Reactions in dry cell: Anode: Zn -> Zn2+ + 2e-; Cathode: 2 + 2NH4+ + 2e- -> Mn2 + 2 + . Minimum mass — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Total Electrical Charge ($Q$) and Moles of Electrons ($n_{e^-}$) Passed The total electrical charge ($Q$) passed through the dry cell is given by the formula: $$Q = I \times t$$ Where: * Current ($I$) = $0.25\text{ A}$ * Time ($t$) = $9.65\text{ h} = 9.65 \times 3,600\text{ s} = 34,740\text{ s}$ Substituting the values with units: $$Q = 0.25\text{ A} \times 34,740\text{ s} = 8,685\text{ C}$$ The number of moles of electrons ($n_{e^-}$) transferred is related to the charge ($Q$) by Faraday's constant ($F = 96,500\text{ C mol}^{-1}$): $$n_{e^-} = \frac{Q}{F}$$ $$n_{e^-} = \frac{8,685\text{ C}}{96,500\text{ C mol}^{-1}} = 0.09\text{ mol}$$ Step 2 - Calculate the Minimum Mass of Zinc ($\ce{Zn}$) Reactant Required The oxidation reaction occurring at the anode is: $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ According to the stoichiometry of the anode half-reaction, $1\text{ mole}$ of solid $\ce{Zn}$ releases $2\text{ moles}$ of electrons. Therefore, the number of moles of $\ce{Zn}$ oxidized is: $$n_{\ce{Zn}} = \frac{n_{e^-}}{2}$$ $$n_{\ce{Zn}} = \frac{0.09\text{ mol}}{2} = 0.045\text{ mol}$$ The minimum mass of zinc required is: $$\text{Mass of }\ce{Zn} = n_{\ce{Zn}} \times M_{\ce{Zn}}$$ $$\text{Mass of }\ce{Zn} = 0.045\text{ mol} \times 65.4\text{ g mol}^{-1} = \boxed{2.943\text{ g}}$$ This confirms that **Option (A)** is correct. Step 3 - Calculate the Minimum Mass of Manganese Dioxide ($\ce{MnO2}$) Reactant Required The reduction reaction occurring at the cathode is: $$\ce{2MnO2(s) + 2NH4^+(aq) + 2e^- -> Mn2O3(s) + 2NH3(g) + H2O(l)}$$ According to the balanced cathode half-reaction, $2\text{ moles}$ of $\ce{MnO2}$ react with $2\text{ moles}$ of electrons, which means a 1:1 stoichiometric ratio: $$n_{\ce{MnO2}} = n_{e^-} = 0.09\text{ mol}$$ The molar mass of manganese dioxide ($\ce{MnO2}$) is: $$M_{\ce{MnO2}} = M_{\ce{Mn}} + 2 \times M_{\ce{O}} = 55 + 2 \times 16 = 87\text{ g mol}^{-1}$$ The minimum mass of manganese dioxide required is: $$\text{Mass of }\ce{MnO2} = n_{\ce{MnO2}} \times M_{\ce{MnO2}}$$ $$\text{Mass of }\ce{MnO2} = 0.09\text{ mol} \times 87\text{ g mol}^{-1} = \boxed{7.83\text{ g}}$$ This confirms that **Option (B)** is correct and **Option (D)** is incorrect. Step 4 - Calculate the Minimum Mass of Ammonium Ions ($\ce{NH4^+}$) Reactant Required From the cathode reaction, the stoichiometry of ammonium ions to electrons is also 2:2, which simplifies to a 1:1 ratio: $$n_{\ce{NH4^+}} = n_{e^-} = 0.09\text{ mol}$$ The molar mass of ammonium ion ($\ce{NH4^+}$) is: $$M_{\ce{NH4^+}} = M_{\ce{N}} + 4 \times M_{\ce{H}} = 14 + 4 \times 1 = 18\text{ g mol}^{-1}$$ The minimum mass of ammonium ions required is: $$\text{Mass of }\ce{NH4^+} = n_{\ce{NH4^+}} \times M_{\ce{NH4^+}}$$ $$\text{Mass of }\ce{NH4^+} = 0.09\text{ mol} \times 18\text{ g mol}^{-1} = \boxed{1.62\text{ g}}$$ This confirms that **Option (C)** is correct. Step 5 - Evaluate and Summarize the Options * **Option (A) is correct:** The calculated minimum mass of zinc reactant required is exactly $2.943\text{ g}$. * **Option (B) is correct:** The calculated minimum mass of manganese dioxide reactant required is exactly $7.83\text{ g}$. * **Option (C) is correct:** The calculated minimum mass of ammonium ions reactant required is exactly $1.62\text{ g}$. * **Option (D) is incorrect:** The calculated minimum mass of manganese dioxide required is $7.83\text{ g}$. The value $3.915\text{ g}$ represents exactly half of the required amount, which is a mathematical error resulting from an incorrect 1:2 stoichiometric ratio. $$\text{Correct Options: } \boxed{\text{A, B, C}}$$