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In a dry cell, MnO_2 occurs in a semisolid electro

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Answer #1

we have formula from 1st law of electrolyis

m = ( i x t /F) x ( M/n)

i = 38.5 A, , F = 96485 C , M = molar mass of MnO2 = 86.937 g/mol , n = number of elctrons per reaction =2

m = 1.4 kg = 1400 g

now 1400 = ( 38.5 x t/96485) x ( 86.937/2)

t = 80715 sec = 80715/3600 hours = 22.42 hours

MnO2 is formed at anode      ( where Mn2+ is oxidised to MnO2)

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