The number of Faradays required to produce 1 g-atom of Mg from is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Definition of "g-atom" (Gram-Atom) The term "gram-atom" (g-atom) is a historical chemical unit used to represent exactly one mole of atoms of a particular element. Therefore, producing $1\text{ g-atom}$ of magnesium ($\ce{Mg}$) is equivalent to producing exactly $1\text{ mole}$ of magnesium: $$\text{1 g-atom of Mg} = 1\text{ mole of Mg}$$ Step 2 - Write the Reduction Half-Reaction In magnesium chloride ($\ce{MgCl2}$), magnesium exists as divalent cations ($\ce{Mg^2+}$). During electrolysis, these cations undergo reduction at the cathode by gaining electrons to deposit as solid magnesium metal ($\ce{Mg}$): $$\ce{Mg^2+(l) + 2e^- -> Mg(s)}$$ Step 3 - Calculate the Number of Faradays Required 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 stoichiometry of the reduction half-reaction: $$\text{Moles of electrons required } (n) = 2\text{ moles of } e^{-} \text{ per mole of Mg}$$ Using the relation between the quantity of electricity ($Q$ in Faradays) and the moles of substance produced: $$Q = n \times (\text{moles of Mg})$$ Substitute $n = 2\text{ F/mol}$ and $\text{moles of Mg} = 1\text{ mol}$: $$Q = 2\text{ F/mol} \times 1\text{ mol} = \boxed{2\text{ F}}$$ Step 4 - Explanation of Options * **Option (A)** is incorrect because $1\text{ Faraday}$ of charge corresponds to the transfer of $1\text{ mole}$ of electrons, which can only reduce $0.5\text{ mole}$ ($0.5\text{ g-atom}$) of divalent $\ce{Mg^2+}$ ions. * **Option (B)** is correct because $2\text{ Faradays}$ correspond to $2\text{ moles}$ of electrons, which are exactly required to reduce $1\text{ mole}$ ($1\text{ g-atom}$) of $\ce{Mg^2+}$ to metallic $\ce{Mg}$. * **Option (C)** is incorrect because $0.5\text{ Faraday}$ would only produce $0.25\text{ g-atom}$ of $\ce{Mg}$. * **Option (D)** is incorrect because $4\text{ Faradays}$ would produce $2\text{ g-atoms}$ of $\ce{Mg}$. $$\text{Correct Option: } \boxed{\text{B}}$$