Two electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chlor — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Chemical Species and Cathodic Reactions In an electrolytic system, when two cells are connected in series, the same quantity of electric charge ($Q$) passes through both cells. We are given two cells containing different iron salt solutions: 1. **Cell 1** contains acidified ferrous chloride ($\ce{FeCl2}$), where iron exists as divalent ferrous cations ($\ce{Fe^{2+}}$). 2. **Cell 2** contains acidified ferric chloride ($\ce{FeCl3}$), where iron exists as trivalent ferric cations ($\ce{Fe^{3+}}$). At the cathodes of the respective cells, these cations undergo reduction to deposit solid metallic iron ($\ce{Fe}$): * **Cathode of Cell 1 (Reduction of Ferrous Ions):** $$\ce{Fe^{2+}(aq) + 2e^- -> Fe(s)}$$ Here, the transfer of $2\text{ moles}$ of electrons is required to deposit $1\text{ mole}$ of iron metal. Thus, the valency factor ($n$-factor) is: $$n_1 = 2$$ * **Cathode of Cell 2 (Reduction of Ferric Ions):** $$\ce{Fe^{3+}(aq) + 3e^- -> Fe(s)}$$ Here, the transfer of $3\text{ moles}$ of electrons is required to deposit $1\text{ mole}$ of iron metal. Thus, the valency factor ($n$-factor) is: $$n_2 = 3$$ Step 2 - Determine the Chemical Equivalent Weights ($E$) of Iron in Both Cells The chemical equivalent weight ($E$) of a substance is defined as its molar mass ($M$) divided by its valency factor ($n$-factor): $$E = \frac{M}{n}$$ Let $M$ represent the atomic mass of iron. * The equivalent weight of iron in the first cell ($E_1$) is: $$E_1 = \frac{M}{2}$$ * The equivalent weight of iron in the second cell ($E_2$) is: $$E_2 = \frac{M}{3}$$ Step 3 - Apply Faraday's Second Law of Electrolysis Faraday's Second Law of Electrolysis states that when the same quantity of electricity is passed through several electrolytic cells connected in series, the masses ($W$) of the substances deposited at the electrodes are directly proportional to their chemical equivalent weights ($E$): $$W \propto E$$ Therefore, the ratio of the mass of iron deposited in Cell 1 ($W_1$) to that in Cell 2 ($W_2$) is equal to the ratio of their chemical equivalent weights: $$\frac{W_1}{W_2} = \frac{E_1}{E_2}$$ Step 4 - Calculate the Mass Ratio of Deposited Iron Substitute the expressions for $E_1$ and $E_2$ into the ratio equation: $$\frac{W_1}{W_2} = \frac{\frac{M}{2}}{\frac{M}{3}}$$ Simplifying this algebraic fraction: $$\frac{W_1}{W_2} = \frac{M}{2} \times \frac{3}{M} = \frac{3}{2}$$ Thus, the mass ratio of the iron deposited at the cathodes in the two cells is: $$\frac{W_1}{W_2} = \boxed{3:2}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This ratio ($3:1$) would occur if the iron in the first cell had a valency factor of $1$ (such as $\ce{Fe^+}$) and the second cell had $3$ ($\ce{Fe^{3+}}$), which does not correspond to the actual ferrous chloride ($\ce{Fe^{2+}}$) system. * **Option (B) is incorrect:** This ratio ($2:3$) is the inverse of the correct ratio. It mistakenly assumes that the mass of metal deposited is directly proportional to its valency factor. In reality, because a higher valency requires more electrons per atom to reduce, the deposited mass is inversely proportional to the valency factor. * **Option (C) is incorrect:** This ratio ($1:1$) would only be correct if both cells contained iron in the same oxidation state (having the same equivalent weight), which is not the case here. * **Option (D) is correct:** As mathematically derived from Faraday's second law of electrolysis, the mass ratio of iron deposited is exactly $3:2$. $$\text{Correct Option: } \boxed{\text{D}}$$