1. Predicting products of electrolysis
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2. Electrolysis of molten salts
Pb2+ ions are
attracted to the
cathode (negative electrode),
where they gain
electrons to form lead.
Br− ions are attracted to the anode (positive electrode), where they lose electrons
to form bromine.
Overall: PbBr2(l) → Pb(l) + Br2(g)
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3. Question Predict the product formed at each electrode during the electrolysis of molten potassium chloride. Write the overall equation, with state symbols, for the electrolysis of molten potassium chloride, KCl. Hint: the negative ions are attracted to the anode and the positive ions are attracted to the cathode. Answer anode – chlorine cathode – potassium 2KCl(l) → 2K(l) + Cl2(g)
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4. Question Write the half equation for the reaction taking place at each electrode during the electrolysis of molten potassium chloride. Classify each half equation as oxidation or reduction. Hint: OIL RIG Answer anode: 2Cl−(l) → Cl2(g) + 2e oxidation cathode: K+(l) + e → K(l) reduction
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5. Electrolysis of salt solutions
Copper chloride solution contains Cu2+ and Cl− ions as well as some H+ and OH− ions from water.
Cl− ions are discharged by losing electrons to form chlorine at the anode.
Cu2+ ions are discharged by gaining electrons to form copper at the cathode.
Overall: CuCl2(aq) → Cu(s) + Cl2(g)
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6. Question Which ion is discharged?
Ions attracted to electrode
Electrode reaction
H+ and ions from an unreactive metal e.g. Cu2+ or Ag+
Metal ions gain electrons so copper or silver is formed at the cathode.
H+ and ions from a reactive metal e.g. Na+ or Mg2+
H+ ions gain electrons so hydrogen gas is formed at the cathode.
OH− and halide ions i.e. Cl−, Br− and I−
Halide ions lose electrons so Cl2, Br2 or I2 is formed at the anode.
OH− and SO42− or NO3−
Hydroxide ions lose electrons so oxygen is formed at the anode.
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7. Question Predict the product formed at each electrode during the electrolysis of potassium bromide solution. Write the overall equation, with state symbols, for the electrolysis of potassium bromide solution, KBr(aq). Answer anode – bromine cathode – hydrogen 2KBr(aq) + 2H2O(l) → H2(g) + Br2(g) + 2KOH(aq)
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8. Question Write the half equation for the reaction taking place at each electrode during the electrolysis of potassium bromide solution. Classify each half equation as oxidation or reduction. Answer anode: 2Br−(aq) → Br2(g) + 2e oxidation cathode: 2H+(aq) + 2e → H2(g) reduction
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9. Question Predict the product formed at each electrode during the electrolysis of silver nitrate solution. Write the overall equation, with state symbols, for the electrolysis of silver nitrate solution, AgNO3(aq). Answer anode – oxygen cathode – silver 4AgNO3(aq) + 2H2O(l) → 4Ag(s) + O2(g) + 4HNO3(aq)
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10. Question Write the half equation for the reaction taking place at each electrode during the electrolysis of silver nitrate solution. Classify each half equation as oxidation or reduction. Answer anode: 4OH−(aq) → O2(g) + 2H2O(l) + 4e oxidation cathode: Ag+(aq) + e → Ag(s) reduction
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11. Question Predict the product formed at each electrode during the electrolysis of magnesium sulfate solution. Write the overall equation, with state symbols, for the electrolysis of magnesium sulfate solution, MgSO4(aq). Answer anode – oxygen cathode – hydrogen 2H2O(l) → 2H2(g) + O2(g)
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12. Question Write the half equation for the reaction taking place at each electrode during the electrolysis of magnesium sulfate solution. Classify each half equation as oxidation or reduction. Answer anode: 4OH−(aq) → O2(g) + 2H2O(l) + 4e oxidation cathode: 2H+(aq) + 2e → H2(g) reduction
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13. Purification of copper
Copper electrodes and copper sulfate solution are used.
Copper atoms from the anode lose electrons to form copper ions (Cu2+) that dissolve. Cu2+ ions are attracted to the cathode where they gain electrons to form copper atoms.
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14. Question Write the half equation for the reaction taking place at each electrode during the electrolysis of copper sulfate solution using copper electrodes. Classify each half equation as oxidation or reduction. Answer anode: Cu(s) → Cu2+(aq) + 2e oxidation cathode: Cu2+(aq) + 2e → Cu(s) reduction
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