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InterviewSolution
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In a zinc manganese dioxide dry cell, the anode is made up of zinc and cathode of a carbon rod surrounded by a mixture of `MnO_(2)`, carbon `NH_(4)Cl and ZnCl_(2)` in aqueous base. The cathodic reaction may be represented as: `2MnO_(2) (s) + Zn^(2+) + 2e^(-) rarr ZnMn_(2) O_(4) (s)` Let there be 8g `MnO_(2)` in the cathodic compartment. How many days will the dry cell continue to give a current of `4 xx 10^(-3)` ampere ? |
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Answer» When `MnO_(2)` will be used up in cathodic process, the dry cell will stop to produce current. Cathodic process: `overset(+4)(2MnO_(2))(s) + Zn^(2+) + 2e^(-) rarr overset(+3)(ZnMn_(2)O_(4))` Equivlanet mass of `MnO_(2) = ("Molecular mass")/("Change in oxidation state")` `= (87)/(1) = 87` From first law of electrolysis, `W = (I tE)/(96500)` `8 = ( 4 xx 10^(-3) xx t xx 87)/(96500)` `t = 2218390.8` second `= (2218390.8)/(3600 xx 24) = 25.675` days |
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