On passing electricity through dilute solution, the mass of substances liberated at the cathode and — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Cathodic Reaction and Calculate the Equivalent Weight of Hydrogen During the electrolysis of dilute sulphuric acid ($\ce{H2SO4}$), water undergoes decomposition. At the cathode (negative electrode), hydrogen ions ($\ce{H^+}$) are reduced to form hydrogen gas ($\ce{H2}$): $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ The valency factor ($z$) of hydrogen gas ($\ce{H2}$) is defined as the number of moles of electrons required to produce $1\text{ mole}$ of $\ce{H2}$ gas, which is: $$z_{\ce{H2}} = 2$$ The equivalent weight ($E_{\ce{H2}}$) of hydrogen gas is calculated by dividing its molar mass ($M_{\ce{H2}} = 2\text{ g/mol}$) by its valency factor ($z_{\ce{H2}}$): $$E_{\ce{H2}} = \frac{M_{\ce{H2}}}{z_{\ce{H2}}}$$ $$E_{\ce{H2}} = \frac{2\text{ g/mol}}{2} = 1\text{ g/eq}$$ Step 2 - Write the Anodic Reaction and Calculate the Equivalent Weight of Oxygen At the anode (positive electrode), water molecules are oxidized to liberate oxygen gas ($\ce{O2}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ The valency factor ($z$) of oxygen gas ($\ce{O2}$) is defined as the number of moles of electrons transferred per mole of $\ce{O2}$ produced, which is: $$z_{\ce{O2}} = 4$$ The equivalent weight ($E_{\ce{O2}}$) of oxygen gas is calculated by dividing its molar mass ($M_{\ce{O2}} = 32\text{ g/mol}$) by its valency factor ($z_{\ce{O2}}$): $$E_{\ce{O2}} = \frac{M_{\ce{O2}}}{z_{\ce{O2}}}$$ $$E_{\ce{O2}} = \frac{32\text{ g/mol}}{4} = 8\text{ g/eq}$$ Step 3 - Apply Faraday's Second Law of Electrolysis to Find the Mass Ratio According to Faraday's Second Law of Electrolysis, when the same quantity of electricity ($Q$) is passed through an electrolytic system, the masses ($w$) of the substances liberated or deposited at the electrodes are directly proportional to their chemical equivalent weights ($E$): $$w \propto E$$ Therefore, the ratio of the mass of hydrogen liberated at the cathode ($w_{\text{cathode}}$) to the mass of oxygen liberated at the anode ($w_{\text{anode}}$) is given by the ratio of their equivalent weights: $$\frac{w_{\text{cathode}}}{w_{\text{anode}}} = \frac{E_{\ce{H2}}}{E_{\ce{O2}}}$$ $$\frac{w_{\text{cathode}}}{w_{\text{anode}}} = \frac{1\text{ g/eq}}{8\text{ g/eq}} = \frac{1}{8}$$ Thus, the ratio of the masses of substances liberated at the cathode and anode is $1:8$. Step 4 - Evaluate the Options * **Option (A) is correct:** As mathematically derived using Faraday's second law of electrolysis, the ratio of masses of hydrogen and oxygen liberated is exactly $1:8$. * **Option (B) is incorrect:** This is the inverse ratio ($8:1$), which represents the ratio of the mass of oxygen at the anode to that of hydrogen at the cathode. * **Option (C) is incorrect:** This ratio ($1:32$) is obtained if one incorrectly uses the molar mass of oxygen gas directly without dividing by its valency factor. * **Option (D) is incorrect:** This ratio ($1:16$) is obtained if the valency factor of oxygen is incorrectly assumed to be $2$ instead of $4$. $$\text{Correct Option: } \boxed{\text{A}}$$