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In a zinc manganese dioxide dry cell, th...

In a zinc manganese dioxide dry cell, the anode is madeup of `Zn1` and cathode of carbon rod surrounded by a mixture of `MnO_(2)`, carbon, `NH_(4)Cl`, and `ZnCl_(2)` in aqueous base. The cathodic reaction is `: `
`ZnMnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4)(s)`
`8.7g` of `MnO_(2)` is taken in the cathodic compartment. How many days will the dry cell continue to give a current of `9.65xx10^(-3)A` ? `(` Atomic weight of `Mn=55)(Mw` of `MnO_(2)=87g mol^(-1))`

Text Solution

Verified by Experts

When `MnO_(2)` will be used up in cathodic porcess, the dry cell will stop to produce current.
`overset(+2)(Zn)overset(+2)(Mn)overset(=4)(O_(2))(s)+Zn^(2+)+2e^(-) rarroverset(+2)(Zn)overset(+3xx2)(Mn)overset(-2xx4)(O_(2)(s))`
`(n` factor for `MnO_(2)=1 mol `of `MnO_(2)=87g`
`(1F=96500C=1 mol ` of `MnO_(2)=87g)`
`:. 87g `of `MnO_(2)-=96500C`
`8.7g ` of `MnO_(2)=9650C=Ixxt`
`=9.5xx10^(-3)Axxt(s)`
`:. t=(9650C)/(9.65xx10^(-3)A)=10^(6)s=(10^(6))/(60xx60xx24)`
`=11.57` days
Alternatively, use the direct formula
`E_(MnO_(2))=(EwxxIxxt)/(96500C)`
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