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: `MnO_(2) + NH_(4)^(+) + e^(-) rarr MnO(OH) + NH_(3)`. 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 ?

Options

A. 32 days
B. 25.675days
C. 48.35 days
D. 18.55 days

Answer

B

Explanation

When `MnO_(2)` will be used up in cathodic process, the dry cell will stop to produce current. Cathodic process `MnO_(2) + NH_(4)^(+) + e^(-) rarr MnO(OH)+ NH_(3)`
Equivalent mass of `MnO_(2) = ("Molecular mass")/("Change in oxidation state") = (87)/(1)= 87`
`W= (I t E)/(96500), 8=(4 xx 10^(-3)xx t xx 87)/(96500)`
t= 2218390.8 seconds `=(2218390.8)/(3600 xx 24)= 25.675` days