Two platinum electrodes were immersed in a solution of and electric current was passed through the s — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Ions Present in the Solution The electrolyte used is an aqueous solution of copper sulphate ($\ce{CuSO4}$). In water, $\ce{CuSO4}$ dissociates completely into its constituent mobile ions: $$\ce{CuSO4(aq) -> Cu^2+(aq) + SO4^2-(aq)}$$ Since this is an aqueous solution, water molecules ($\ce{H2O}$) are also present and undergo autoionization to a very small extent: $$\ce{H2O(l) <=> H^+(aq) + OH^-(aq)}$$ Thus, we have two types of cations ($\ce{Cu^2+}$ and $\ce{H^+}$) and two types of anions ($\ce{SO4^2-}$ and $\ce{OH^-}$) in the solution, along with neutral water molecules. Step 2 - Analyze the Cathodic Reduction Process During electrolysis, cations migrate toward the negatively charged cathode (which is made of inert platinum). Here, they compete to accept electrons and undergo reduction: * **Competition:** $\ce{Cu^2+(aq)}$ vs. $\ce{H^+(aq)}$ (or water molecules) * The standard reduction potentials at $298\text{ K}$ are: $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \quad E^\circ = +0.34\text{ V}$$ $$\ce{2H^+(aq) + 2e^- -> H2(g)} \quad E^\circ = 0.00\text{ V}$$ Since the standard reduction potential of copper ($E^\circ_{\ce{Cu^2+/Cu}} = +0.34\text{ V}$) is higher (more positive) than that of hydrogen, $\ce{Cu^2+}$ ions are preferentially reduced at the cathode: $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$$ As a result, metallic copper is deposited on the surface of the platinum cathode. As $\ce{Cu^2+}$ ions are continuously discharged and removed from the solution, the characteristic blue colour of the solution (which is due to the presence of hydrated $\ce{Cu^2+}$ ions) gradually fades and eventually disappears. Step 3 - Analyze the Anodic Oxidation Process Anions migrate toward the positively charged anode (which is also made of inert platinum). Here, they compete to release electrons and undergo oxidation: * **Competition:** $\ce{SO4^2-(aq)}$ vs. water molecules ($\ce{H2O}$) * Sulphate ions ($\ce{SO4^2-}$) are extremely stable because sulphur is already in its highest oxidation state of $+6$. Therefore, the oxidation of sulphate ions is thermodynamically very difficult. * Water molecules are oxidized much more easily at the anode in preference to sulphate ions: $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}$$ This oxidation process results in the evolution of oxygen gas ($\ce{O2}$) at the anode and continuously releases hydrogen ions ($\ce{H^+}$) into the solution. Step 4 - Determine the Remaining Solutes in the Colourless Solution By combining the processes occurring in the cell, we observe that: 1. All $\ce{Cu^2+}$ ions are discharged and deposited as solid copper at the cathode. 2. $\ce{SO4^2-}$ ions remain unreacted in the solution. 3. $\ce{H^+}$ ions are continuously produced at the anode and remain in the solution. The ions left behind in the beaker are hydrogen ions ($\ce{H^+}$) and sulphate ions ($\ce{SO4^2-}$). Together, they form an aqueous solution of **sulphuric acid** ($\ce{H2SO4}$): $$\ce{2H^+(aq) + SO4^2-(aq) -> H2SO4(aq)}$$ Since sulphuric acid is colourless, the solution becomes completely colourless. Step 5 - Evaluate the Options * **Option (A) is incorrect:** Platinum is an inert electrode. It does not undergo oxidation or dissolve to form platinum sulphate under these conditions. * **Option (B) is incorrect:** Copper sulphate has been completely consumed, which is why the blue colour of the solution disappeared. * **Option (C) is incorrect:** Copper hydroxide is a light blue precipitate. It does not form here because the solution is highly acidic due to the accumulation of $\ce{H^+}$ ions. * **Option (D) is correct:** As demonstrated, the remaining colourless solution is an aqueous solution of sulphuric acid ($\ce{H2SO4}$). $$\text{Correct Option: } \boxed{\text{D}}$$