[Four-digit Integer] On passing electricity through nitrobenzene solution, it is converted into azob — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Moles of Fullerene ($\ce{C60}$) Fullerene ($\ce{C60}$) is an allotrope of carbon consisting of 60 carbon atoms. Its molar mass is: $$M(\ce{C60}) = 60 \times 12\text{ g mol}^{-1} = 720\text{ g mol}^{-1}$$ Given the mass of fullerene as $96\text{ g}$, we compute the number of moles: $$n(\ce{C60}) = \frac{96\text{ g}}{720\text{ g mol}^{-1}} = \frac{1}{7.5}\text{ mol} \approx 0.1333\text{ mol}$$ Step 2 - Calculate the Moles of Oxygen ($\ce{O2}$) Required for Combustion The balanced chemical equation for the complete combustion of fullerene is: $$\ce{C60(s) + 60O2(g) -> 60CO2(g)}$$ From the stoichiometry of this reaction, $1\text{ mole}$ of $\ce{C60}$ requires $60\text{ moles}$ of $\ce{O2}$ gas for complete combustion. Using the moles of fullerene from Step 1, the moles of oxygen required are: $$n(\ce{O2}) = 60 \times n(\ce{C60}) = 60 \times \frac{1}{7.5}\text{ mol} = 8\text{ mol}$$ Step 3 - Determine the Moles of Electrons (Faradays) Passed during Electrolysis Oxygen gas is produced at the anode during the electrolysis of water (or an aqueous solution) via the following oxidation half-reaction: $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ According to this reaction, the production of $1\text{ mole}$ of $\ce{O2}$ gas requires the transfer of $4\text{ moles}$ of electrons ($4\text{ F}$ of electricity). Since $8\text{ moles}$ of $\ce{O2}$ are produced, the total moles of electrons passed through the circuit are: $$n(e^-) = 4 \times n(\ce{O2}) = 4 \times 8\text{ mol} = 32\text{ mol of electrons}$$ Step 4 - Understand the Electrolytic Reduction of Nitrobenzene to Azobenzene The electrolytic reduction of nitrobenzene ($\ce{C6H5NO2}$) to azobenzene ($\ce{C12H10N2}$) in an alkaline medium is represented by the following balanced equation: $$\ce{2C6H5NO2 + 8H^+ + 8e^- -> C12H10N2 + 4H2O}$$ From the stoichiometry of this reduction reaction, the formation of $1\text{ mole}$ of azobenzene requires $8\text{ moles}$ of electrons ($8\text{ F}$ of electricity). Step 5 - Calculate the Mass of Azobenzene and Address the Textbook Scale Typo Using the $32\text{ mol}$ of electrons passed through the solution (determined in Step 3), the moles of azobenzene produced are: $$\text{Moles of azobenzene} = \frac{32\text{ mol of electrons}}{8\text{ electrons per mole of azobenzene}} = 4\text{ mol}$$ The molar mass of azobenzene ($\ce{C12H10N2}$) is: $$M(\ce{C12H10N2}) = 12 \times 12 + 10 \times 1 + 2 \times 14 = 144 + 10 + 28 = 182\text{ g mol}^{-1}$$ The mass of azobenzene produced in grams is: $$\text{Mass} = 4\text{ mol} \times 182\text{ g mol}^{-1} = 728\text{ g}$$ **Addressing the Textbook Typographical Error:** The textbook question asks for the mass of azobenzene in **milligrams (mg)** but provides the mass of fullerene as **$96\text{ g}$**. This scale mismatch leads to a theoretical value of $728\text{ g}$ ($728,000\text{ mg}$). To obtain the textbook's correct answer of **$728\text{ mg}$**, the mass of fullerene in the question should be **$96\text{ mg}$** (or the units of the final mass should be in grams). With $96\text{ mg}$ of fullerene: $$n(\ce{C60}) = \frac{96 \times 10^{-3}\text{ g}}{720\text{ g mol}^{-1}} = \frac{1}{7.5}\text{ mmol}$$ $$n(\ce{O2}) = 8\text{ mmol}$$ $$n(e^-) = 32\text{ mmol}$$ $$\text{Moles of azobenzene} = 4\text{ mmol}$$ $$\text{Mass of azobenzene} = 4\text{ mmol} \times 182\text{ g mol}^{-1} = 728\text{ mg}$$ Thus, the numerical value is **$728$**. Step 6 - Format as a Four-digit Integer Representing the value $728$ as a four-digit integer as required: $$\boxed{0728}$$