How much of nickel is plated on the cathode per hour? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Cathode Reactions and Current Efficiency During the electroplating process of nickel from a nickel sulfate (\ce{NiSO4}) solution, two reduction reactions compete at the cathode: * The reduction of nickel(II) ions to metallic nickel: $$\ce{Ni^2+(aq) + 2e^- -> Ni(s)}$$ * The reduction of hydrogen ions (or water) to release hydrogen gas: $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ The current efficiency ($\eta$) with respect to the deposition of nickel is given as $60\%$. This means that only $60\%$ of the total electrical charge passed through the cell is actively consumed to reduce $\ce{Ni^2+}$ ions to solid nickel ($\ce{Ni}$), while the remaining $40\%$ of the charge is wasted in producing hydrogen gas. Step 2 - Calculate the Total Electrical Charge Passed in One Hour We calculate the total charge ($Q_{\text{total}}$) passed through the cell using the current ($I = 15.0\text{ A}$) and time ($t = 1\text{ hour} = 3600\text{ s}$): $$Q_{\text{total}} = I \times t$$ $$Q_{\text{total}} = 15.0\text{ A} \times 3600\text{ s} = 54,000\text{ C}$$ Step 3 - Determine the Active Charge for Nickel Deposition Using the current efficiency of $60\%$, we calculate the active charge ($Q_{\text{active}}$) that contributes solely to the deposition of nickel: $$Q_{\text{active}} = \eta \times Q_{\text{total}}$$ $$Q_{\text{active}} = 0.60 \times 54,000\text{ C} = 32,400\text{ C}$$ Step 4 - Determine the Equivalent Weight of Nickel The equivalent weight ($E_{\text{Ni}}$) of a metal is its atomic mass divided by its valency factor (the number of electrons transferred per atom): $$E_{\text{Ni}} = \frac{\text{Atomic Weight of Nickel}}{\text{Valency factor } (n)}$$ Since nickel is present as divalent ions ($\ce{Ni^2+}$) in a $\ce{NiSO4}$ solution, the valency factor $n$ is $2$. Given the atomic weight of nickel as $58.7\text{ g mol}^{-1}$: $$E_{\text{Ni}} = \frac{58.7\text{ g mol}^{-1}}{2} = 29.35\text{ g eq}^{-1}$$ Step 5 - Calculate the Mass of Nickel Plated per Hour Using Faraday's first law of electrolysis, the mass of nickel deposited ($W$) is calculated by substituting our active charge and equivalent weight, using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$): $$W = \frac{Q_{\text{active}} \times E_{\text{Ni}}}{F}$$ $$W = \frac{32,400\text{ C} \times 29.35\text{ g eq}^{-1}}{96,500\text{ C eq}^{-1}}$$ $$W = \frac{950,940}{96,500}\text{ g} \approx 9.854\text{ g}$$ Step 6 - Evaluate and Explain Each Option * **Option (a) is incorrect:** This value ($16.43\text{ g}$) would be the mass of nickel deposited if the current efficiency were $100\%$ instead of $60\%$. * **Option (b) is incorrect:** This value is significantly higher and does not align with the stoichiometry of a $2$-electron reduction under these conditions. * **Option (c) is incorrect:** This value corresponds to twice the correct mass ($19.7\text{ g}$), which would be obtained if the $n$-factor of nickel were incorrectly taken as $1$. * **Option (d) is correct:** Our calculated mass of nickel plated per hour is approximately $9.85\text{ g}$, which matches option (d) perfectly. $$\text{Correct Option: } \boxed{D}$$