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Shweta thakur Asked a Question June 6, 2021 4:31 pm 30 pts to Tedue he process is Electr lysis and Electrolytic ductance of mitrobernzene to aniline, y the c efficiency for ired energy will be consu potential drop across the cell isvolts, how much 18 96500 10 0-620.4 ie, 78839-86 coulomt 35 Solution: No. of eq. of nitrobenzene to aniline time required =0 of coulombs wt. in g current in amp eq. wt. of CH,NO, 1230-6. 78839-86 = 39419-93 seconds 2 39420 seconds 39420 123/6 60 x 60 nours C.HNO C,HNH,; change in ON = 6 3 = 10-95 hours. +3 mol. wt. ar After electrolysis of a sodium chloride (NaCl) solution with inert for a certain period of time, b00 mu of the solution was left which he 1 N in sodium hydroxide. Durmg the same time 31-8 g of Cu we aCu voltameter in series with the electrolytic cell. Calculate the of theoretical yield of the sodium hydroxide obtained. eq. wt. of CgHsNO2 change in ON 6 mole of electricity for 100% current efficiency = 06 F.) But the current efficiency is 50%. mole of electricity used = 0-6 x2= 1-2 F =12x 96500 coulombs = 115800 coulombs Solution: *No. of equivalent of NaOH that can be produced theoretically for 100%% current efficiency = no. of eq. of NaCl decomposed The energy consumed = electricity in coulomb xpot. drop in volt = no. of eq. of Cu deposited wt. of Cu eq. wt. of Cu 63-52 Farada = 115800 x 3J 31-8 Ean 347400 J 347.40 kJ. Now, no. of eq. of NaOH produced experimentally m.e. of NaOH 1000 Ex. 14. Potassium chlorate is prepared by electrolysis of KCI in basic solution: (E 6OH+CI CIO, +3H,O + 6e fonly 60% of the current is utilised in the reaction, what time will be required to produce 10 g of KCIO, using a current of2 amp? normality x volume 1x600 1000 1000 06. Solution: According to the given equation, percentage yield = x100 = 60%. eq. wt. of KCIO, =mol. wt. 122-5 204. Ex. 16. Thè density of Cu is 8.94 g per mL. Find out the charg to plate an area 10 cm x 10 cm to a thickmess of 10 cm as electrolyte. 6 equivalent of KCIO, to be produced = 10 20-4 Since current efficiency is 60%; hence, 1 faraday (i.e, 96500 coulombs) shall produce 0-6 equivalent instead of 1 equivalent. Solution: Wt. of Cu to be coated = volume of the dep -10 x 10 v 10-2 .0 10 production of eq. of KCIO, shall requir 20-4 eq. of KClO, shall reaui 1 joule = oll wce

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