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Electrochemistry - Electrolysis of molten compounds

Grade 11A LevelChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electrolysis is the process by which electrical energy is used to cause a non-spontaneous chemical reaction, breaking down an ionic compound into its constituent elements.

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For electrolysis to occur in ionic compounds, they must be in a molten state or in an aqueous solution so that the ions (Mn+M^{n+} and Xn−X^{n-}) are free to move and carry the charge.

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The electrode connected to the positive terminal of the power supply is the anode (++), and the electrode connected to the negative terminal is the cathode (−-). A useful mnemonic is PANIC: Positive Anode, Negative Is Cathode.

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During electrolysis, cations (positive ions) migrate toward the cathode where they gain electrons (Reduction). Anions (negative ions) migrate toward the anode where they lose electrons (Oxidation).

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In the electrolysis of molten binary salts, the metal is always produced at the cathode and the non-metal is produced at the anode.

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The state symbols are crucial: in molten electrolysis, the electrolyte is liquid (l)(l), and products can be liquid (l)(l) or gas (g)(g) depending on the operating temperature.

📐Formulae

Cathode (Reduction): Mn++ne−→M\text{Cathode (Reduction): } M^{n+} + ne^- \rightarrow M

Anode (Oxidation): 2Xn−→X2+2ne−\text{Anode (Oxidation): } 2X^{n-} \rightarrow X_2 + 2ne^-

Q=I×tQ = I \times t

Overall Reaction: PbBr2(l)→Pb(l)+Br2(g)\text{Overall Reaction: } PbBr_2(l) \rightarrow Pb(l) + Br_2(g)

💡Examples

Problem 1:

Describe the electrolysis of molten lead(II) bromide (PbBr2PbBr_2). State the observations and the half-equations for the reactions at each electrode.

Solution:

At the cathode (−-), lead ions are reduced: Pb2+(l)+2e−→Pb(l)Pb^{2+}(l) + 2e^- \rightarrow Pb(l). At the anode (++), bromide ions are oxidized: 2Br−(l)→Br2(g)+2e−2Br^-(l) \rightarrow Br_2(g) + 2e^-.

Explanation:

Observation at the cathode: A silvery-grey bead of molten lead forms. Observation at the anode: Red-brown pungent fumes of bromine gas are evolved. The molten PbBr2PbBr_2 acts as the electrolyte because the ions Pb2+Pb^{2+} and Br−Br^- are mobile.

Problem 2:

In the industrial extraction of aluminum, molten aluminum oxide (Al2O3Al_2O_3) is electrolyzed. Provide the half-equation for the formation of aluminum at the cathode.

Solution:

Al3++3e−→Al(l)Al^{3+} + 3e^- \rightarrow Al(l)

Explanation:

Aluminum ions (Al3+Al^{3+}) are attracted to the negative cathode. Each ion gains three electrons to be reduced to a neutral aluminum atom. Because the process occurs at high temperatures, the aluminum is produced in liquid form.

Problem 3:

Predict the products of the electrolysis of molten sodium chloride (NaClNaCl) and write the overall balanced chemical equation.

Solution:

Products: Sodium metal (NaNa) at the cathode and Chlorine gas (Cl2Cl_2) at the anode. Overall equation: 2NaCl(l)→2Na(l)+Cl2(g)2NaCl(l) \rightarrow 2Na(l) + Cl_2(g).

Explanation:

The sodium ions (Na+Na^+) receive electrons at the cathode to form sodium. The chloride ions (Cl−Cl^-) release electrons at the anode to form diatomic chlorine gas molecules.

Electrolysis of molten compounds Grade 11 Notes & Examples