Which process occurs in the electrolysis of an aqueous solution of nickel chloride at nickel anode? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Components of the Electrolytic Cell We are given an aqueous solution of nickel chloride ($\ce{NiCl2}$). In this system, the chemical species present near the electrodes are: * Cations: Nickel ions ($\ce{Ni^2+(aq)}$) and hydrogen ions ($\ce{H+(aq)}$) from the dissociation of water. * Anions: Chloride ions ($\ce{Cl^-(aq)}$) and hydroxide ions ($\ce{OH^-(aq)}$) from the dissociation of water. * Solvent: Water molecules ($\ce{H2O(l)}$). * Anode: Active nickel metal ($\ce{Ni(s)}$). Step 2 - Analyze the Competitive Oxidation Reactions at the Anode The anode is the positive electrode where oxidation (loss of electrons) occurs. Since the anode is made of active nickel metal rather than an inert material like platinum or graphite, the nickel metal itself can participate in the oxidation process. The three competing oxidation half-reactions at the anode are: 1. **Oxidation of the active nickel metal anode (Anodic dissolution):** $$\ce{Ni(s) -> Ni^2+(aq) + 2e^-} \quad E^\circ_{\text{ox}} = +0.25\text{ V}$$ 2. **Oxidation of water molecules:** $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{\text{ox}} = -1.23\text{ V}$$ 3. **Oxidation of chloride ions:** $$\ce{2Cl^-(aq) -> Cl2(g) + 2e^-} \quad E^\circ_{\text{ox}} = -1.36\text{ V}$$ Step 3 - Apply the Principle of Preferential Discharge The species with the highest (most positive or least negative) standard oxidation potential ($E^\circ_{\text{ox}}$) will be preferentially oxidized at the anode because it requires the least amount of energy to lose electrons. Comparing the standard oxidation potentials: $$E^\circ_{\text{ox}}(\ce{Ni/Ni^2+}) = +0.25\text{ V} > E^\circ_{\text{ox}}(\ce{H2O/O2}) = -1.23\text{ V} > E^\circ_{\text{ox}}(\ce{Cl^-/Cl2}) = -1.36\text{ V}$$ Since the oxidation potential of active nickel metal is significantly higher than that of both water and chloride ions, the nickel anode undergoes oxidation and dissolves into the solution, yielding nickel ions ($\ce{Ni^2+}$) and releasing electrons. Therefore, the preferred anodic process is: $$\ce{Ni(s) -> Ni^2+(aq) + 2e^-}$$ Step 4 - Evaluate the Options * **Option (A) is correct:** The oxidation of the active nickel anode ($\ce{Ni -> Ni^2+ + 2e^-}$) occurs preferentially due to its high standard oxidation potential. * **Option (B) is incorrect:** This represents a reduction half-reaction ($\ce{Ni^2+ + 2e^- -> Ni}$), which can only occur at the cathode. * **Option (C) is incorrect:** This represents the oxidation of chloride ions ($\ce{2Cl^- -> Cl2 + 2e^-}$). This reaction would only occur if an inert electrode (such as platinum or carbon) was used. * **Option (D) is incorrect:** This represents the reduction of hydrogen ions ($\ce{2H^+ + 2e^- -> H2}$), which is a cathodic reduction process. $$\text{Correct Option: } \boxed{\text{A}}$$