Passage of 96,500 coulomb of electricity liberates ______ L of at 273°C and 2 atm during electrolysi — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the Anodic Half-Reaction for the Oxidation of Water During the electrolysis of an aqueous solution, water ($\ce{H2O}$) undergoes oxidation at the anode to liberate oxygen gas ($\ce{O2}$): $$\ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}$$ This stoichiometric equation shows that the liberation of $1\text{ mole}$ of oxygen gas ($\ce{O2}$) requires the transfer of $4\text{ moles}$ of electrons. Step 2 - Calculate the Moles of Electrons Transferred According to Faraday's first law of electrolysis, the charge of $1\text{ mole}$ of electrons is equal to $1\text{ Faraday}$ ($F$), which is approximately $96500\text{ Coulombs}$: $$Q = n_{\ce{e-}} \times F$$ Where: * $Q = 96500\text{ C}$ (the quantity of electricity passed). * $F = 96500\text{ C/mol}$ (Faraday's constant). Substituting these values to find the moles of electrons ($n_{\ce{e-}}$): $$n_{\ce{e-}} = \frac{Q}{F} = \frac{96500\text{ C}}{96500\text{ C/mol}} = 1\text{ mole of electrons}$$ Step 3 - Determine the Moles of Oxygen Gas Liberated From the stoichiometry of the anode half-reaction in Step 1: $$4\text{ moles of electrons} \implies 1\text{ mole of }\ce{O2}$$ Therefore, the number of moles of oxygen gas ($n_{\ce{O2}}$) produced by $1\text{ mole of electrons}$ is: $$n_{\ce{O2}} = \frac{1\text{ mole of electrons}}{4} = 0.25\text{ moles of }\ce{O2}$$ Step 4 - Calculate the Volume of Oxygen Gas using the Ideal Gas Equation The volume ($V$) of a gas under non-standard conditions of temperature and pressure is calculated using the Ideal Gas Equation: $$P V = n R T \implies V = \frac{n R T}{P}$$ Where: * $n = n_{\ce{O2}} = 0.25\text{ moles}$ * $P = 2\text{ atm}$ (given pressure) * $T = 273^\circ\text{C}$ (given temperature) Convert the temperature from Celsius to Kelvin: $$T = 273 + 273 = 546\text{ K}$$ * $R = 0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ (universal gas constant) Now, substitute these values into the formula to compute the volume: $$V = \frac{0.25\text{ mol} \times 0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \times 546\text{ K}}{2\text{ atm}}$$ $$V = \frac{11.20665\text{ L}}{2}$$ $$V = \boxed{5.6\text{ L}}$$ Step 5 - Explanation of Each Option * **Option (A) is correct:** Our calculated volume is $5.6\text{ L}$, which matches this option exactly. * **Option (B) is incorrect:** A volume of $16.8\text{ L}$ is incorrect because it would correspond to the liberation of $0.75\text{ moles}$ of gas. * **Option (C) is incorrect:** A volume of $22.4\text{ L}$ represents the volume of $1\text{ mole}$ of an ideal gas at STP ($0^\circ\text{C}$ and $1\text{ atm}$), not the volume of $0.25\text{ moles}$ of $\ce{O2}$ at $273^\circ\text{C}$ and $2\text{ atm}$. * **Option (D) is incorrect:** A volume of $11.2\text{ L}$ represents the volume of $0.25\text{ moles}$ at STP, but not under the specified conditions of $273^\circ\text{C}$ and $2\text{ atm}$. $$\text{Correct Option: } \boxed{\text{A}}$$