How long will it take to produce 0.10 mole of HNO2 by this reaction if a current of 10 A passes thro — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write and Balance the Half-Reaction for \ce{HNO2} Production To determine the relationship between the quantity of electricity passed and the moles of nitrous acid (\ce{HNO2}) produced, we must first find the number of electrons ($n$) involved in the reduction of the nitrate ion (\ce{NO3^-}) to nitrous acid. Let us calculate the oxidation states of nitrogen in the reactant and the product: * In the nitrate ion (\ce{NO3^-}): $$\text{Oxidation state of N} + 3(-2) = -1 \implies \text{Oxidation state of N} = +5$$ * In nitrous acid (\ce{HNO2}): $$+1 + \text{Oxidation state of N} + 2(-2) = 0 \implies \text{Oxidation state of N} = +3$$ The change in the oxidation state of nitrogen is from $+5$ to $+3$, which corresponds to a gain of $2$ electrons. The balanced reduction half-reaction involving hydronium ions ($\ce{H3O^+}$) is: $$\ce{NO3^-(aq) + 3H3O^+(aq) + 2e^- -> HNO2(aq) + 4H2O(l)}$$ From the stoichiometry of this half-reaction, producing $1\text{ mole}$ of $\ce{HNO2}$ requires $2\text{ moles}$ of electrons. Therefore, the electron change per mole of product ($n$-factor) is: $$n = 2$$ Step 2 - Calculate the Total Electric Charge ($Q$) Required According to Faraday's First Law of Electrolysis, the total electrical charge ($Q$) in Coulombs required to produce a specific amount of substance is: $$Q = n \times F \times \text{moles of substance}$$ Where: * $n = 2$ (moles of electrons per mole of \ce{HNO2}) * $F = 96,500\text{ C mol}^{-1}$ (Faraday's constant) * Moles of \ce{HNO2} = $0.10\text{ mol}$ Substituting these values into the formula: $$Q = 2 \times 96,500\text{ C mol}^{-1} \times 0.10\text{ mol}$$ $$Q = 19,300\text{ C}$$ Step 3 - Calculate the Time ($t$) Required The electrical charge ($Q$) is related to current ($I$) and time ($t$) by the equation: $$Q = I \times t$$ Rearranging the formula to solve for time ($t$): $$t = \frac{Q}{I}$$ Substitute the calculated charge $Q = 19,300\text{ C}$ and the given current $I = 10\text{ A}$ (where $1\text{ A} = 1\text{ C s}^{-1}$): $$t = \frac{19,300\text{ C}}{10\text{ A}}$$ $$t = \boxed{1,930\text{ s}}$$ Step 4 - Evaluate and Explain the Options The given options for this sub-question are: * **(a) 965 s** * **(b) 193 s** * **(c) 1930 s** * **(d) 482.5 s** * **Option (A) is incorrect:** $965\text{ s}$ is obtained if the $n$-factor is incorrectly assumed to be $1$ instead of $2$. * **Option (B) is incorrect:** $193\text{ s}$ is obtained if one forgets to multiply by the $n$-factor of $2$ and also makes a decimal-place calculation error. * **Option (C) is correct:** As mathematically demonstrated, the time required to produce $0.10\text{ mole}$ of \ce{HNO2} is exactly $1,930\text{ s}$. * **Option (D) is incorrect:** $482.5\text{ s}$ is obtained if the $n$-factor is incorrectly taken as $0.5$ instead of $2$. $$\text{Correct Option: } \boxed{\text{C}}$$