In a Cell, the anode is made up of Zn and cathode of carbon rod surrounded by a mixture of , Carbon, — Electrochemistry Chemistry Question
Question
In a $Zn-MnO_2$ 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. If 8.7 g $MnO_2$ is present in cathodic compartment, how many days the dry cell will continue to give a current of $3.99 \times 10^{-3}$ A?
💡 Solution & Explanation
Moles of $MnO_2$ = $8.7 / (55 + 32) = 8.7 / 87 = 0.1$ mol. The cathode reduction reaction is $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$, meaning 1 mole of $MnO_2$ requires 1 Faraday of charge. Total charge capacity $Q = 0.1 \text{ F} = 0.1 \times 96500 \text{ C} = 9650$ C. Time $t = Q / I = 9650 / (3.99 \times 10^{-3}) = 2,418,546$ seconds. Converting to days: $2,418,546 / (24 \times 3600) = 2,418,546 / 86400 \approx 27.99$ days. Rounding to the nearest integer gives 28. Therefore, correct answer is 0028.