The quantity of electricity required for the reduction of 1 mole of to Fe is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Oxidation State of Iron in $\ce{Fe2O3}$ First, let us calculate the oxidation state of iron ($\ce{Fe}$) in iron(III) oxide ($\ce{Fe2O3}$). We know that oxygen typically has an oxidation state of $-2$ in oxides. Let the oxidation state of iron be $x$: $$2(x) + 3(-2) = 0$$ $$2x - 6 = 0 \implies x = +3$$ Thus, iron exists as trivalent ferric cations ($\ce{Fe^{3+}}$) in $\ce{Fe2O3}$. Step 2 - Analyze the Molar Composition of $\ce{Fe2O3}$ From the chemical formula $\ce{Fe2O3}$, each formula unit contains $2$ iron atoms. Therefore, $1\text{ mole}$ of $\ce{Fe2O3}$ contains exactly $2\text{ moles}$ of $\ce{Fe^{3+}}$ ions: $$1\text{ mole of }\ce{Fe2O3} \implies 2\text{ moles of }\ce{Fe^{3+}}\text{ ions}$$ Step 3 - Write the Balanced Reduction Half-Reaction The reduction of a single $\ce{Fe^{3+}}$ ion to metallic iron ($\ce{Fe}$) requires $3$ electrons: $$\ce{Fe^{3+} + 3e^- -> Fe}$$ To reduce the $2\text{ moles}$ of $\ce{Fe^{3+}}$ ions present in $1\text{ mole}$ of $\ce{Fe2O3}$, the reaction stoichiometry is scaled by a factor of $2$: $$\ce{2Fe^{3+} + 6e^- -> 2Fe}$$ Step 4 - Calculate the Quantity of Electricity in Faradays According to Faraday's laws of electrolysis, the electrical charge carried by $1\text{ mole}$ of electrons is defined as $1\text{ Faraday}$ ($1\text{ F}$): $$\text{Charge of 1 mole of electrons} = 1\text{ F}$$ From the balanced reduction half-reaction, the total number of moles of electrons required to reduce $2\text{ moles}$ of $\ce{Fe^{3+}}$ is: $$n = 6\text{ moles of } e^-$$ Thus, the total quantity of electricity ($Q$) required is: $$Q = n \times F = 6\text{ moles of } e^- \times 1\text{ F/mol} = \boxed{6\text{ F}}$$ Step 5 - Explanation of Options * **Option (A) is incorrect:** $1\text{ F}$ corresponds to $1\text{ mole}$ of electrons, which can only reduce $\frac{1}{3}\text{ mole}$ of $\ce{Fe^{3+}}$ ions (or $\frac{1}{6}\text{ mole}$ of $\ce{Fe2O3}$). * **Option (B) is incorrect:** $0.33\text{ F}$ is mathematically incorrect and would only reduce a tiny fraction of the iron oxide. * **Option (C) is incorrect:** $3\text{ F}$ corresponds to $3\text{ moles}$ of electrons. This is the quantity required to reduce $1\text{ mole}$ of $\ce{Fe^{3+}}$ ions to metallic iron, which would only suffice for $0.5\text{ mole}$ of $\ce{Fe2O3}$. * **Option (D) is correct:** $6\text{ F}$ of electricity provides the $6\text{ moles}$ of electrons required to completely reduce both moles of $\ce{Fe^{3+}}$ ions present in $1\text{ mole}$ of $\ce{Fe2O3}$. $$\text{Correct Option: } \boxed{\text{D}}$$