The same quantity of electricity is passed through one molar solution of and one molar solution of . — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cathodic Reaction in Both Solutions In both aqueous solutions, namely $1\text{ M}$ sulphuric acid ($\ce{H2SO4}$) and $1\text{ M}$ hydrochloric acid ($\ce{HCl}$), the predominant reaction occurring at the cathode (negative electrode) is the reduction of hydrogen ions ($\ce{H^+}$) to liberate hydrogen gas ($\ce{H2}$): $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ For this reaction, the valency factor ($z$ or $n$-factor) of hydrogen gas is defined as the number of moles of electrons required to produce $1\text{ mole}$ of $\ce{H2}$ gas: $$z_{\ce{H2}} = 2$$ Step 2 - Apply Faraday's First Law of Electrolysis Faraday's First Law of Electrolysis states that the mass ($w$) of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte: $$w = Z \cdot Q = \frac{E}{F} \cdot Q$$ Where: * $w$ is the mass of the substance liberated. * $Q$ is the quantity of electricity passed (in coulombs). * $E$ is the chemical equivalent weight of the substance. * $F$ is Faraday's constant ($\approx 96,500\text{ C/eq}$). The chemical equivalent weight of hydrogen gas ($E_{\ce{H2}}$) is calculated by dividing its molar mass ($M_{\ce{H2}} \approx 2\text{ g/mol}$) by its valency factor ($z_{\ce{H2}} = 2$): $$E_{\ce{H2}} = \frac{M_{\ce{H2}}}{z_{\ce{H2}}} = \frac{2\text{ g/mol}}{2} = 1\text{ g/eq}$$ Step 3 - Compare the Amounts of Hydrogen Evolved We are given that the same quantity of electricity ($Q$) is passed through both the $1\text{ M}$ $\ce{H2SO4}$ and $1\text{ M}$ $\ce{HCl}$ solutions. Let $w_1$ be the mass of hydrogen evolved from $\ce{H2SO4}$ and $w_2$ be the mass of hydrogen evolved from $\ce{HCl}$: $$w_1 = \frac{E_{\ce{H2}} \cdot Q}{F}$$ $$w_2 = \frac{E_{\ce{H2}} \cdot Q}{F}$$ Since both $Q$ and $E_{\ce{H2}}$ are identical for both systems: $$w_1 = w_2$$ Similarly, the number of moles of $\ce{H2}$ gas evolved is given by: $$n_{\ce{H2}} = \frac{Q}{2F}$$ Since $Q$ is the same, the moles and hence the volume (at a given temperature and pressure) of hydrogen evolved in both cases are exactly identical. Therefore, the amount of hydrogen evolved from $\ce{H2SO4}$ is the same as that from $\ce{HCl}$. Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As mathematically shown, because the cathodic reduction reaction, the equivalent weight of hydrogen, and the passed electrical charge are identical in both cases, the amount of hydrogen evolved is the same. * **Option (B) is incorrect:** Although $\ce{H2SO4}$ is a diprotic acid and can potentially furnish twice as many $\ce{H^+}$ ions per mole of acid compared to the monoprotic $\ce{HCl}$ in solution, the amount of substance electrolyzed is limited solely by the quantity of electricity passed ($Q$), not the total concentration of available ions in the bulk solution (as both are in excess). * **Option (C) is incorrect:** This is the inverse of the misconception in Option (B), and is incorrect. * **Option (D) is incorrect:** The size of the electrode affects the current density and rate of electrolysis, but the total stoichiometric amount of substance liberated depends strictly on the total charge passed ($Q = I \cdot t$), according to Faraday's laws. $$\text{Correct Option: } \boxed{\text{A}}$$