In the electrolysis of acidified solution using Pt-electrodes, the anode reaction is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Chemical Species in the Electrolytic System We are analyzing the electrolysis of an aqueous, acidified silver nitrate ($\ce{AgNO3}$) solution using inert platinum ($\text{Pt}$) electrodes. When silver nitrate dissolves in water, it dissociates completely into its constituent ions: $$\ce{AgNO3(s) ->[\ce{H2O}] Ag^+(aq) + NO3^-(aq)}$$ Additionally, because the solution is acidified, there is a high concentration of hydrogen ions ($\ce{H^+}$). Water ($\ce{H2O}$) is the solvent and is also present in abundance. The electrodes are made of platinum ($\text{Pt}$), which is a noble metal and acts as an inert electrode. It does not actively participate in the chemical redox reactions but simply provides a surface for the transfer of electrons. Step 2 - Analyze the Competing Species at the Anode The anode is the positive electrode connected to the positive terminal of the external power supply. Oxidation (loss of electrons) always takes place at the anode. The negative ions (anions) and neutral molecules migrate toward or are present at the anode surface: 1. **Nitrate Ions ($\ce{NO3^-}$):** These are the primary anions in the solution. 2. **Water Molecules ($\ce{H2O}$):** These are the solvent molecules present at the electrode-solution interface. 3. **Platinum Electrode ($\text{Pt}$):** The physical anode material. To determine which species will undergo oxidation, we compare their thermodynamic ease of oxidation (standard oxidation potentials, $E^\circ_{\text{ox}}$): * **Oxidation of Platinum ($\text{Pt}$):** Platinum is highly inert and has an extremely negative oxidation potential, meaning it resists dissolving or oxidizing to $\ce{Pt^{n+}}$ under normal conditions: $$\ce{Pt(s) \not\to Pt^3+(aq) + 3e^-}$$ * **Oxidation of Nitrate Ions ($\ce{NO3^-}$):** In the nitrate anion, the central nitrogen atom is already in its maximum possible oxidation state of $+5$. Because nitrogen cannot be further oxidized under these conditions, the nitrate ion is exceptionally stable and possesses an extremely low tendency to lose electrons. Its oxidation is thermodynamically unfavorable in aqueous media. * **Oxidation of Water ($\ce{H2O}$):** Water molecules can undergo oxidation to produce oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H^+}$) according to the following half-cell reaction: $$\ce{2H2O(l) -> 4H^+(aq) + O2(g) + 4e^-} \quad E^\circ_{\text{ox}} = -1.23\text{ V}$$ Step 3 - Determine the Preferred Anodic Reaction Comparing the competing species, the standard oxidation potential of water ($-1.23\text{ V}$) is significantly less negative (more favorable) than the oxidation potential required to decompose the highly stable nitrate ion ($\ce{NO3^-}$). Furthermore, because the platinum electrode is completely inert, it does not react. Therefore, water undergoes oxidation at the anode in preference to both the nitrate ions and the platinum metal. The resulting anode reaction is: $$\ce{2H2O -> 4H^+ + O2 + 4e^-}$$ This reaction leads to the continuous evolution of oxygen gas ($\ce{O2}$) bubbles at the anode and causes the solution in the anode compartment to become increasingly acidic due to the accumulation of hydronium ($\ce{H^+}$) ions. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** This reaction shows the oxidation of nitrate ions to nitrogen dioxide and oxygen gas. This reaction does not occur because the nitrate ion is thermodynamically too stable to be oxidized in the presence of water. * **Option (B) is incorrect:** This represents another hypothetical oxidation of nitrate ions to nitric oxide and oxygen gas, which is also incorrect due to the high stability of nitrogen in its $+5$ oxidation state. * **Option (C) is correct:** As derived, water is preferentially oxidized at the inert platinum anode, releasing oxygen gas and producing hydrogen ions. * **Option (D) is incorrect:** Since platinum is an inert electrode, it does not oxidize or dissolve to form platinum cations in this electrolytic system. $$\text{Correct Option: } \boxed{\text{C}}$$