A current of 3.7 A is passed for 6 h between nickel electrodes in 0.50 l of 2 M solution of Ni(NO3)2 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Nature of the Electrodes and Solution Components In this electrolysis setup, we have: * Electrolyte: An aqueous solution of nickel nitrate, $\ce{Ni(NO3)2}$, containing nickel ions ($\ce{Ni^2+}$) and nitrate ions ($\ce{NO3^-}$). * Initial volume of the solution ($V$) = $0.50\text{ L}$ * Initial concentration of $\ce{Ni^2+}$ ($C_{\text{initial}}$) = $2\text{ M}$ * Electrodes: Both the anode and cathode are made of **active nickel metal ($\ce{Ni}$)**. Step 2 - Analyze the Cathode Reaction (Reduction) At the cathode (the negative electrode), reduction occurs. Dissolved nickel ions ($\ce{Ni^2+}$) from the solution migrate to the cathode, gain electrons, and deposit as metallic nickel: $$\ce{Ni^2+(aq) + 2e^- -> Ni(s)}$$ This cathodic reduction process consumes $\ce{Ni^2+}$ ions from the solution, tending to decrease their concentration. Step 3 - Analyze the Anode Reaction (Oxidation) At the anode (the positive electrode), oxidation occurs. Since active nickel electrodes are used rather than inert electrodes (such as platinum or graphite), the nickel metal of the anode itself is oxidized preferentially because of its high standard oxidation potential: $$\ce{Ni(s) -> Ni^2+(aq) + 2e^-}$$ This anodic oxidation process produces $\ce{Ni^2+}$ ions and releases them into the solution, tending to increase their concentration. Step 4 - Calculate the Net Change in $\ce{Ni^2+}$ Concentration Since the same electric current ($I = 3.7\text{ A}$) flows through the entire series circuit for the same duration ($t = 6\text{ h}$), the rate of electron transfer at both electrodes is identical. * According to Faraday's Laws of Electrolysis, the moles of $\ce{Ni^2+}$ ions reduced at the cathode must be exactly equal to the moles of $\ce{Ni^2+}$ ions produced by the dissolution of the nickel anode: $$\text{Rate of consumption of } \ce{Ni^2+} = \text{Rate of production of } \ce{Ni^2+}$$ * Therefore, the net change in the number of moles of $\ce{Ni^2+}$ ions in the electrolyte solution is zero: $$\Delta n_{\ce{Ni^2+}} = n_{\text{produced}} - n_{\text{consumed}} = 0\text{ mol}$$ Since the volume of the solution remains virtually constant, the molar concentration of $\ce{Ni^2+}$ at the end of the electrolysis is identical to its initial value: $$C_{\text{final}} = C_{\text{initial}} = \mathbf{2\text{ M}}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This value ($1.172\text{ M}$) is obtained if one incorrectly assumes that $\ce{Ni^2+}$ is only consumed at the cathode (as would happen with an inert anode like platinum where water or nitrate is oxidized instead of nickel) and calculates the loss using Faraday's laws: $$Q = I \times t = 3.7\text{ A} \times (6 \times 3600\text{ s}) = 79,920\text{ C}$$ $$\text{Moles of electrons} = \frac{79,920}{96,500} \approx 0.828\text{ mol}$$ $$\text{Moles of }\ce{Ni^2+}\text{ lost} = \frac{0.828}{2} \approx 0.414\text{ mol}$$ $$\text{Initial moles of }\ce{Ni^2+} = 2\text{ M} \times 0.50\text{ L} = 1.0\text{ mol}$$ $$\text{Remaining moles of }\ce{Ni^2+} = 1.0 - 0.414 = 0.586\text{ mol}$$ $$\text{Remaining concentration} = \frac{0.586\text{ mol}}{0.50\text{ L}} = 1.172\text{ M}$$ This calculation is physically incorrect because it ignores the oxidation of the active nickel anode. * **Option (B) is incorrect:** This value ($0.172\text{ M}$) represents another stoichiometric error based on an incorrect calculation of ion consumption. * **Option (C) is incorrect:** This value ($0.586\text{ M}$) represents the remaining moles of nickel ions in the solution, rather than the molarity, when the anodic reaction is ignored. * **Option (D) is correct:** Because active nickel electrodes are used, the concentration of nickel ions in the solution does not change, and the molarity remains exactly $2\text{ M}$ throughout the electrolysis. $$\text{Correct Option: } \boxed{\text{D}}$$