[Four-digit Integer] In a Zn- Cell, the anode is made up of Zn and cathode of carbon rod surrounded — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Cathodic Reaction and Electron Transfer In a standard $\ce{Zn-MnO2}$ dry cell (Leclanché cell), zinc acts as the anode and is oxidized, while a carbon (graphite) rod surrounded by manganese dioxide ($\ce{MnO2}$) and carbon black acts as the cathode. The cathode compartment is filled with an electrolyte mixture containing $\ce{NH4Cl}$ and $\ce{ZnCl2}$. The reduction half-reaction occurring at the cathode is: $$\ce{MnO2(s) + NH4+(aq) + e^- -> MnO(OH)(s) + NH3(aq)}$$ In this reaction: * Manganese is reduced from an oxidation state of $+4$ in $\ce{MnO2}$ to an oxidation state of $+3$ in manganese oxyhydroxide, $\ce{MnO(OH)}$. * This stoichiometric relationship shows that $1\text{ mole}$ of electrons ($1\ e^-$) is consumed for every $1\text{ mole}$ of $\ce{MnO2}$ reduced. Step 2 - Calculate the Moles of Manganese Dioxide ($\ce{MnO2}$) First, we find the molar mass of $\ce{MnO2}$ using the atomic masses of Manganese ($\text{Mn} = 55\text{ g mol}^{-1}$) and Oxygen ($\text{O} = 16\text{ g mol}^{-1}$): $$\text{Molar Mass of }\ce{MnO2} = 55 + (2 \times 16) = 87\text{ g mol}^{-1}$$ Given that the mass of $\ce{MnO2}$ in the cathode compartment is $8.7\text{ g}$, we calculate the number of moles ($n_{\ce{MnO2}}$): $$n_{\ce{MnO2}} = \frac{\text{Given Mass}}{\text{Molar Mass}}$$ $$n_{\ce{MnO2}} = \frac{8.7\text{ g}}{87\text{ g mol}^{-1}} = 0.10\text{ mol}$$ Step 3 - Calculate the Total Electrical Charge ($Q$) Stored in the Cell Since the reduction of $1\text{ mole}$ of $\ce{MnO2}$ requires $1\text{ mole}$ of electrons: $$\text{Moles of electrons transferred } (n_{e^-}) = n_{\ce{MnO2}} = 0.10\text{ mol}$$ Using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$), we find the total charge ($Q$) in coulombs that can be drawn from the cathode compartment: $$Q = n_{e^-} \times F$$ $$Q = 0.10\text{ mol} \times 96,500\text{ C mol}^{-1} = 9650\text{ C}$$ Step 4 - Calculate the Total Time in Seconds The relationship between charge ($Q$), current ($I$), and time ($t$) in seconds is: $$Q = I \times t \implies t = \frac{Q}{I}$$ Substitute the total charge $Q = 9650\text{ C}$ and the constant current drawn $I = 3.99 \times 10^{-3}\text{ A}$: $$t = \frac{9650\text{ C}}{3.99 \times 10^{-3}\text{ A}}$$ $$t \approx 2,418,546.37\text{ s}$$ Step 5 - Convert Time to Days and Format the Final Answer To convert the time from seconds to days, we divide by the total number of seconds in a single day ($1\text{ day} = 24\text{ hours} \times 3600\text{ s hour}^{-1} = 86,400\text{ s}$): $$t_{\text{days}} = \frac{t}{86,400}$$ $$t_{\text{days}} = \frac{2,418,546.37\text{ s}}{86,400\text{ s day}^{-1}} \approx 27.992\text{ days} \approx 28\text{ days}$$ As the question specifies a four-digit integer format: $$\boxed{0028}$$