What is the thickness of the plating if the cathode consists of a sheet of metal 4.0 cm^2, which is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Cathode Reactions and Current Efficiency During the electroplating of nickel from a nickel sulfate (\ce{NiSO4}) solution, two reduction reactions occur simultaneously at the cathode: * The reduction of nickel(II) ions to deposit metallic nickel: $$\ce{Ni^{2+}(aq) + 2e^- -> Ni(s)}$$ * The reduction of hydronium ions to produce hydrogen gas: $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ The current efficiency ($\eta$) with respect to nickel formation is $60\%$. This indicates that only $60\%$ of the total electrical charge passed through the cell is used to reduce $\ce{Ni^2+}$ ions to metallic nickel, while the remaining $40\%$ is consumed in the parallel evolution of hydrogen gas. Step 2 - Calculate the Mass of Nickel Deposited Per Hour First, we write down the formula for the mass of nickel deposited ($w$) during electrolysis using Faraday's laws of electrolysis, incorporating the current efficiency ($\eta$): $$w = \frac{I \times t \times \eta}{n \times F} \times M$$ Where: * $I$ is the electric current = $15.0\text{ A}$ * $t$ is the electrolysis time = $1\text{ hour} = 3600\text{ s}$ * $\eta$ is the current efficiency = $60\% = 0.60$ * $n$ is the number of electrons transferred per nickel atom = $2$ * $F$ is Faraday's constant = $96,500\text{ C mol}^{-1}$ * $M$ is the molar mass of nickel = $58.7\text{ g mol}^{-1}$ Substitute the given values into the formula: $$w = \frac{15.0\text{ A} \times 3600\text{ s} \times 0.60}{2 \times 96,500\text{ C mol}^{-1}} \times 58.7\text{ g mol}^{-1}$$ $$w = \frac{32,400\text{ C}}{193,000\text{ C mol}^{-1}} \times 58.7\text{ g mol}^{-1}$$ $$w = 0.16788\text{ mol} \times 58.7\text{ g mol}^{-1}$$ $$w \approx 9.854\text{ g}$$ Step 3 - Calculate the Volume of Deposited Nickel To find the physical volume ($V$) of the deposited nickel layer, we use the density of nickel ($\rho$): $$V = \frac{w}{\rho}$$ Substituting the mass of deposited nickel ($w \approx 9.854\text{ g}$) and the density of nickel ($\rho = 8.9\text{ g cm}^{-3}$): $$V = \frac{9.854\text{ g}}{8.9\text{ g cm}^{-3}}$$ $$V \approx 1.107\text{ cm}^3$$ Step 4 - Calculate the Total Area to be Plated The cathode consists of a flat sheet of metal with a face area of $4.0\text{ cm}^2$. Since the sheet is to be plated on both sides, the total surface area ($A_{\text{total}}$) on which the nickel is deposited is: $$A_{\text{total}} = 2 \times A_{\text{one side}}$$ $$A_{\text{total}} = 2 \times 4.0\text{ cm}^2 = 8.0\text{ cm}^2$$ Step 5 - Calculate the Thickness of the Nickel Plating The thickness ($d$) of the plating is the ratio of the volume of deposited nickel to the total surface area: $$d = \frac{V}{A_{\text{total}}}$$ Substitute the calculated values: $$d = \frac{1.107\text{ cm}^3}{8.0\text{ cm}^2}$$ $$d = 0.1384\text{ cm}$$ Convert this thickness to millimeters: $$d = 0.1384\text{ cm} \times 10\text{ mm cm}^{-1} = 1.384\text{ mm}$$ Step 6 - Evaluate and Explain the Options The options given in this multiple-choice question are: * (A) $13.8\text{ mm}$ * (B) $27.6\text{ mm}$ * (C) $6.9\text{ mm}$ * (D) $23.0\text{ mm}$ Analyzing the options: * **Option (A) is correct:** Our calculated thickness is $1.384\text{ mm}$. The options listed in the question contain a minor typographical decimal-point error, where the decimal point has been shifted one place to the right, showing $13.8\text{ mm}$ instead of $1.38\text{ mm}$. Thus, Option (A) is the intended correct option. * **Option (B) is incorrect:** This value ($27.6\text{ mm}$) corresponds to a thickness of $2.76\text{ mm}$ (with the same decimal error), which would occur if the sheet were coated on only one side ($4.0\text{ cm}^2$ instead of $8.0\text{ cm}^2$). * **Option (C) is incorrect:** This value ($6.9\text{ mm}$) represents half of Option (A), corresponding to $0.69\text{ mm}$. * **Option (D) is incorrect:** This value ($23.0\text{ mm}$) corresponds to $2.3\text{ mm}$. $$\text{Correct Option: } \boxed{A}$$