[Four-digit Integer] An object whose surface area is 80 cm^2 is to be plated with an even layer of g — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Volume of the Gold Layer to be Deposited To find the total volume ($V$) of gold required for electroplating, we multiply the surface area ($A$) of the object by the desired thickness ($d$) of the plating layer: * $$\text{Surface Area } (A) = 80\text{ cm}^2$$ * $$\text{Thickness } (d) = 8.0 \times 10^{-4}\text{ cm}$$ $$V = A \times d$$ $$V = 80\text{ cm}^2 \times (8.0 \times 10^{-4}\text{ cm}) = 0.064\text{ cm}^3$$ Step 2 - Calculate the Mass of Gold Required Using the given density of gold ($\rho_{\ce{Au}} = 19.7\text{ g cm}^{-3}$), we calculate the mass ($m$) of gold that needs to be electroplated: $$m = V \times \rho_{\ce{Au}}$$ $$m = 0.064\text{ cm}^3 \times 19.7\text{ g cm}^{-3} = 1.2608\text{ g}$$ Step 3 - Calculate the Moles of Gold Deposited Using the atomic mass of gold ($\text{Au} = 197\text{ g mol}^{-1}$), we find the number of moles ($n_{\ce{Au}}$) of gold atoms: $$n_{\ce{Au}} = \frac{m}{\text{Atomic mass of Au}}$$ $$n_{\ce{Au}} = \frac{1.2608\text{ g}}{197\text{ g mol}^{-1}} = 0.0064\text{ mol}$$ Step 4 - Formulate the Cathode Reduction Reaction and Find Moles of Electrons Required The electrolyte used is gold(III) nitrate, $\ce{Au(NO3)3}$, in which gold exists as trivalent cations ($\ce{Au^3+}$). At the cathode, these cations undergo reduction to form solid gold metal: $$\ce{Au^3+(aq) + 3e^- -> Au(s)}$$ This stoichiometry indicates that depositing $1\text{ mole}$ of gold metal requires $3\text{ moles}$ of electrons ($n\text{-factor} = 3$). Thus, the total moles of electrons ($n_{e^-}$) required is: $$n_{e^-} = 3 \times n_{\ce{Au}}$$ $$n_{e^-} = 3 \times 0.0064\text{ mol} = 0.0192\text{ mol}$$ Step 5 - Calculate the Total Electric Charge ($Q$) and the Time ($t$) in Seconds Using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$), we calculate the total electrical charge ($Q$) in coulombs: $$Q = n_{e^-} \times F$$ $$Q = 0.0192\text{ mol} \times 96,500\text{ C mol}^{-1} = 1852.8\text{ C}$$ Since the electrical charge is also defined as the product of current ($I = 2.4\text{ A}$) and time ($t$) in seconds, we rearrange the formula to solve for $t$: $$Q = I \times t \implies t = \frac{Q}{I}$$ $$t = \frac{1852.8\text{ C}}{2.4\text{ A}} = 772\text{ s}$$ Since the question requires a four-digit integer format: $$\boxed{0772}$$