In a zinc-manganese dioxide cell- the anode is made up of Zn and cathode of carbon rod surrounded by a mixture of MnO 2- carbon- NH 4Cl and ZnCl 2 in aqueous base- The cathodic reaction is- 2MnO2 s - Zn2- -2e - ZnMn2O4sIf 8 g MnO 2 is present in cathodic compartment- determine the number of days the dry cell will continue to give a current of 4 - 10-3 ampere- Write your answer to the nearest integer

The molar mass of MnO_2 is 86.9 g/mol. 8 g of MnO_2 corresponds to 886.9=0.092 moles.They will give 0.092 faraday of electricity.Hence, the ti ...

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Standard XII

Chemistry

Faraday's Second Law

In a zinc-man...

Question

In a zinc-manganese dioxide cell, the anode is made up of Zn 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:
$2 M n O_{2} (s) + Z n^{2 +} + 2 e \to Z n M n_{2} O_{4} (s)$
If 8 g MnO $_{2}$ is present in cathodic compartment, determine the number of days the dry cell will continue to give a current of $4 \times 10^{- 3}$ ampere. (Write your answer to the nearest integer)

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Solution

The molar mass of $M n O_{2}$ is 86.9 g/mol. 8 g of $M n O_{2}$ corresponds to $\frac{8}{86.9} = 0.092$ moles.

They will give 0.092 faraday of electricity.

Hence, the time the dry cell will continue is $t = \frac{0.092 \times 96500}{4 \times 10^{- 3}} = 2.2 \times 10^{6}$ s.

Hence, the number of days, the dry cell will continue is $\frac{2.2 \times 10^{6}}{3600 \times 24} = 26$ days.

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The molar mass of MnO2 is 86.9 g/mol. 8 g of MnO2 corresponds to 886.9=0.092 moles.They will give 0.092 faraday of electricity.Hence, the time the dry cell will continue is t=0.092×965004×10−3=2.2×106s.Hence, the number of days, the dry cell will continue is 2.2×1063600×24=26 days.