How many moles of electrons pass through the circuit when 0.30 mole of HNO2 are produced? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Cathode Half-Reaction and Determine the Nitrogen Oxidation States At the cathode of the electrochemical cell, the nitrate ion ($\ce{NO3^-}$) undergoes reduction to nitrous acid ($\ce{HNO2}$). To determine the number of electrons transferred in this half-reaction, we calculate the oxidation states of nitrogen in both the reactant and product: * **In nitrate ion ($\ce{NO3^-}$):** Let $x$ be the oxidation state of Nitrogen. $$x + 3 \times (-2) = -1 \implies x = +5$$ * **In nitrous acid ($\ce{HNO2}$):** Let $y$ be the oxidation state of Nitrogen. $$+1 + y + 2 \times (-2) = 0 \implies y = +3$$ The net change in the oxidation state of Nitrogen per nitrogen atom is: $$\Delta\text{O.S.} = +5 - (+3) = 2$$ Therefore, the reduction of $1\text{ mole}$ of $\ce{NO3^-}$ to $\ce{HNO2}$ requires the transfer of exactly $2\text{ moles}$ of electrons. Balancing this reduction half-reaction in an acidic medium yields: $$\ce{NO3^-(aq) + 3H^+(aq) + 2e^- -> HNO2(aq) + H2O(l)}$$ Step 2 - Analyze the Anode Half-Reaction and Overall Cell Stoichiometry At the anode of the electrochemical cell, the mercurous ion ($\ce{Hg2^2+}$) undergoes oxidation to form the mercuric ion ($\ce{Hg^2+}$): $$\ce{Hg2^2+(aq) -> 2Hg^2+(aq) + 2e^-}$$ This half-reaction shows that the oxidation of $1\text{ mole}$ of mercurous ions ($\ce{Hg2^2+}$) releases $2\text{ moles}$ of electrons. Combining the oxidation and reduction half-reactions gives the overall balanced cell reaction: $$\ce{Hg2^2+(aq) + NO3^-(aq) + 3H^+(aq) -> 2Hg^2+(aq) + HNO2(aq) + H2O(l)}$$ Using hydronium ions ($\ce{H3O^+}$) for the acidic medium, the balanced equation matches the passage: $$\ce{Hg2^2+(aq) + NO3^-(aq) + 3H3O^+(aq) -> 2Hg^2+(aq) + HNO2(aq) + 4H2O(l)}$$ Step 3 - Verify the Feasibility of the Reaction with Initial Quantities The cell initially contains: * Initial moles of $\ce{Hg2^2+}$ = $0.50\text{ mol}$ * Initial moles of $\ce{NO3^-}$ = $0.40\text{ mol}$ According to the balanced overall reaction stoichiometry: * To produce $0.30\text{ mole}$ of $\ce{HNO2}$, the reaction requires the consumption of $0.30\text{ mole}$ of $\ce{NO3^-}$. Since $0.40\text{ mol}$ of $\ce{NO3^-}$ is initially present, there is a sufficient quantity (with $0.10\text{ mol}$ remaining in excess). * Similarly, producing $0.30\text{ mole}$ of $\ce{HNO2}$ requires the oxidation of $0.30\text{ mole}$ of $\ce{Hg2^2+}$. Since $0.50\text{ mol}$ of $\ce{Hg2^2+}$ is initially present, there is a sufficient quantity (with $0.20\text{ mol}$ remaining in excess). Thus, the reaction is fully feasible and is not limited by the initial quantities of the reactants. Step 4 - Calculate the Moles of Electrons Transferred Since the reduction of nitrate to nitrous acid requires $2\text{ moles}$ of electrons for every $1\text{ mole}$ of $\ce{HNO2}$ produced: $$\text{Moles of } e^- = 2 \times \text{Moles of } \ce{HNO2} \text{ produced}$$ Substituting the given value of $0.30\text{ mole}$ of $\ce{HNO2}$ produced: $$\text{Moles of } e^- = 2 \times 0.30\text{ mol}$$ $$\text{Moles of } e^- = \boxed{0.60\text{ mol}}$$ Step 5 - Evaluate and Explain each Option * **Option (A) is incorrect:** This value ($0.30\text{ mol}$) assumes a $1:1$ molar ratio of electrons to nitrous acid, which incorrectly implies a change in the oxidation state of nitrogen of only $1$. * **Option (B) is correct:** As mathematically proven in Step 4, the production of $0.30\text{ mol}$ of $\ce{HNO2}$ requires exactly $0.60\text{ mol}$ of electrons to pass through the circuit. * **Option (C) is incorrect:** This value ($0.15\text{ mol}$) is obtained by incorrectly dividing the moles of nitrous acid by $2$ instead of multiplying. * **Option (D) is incorrect:** This value ($1.20\text{ mol}$) is double the required amount, corresponding to an incorrect transfer of $4$ electrons per nitrous acid molecule. $$\text{Correct Option: } \boxed{B}$$