dry cell which of the following is an electrolyte that is used so in dry cell the electrolyte used is your ammonium chloride so the correct answer for this question is option number c next cell in a fuel cell which of the following can be utilized as a fuel so you know hydrogen and oxygen can be used so the correct answer for this question is option number c okay and the last question an increase in the mol conductivity in the conductivity equivalent of a solid electrolyte with dilution is primarily due to so there is two things whenever we dilute the electrolyte mol conductivity always increases right so for strong electrolyte for strong electrolyte it increases because the mobility of the ions increase and for weak electrolyte it increases because dissociation of the ions increases right so here it is given for a solid electrolyte what happens when you add the mobility of the ions increases and there and thereby the M conductivity also increases so the correct answer for this question will be option number a okay increased ionic mobility of the ions okay so these were the top 30 questions from your electrochemistry chapter so please revise the concept and then do these questions uh just if you have any doubt in any question you can comment down so I'll be taking it my in my next video and if you have found this video helpful please like share and subscribe our Channel thank you

Channel: Deeksha Karnataka

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Length: 38min 57sec (2337 seconds)

Published: Tue Apr 30 2024