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(Go to anode) (Primary reaction) ... ( Molarity of Na2S2O3 solution = Normality. meq. of I2 liberated = m equiv. of Na2S2O3 = 21.75 ( 0.0831 = 1.807. Thus, = 1.807 = 1.87 ( 10–3 = ( i = 0.0242A. Applications of electrolysis: (1) Determination of equivalent masses of element (2) Electrometallurgy (3) Manufacturing of non – metals (By electrolysis process) (4) Electro – refining of metals ... ................
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