The current required to produce oxygen at the rate of 2.8 ml per second during electrolysis of acidu — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Anodic Oxidation Reaction for Water Electrolysis The electrolysis of acidulated water involves the decomposition of water molecules into hydrogen and oxygen gases. The oxidation of water occurs at the anode to produce oxygen gas ($\ce{O2}$): $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ From this balanced half-reaction, the production of $1\text{ mole}$ of diatomic oxygen gas ($\ce{O2}$) requires the transfer of $4\text{ moles}$ of electrons. Therefore, the valency factor ($n$-factor) for $\ce{O2}$ gas is: $$n = 4$$ Step 2 - Calculate the Molar Rate of Oxygen Production At standard temperature and pressure (STP), the molar volume of an ideal gas ($V_m$) is: $$V_m = 22400\text{ mL mol}^{-1}$$ The given volumetric rate of oxygen production ($r_v$) is: $$r_v = 2.8\text{ mL s}^{-1}$$ The molar rate of oxygen production ($r_m$) in moles per second is calculated by dividing the volumetric rate by the molar volume: $$r_m = \frac{r_v}{V_m}$$ Substitute the values: $$r_m = \frac{2.8\text{ mL s}^{-1}}{22400\text{ mL mol}^{-1}}$$ $$r_m = 1.25 \times 10^{-4}\text{ mol s}^{-1}$$ Step 3 - Calculate the Required Electric Current ($I$) According to Faraday's First Law of Electrolysis, the total electric charge ($Q$) passed is related to the moles of electrons transferred by Faraday's constant ($F \approx 96500\text{ C mol}^{-1}$): $$Q = I \times t = n \times F \times \text{moles of product}$$ Thus, the electric current ($I$), which represents the rate of charge flow per second ($I = \frac{Q}{t}$), is calculated as: $$I = n \times F \times r_m$$ Substitute the values: $$I = 4 \times (96500\text{ C mol}^{-1}) \times (1.25 \times 10^{-4}\text{ mol s}^{-1})$$ $$I = 5.0 \times 10^{-4}\text{ mol s}^{-1} \times 96500\text{ C mol}^{-1}$$ $$I = \boxed{48.25\text{ A}}$$ *(Note: Although the options in the printed question contain the unit "A/s", this is a typographical error for electric current, which is measured in Amperes "A".)* Step 4 - Evaluate the Options * **Option (A) is correct:** As mathematically calculated, the electric current required to produce oxygen at the specified rate is exactly $48.25\text{ A}$. * **Option (B) is incorrect:** This value ($24.12\text{ A}$) would be obtained if the $n$-factor of oxygen gas were incorrectly taken as $2$ instead of $4$. * **Option (C) is incorrect:** This value ($96.5\text{ A}$) represents double the required current, which is a calculation error. * **Option (D) is incorrect:** This value ($0.0048\text{ A}$) represents a decimal place error. $$\text{Correct Option: } \boxed{\text{A}}$$