What volume of at 0°C and 1 atm is formed per hour? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Total Electric Charge Passed in One Hour The total electrical charge ($Q_{\text{total}}$) passed through the electrolytic cell in one hour ($1\text{ hour} = 3600\text{ s}$) at a constant current of $15.0\text{ A}$ is calculated using the formula: $$Q_{\text{total}} = I \times t$$ Substituting the given values: $$Q_{\text{total}} = 15.0\text{ A} \times 3600\text{ s} = 54,000\text{ C}$$ Step 2 - Determine the Charge Consumed by Hydrogen Evolution Both nickel ($\ce{Ni}$) and hydrogen ($\ce{H2}$) are formed at the cathode. We are given that the current efficiency ($\eta_{\ce{Ni}}$) with respect to the deposition of nickel is $60\%$. Since the remaining current is consumed entirely by the parallel reduction of hydrogen ions, the current efficiency with respect to the formation of hydrogen gas ($\eta_{\ce{H2}}$) is: $$\eta_{\ce{H2}} = 100\% - 60\% = 40\% = 0.40$$ The active electrical charge utilized in the evolution of $\ce{H2}$ gas ($Q_{\ce{H2}}$) is: $$Q_{\ce{H2}} = \eta_{\ce{H2}} \times Q_{\text{total}}$$ $$Q_{\ce{H2}} = 0.40 \times 54,000\text{ C} = 21,600\text{ C}$$ Step 3 - Write the Cathode Reduction Reaction and Calculate the Moles of \ce{H2} At the cathode, hydrogen ions ($\ce{H^+}$) undergo reduction to form gaseous hydrogen molecules ($\ce{H2}$): $$\ce{2H^+(aq) + 2e^- -> H2(g)}$$ This stoichiometric relationship shows that the reduction of $2\text{ moles}$ of protons requires $2\text{ moles}$ of electrons ($n = 2$) to produce $1\text{ mole}$ of $\ce{H2}$ gas. Using Faraday's constant ($F \approx 96,500\text{ C mol}^{-1}$), we calculate the total moles of hydrogen gas ($n_{\ce{H2}}$) produced: $$n_{\ce{H2}} = \frac{Q_{\ce{H2}}}{n \times F}$$ $$n_{\ce{H2}} = \frac{21,600\text{ C}}{2 \times 96,500\text{ C mol}^{-1}}$$ $$n_{\ce{H2}} = \frac{21,600}{193,000}\text{ mol} \approx 0.11192\text{ mol}$$ Step 4 - Calculate the Volume of \ce{H2} at STP ($0^\circ\text{C}$ and $1\text{ atm}$) Under standard temperature and pressure conditions (STP: $0^\circ\text{C}$ and $1\text{ atm}$), the molar volume of an ideal gas is $22.4\text{ L mol}^{-1}$. The volume of $\ce{H2}$ gas ($V_{\ce{H2}}$) formed per hour is: $$V_{\ce{H2}} = n_{\ce{H2}} \times 22.4\text{ L mol}^{-1}$$ $$V_{\ce{H2}} = 0.11192\text{ mol} \times 22.4\text{ L mol}^{-1} \approx 2.507\text{ L} \approx 2.5\text{ L}$$ Step 5 - Evaluate the Multiple-Choice Options The options given in the problem are: * (A) $6.27\text{ L}$ * (B) $3.76\text{ L}$ * (C) $2.5\text{ L}$ * (D) $5.01\text{ L}$ Our calculated value of $2.5\text{ L}$ matches **Option (C)** perfectly. $$\text{Correct Option: } \boxed{C}$$