The reactions taking place in the dry cell are: Anode: Zn → Zn^2+ + 2e^-; Cathode: 2 + 2NH4^+ + 2e^- — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Total Electrical Charge Transferred We find the total electrical charge ($Q$) passed through the dry cell using the relationship between electric current ($I$) and time ($t$): $$Q = I \times t$$ First, convert the time from hours to seconds: $$t = 9.65\text{ h} \times 3600\text{ s/h} = 34,740\text{ s}$$ Now, substitute the current $I = 0.25\text{ A}$ and time $t = 34,740\text{ s}$ into the formula: $$Q = 0.25\text{ A} \times 34,740\text{ s}$$ $$Q = 8,685\text{ C}$$ Step 2 - Determine the Moles of Electrons Transferred Using Faraday's constant ($F \approx 96,500\text{ C/mol}$), we calculate the total moles of electrons ($n_{e^-}$) passed through the cell: $$n_{e^-} = \frac{Q}{F}$$ $$n_{e^-} = \frac{8,685\text{ C}}{96,500\text{ C/mol}}$$ $$n_{e^-} = 0.09\text{ mol}$$ Step 3 - Calculate the Minimum Mass of Zinc (\ce{Zn}) Reactant The oxidation half-reaction taking place at the anode is: $$\ce{Zn(s) -> Zn^2+(aq) + 2e^-}$$ According to this balanced half-reaction, $1\text{ mole of Zn}$ releases $2\text{ moles of electrons}$. Therefore, the moles of zinc consumed ($n_{\ce{Zn}}$) is: $$n_{\ce{Zn}} = \frac{n_{e^-}}{2}$$ $$n_{\ce{Zn}} = \frac{0.09\text{ mol}}{2}$$ $$n_{\ce{Zn}} = 0.045\text{ mol}$$ Using the molar mass of zinc ($65.4\text{ g/mol}$), we calculate the minimum required mass of zinc: $$\text{Mass of Zn} = n_{\ce{Zn}} \times \text{Molar Mass of Zn}$$ $$\text{Mass of Zn} = 0.045\text{ mol} \times 65.4\text{ g/mol}$$ $$\text{Mass of Zn} = \boxed{2.943\text{ g}}$$ This calculation confirms that **Option (A) is correct**. Step 4 - Calculate the Minimum Mass of Manganese Dioxide (\ce{MnO2}) Reactant The reduction half-reaction taking place at the cathode is: $$\ce{2MnO2(s) + 2NH4^+(aq) + 2e^- -> Mn2O3(s) + 2NH3(g) + H2O(l)}$$ According to this balanced half-reaction, $2\text{ moles of MnO2}$ react for every $2\text{ moles of electrons}$ consumed, establishing a $1:1$ stoichiometric ratio between $\ce{MnO2}$ and the electrons. Therefore, the moles of manganese dioxide consumed ($n_{\ce{MnO2}}$) is: $$n_{\ce{MnO2}} = n_{e^-} = 0.09\text{ mol}$$ The molar mass of manganese dioxide is: $$\text{Molar Mass of MnO2} = 55\text{ g/mol} + (2 \times 16\text{ g/mol}) = 87\text{ g/mol}$$ Now, we calculate the minimum required mass of manganese dioxide: $$\text{Mass of MnO2} = n_{\ce{MnO2}} \times \text{Molar Mass of MnO2}$$ $$\text{Mass of MnO2} = 0.09\text{ mol} \times 87\text{ g/mol}$$ $$\text{Mass of MnO2} = \boxed{7.83\text{ g}}$$ This calculation confirms that **Option (B) is correct** and Option (D) is incorrect. Step 5 - Calculate the Minimum Mass of Ammonium Ion (\ce{NH4^+}) Reactant From the reduction half-reaction shown in Step 4, $2\text{ moles of NH4^+}$ react for every $2\text{ moles of electrons}$ consumed, establishing a $1:1$ stoichiometric ratio between $\ce{NH4^+}$ and the electrons. Therefore, the moles of ammonium ions consumed ($n_{\ce{NH4^+}}$) is: $$n_{\ce{NH4^+}} = n_{e^-} = 0.09\text{ mol}$$ The molar mass of ammonium ion is: $$\text{Molar Mass of NH4^+} = 14\text{ g/mol} + (4 \times 1\text{ g/mol}) = 18\text{ g/mol}$$ Now, we calculate the minimum required mass of ammonium ions: $$\text{Mass of NH4^+} = n_{\ce{NH4^+}} \times \text{Molar Mass of NH4^+}$$ $$\text{Mass of NH4^+} = 0.09\text{ mol} \times 18\text{ g/mol}$$ $$\text{Mass of NH4^+} = \boxed{1.62\text{ g}}$$ This calculation confirms that **Option (C) is correct**. Step 6 - Final Verdict Evaluating the calculations, the required minimum masses are: - Zinc: $2.943\text{ g}$ - Manganese dioxide: $7.83\text{ g}$ - Ammonium ion: $1.62\text{ g}$ This perfectly matches options A, B, and C. $$\text{Correct Options: } \boxed{A, B, C}$$