The cathodic reaction of a dry cell is represented as `2MnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4) (s)` If there are 8 g `MnO_(2)` in the cathodic compartment then the time for which the dry cell will continue to give current of 2 milliampere, isA. 25.675 dayB. 51.35 dayC. 12.8 dayD. 6.423 day

The cathodic reaction of a dry cell is represented as `2MnO_(2)(s)+Zn^

1.	The cathodic reaction of a dry cell is represented as `2MnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4) (s)` If there are 8 g `MnO_(2)` in the cathodic compartment then the time for which the dry cell will continue to give current of 2 milliampere, isA. 25.675 dayB. 51.35 dayC. 12.8 dayD. 6.423 day
Answer» Correct Answer - B According to Faraday,s law of electrolysis `m=((it)/F)M/Z` or, `t=(mZF)/(IM)` the factor of equivalence can be found dividing the number of electrons by the number of reacted particles. In case of `MnO_(2)`, the factor is `Z=2/2=1` the molar mass of `MnO_(2)` is M `(MnO_(2))=55+16xx2` `=87 g//mol` Now, `t=(8xx1xx96500)/(0.002xx87) =4.436xx10^(6) s=1232=51.3` days

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The cathodic reaction of a dry cell is represented as `2MnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4) (s)` If there are 8 g `MnO_(2)` in the cathodic compartment then the time for which the dry cell will continue to give current of 2 milliampere, isA. 25.675 dayB. 51.35 dayC. 12.8 dayD. 6.423 day

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