Question 1:

The correct answer is A. potassium at the negative electrode and iodine at the positive electrode.

Explanation:

• Negative electrode (cathode): Positively charged ions (cations) are attracted here. In molten potassium iodide (KI), the cations are K⁺ ions. They gain electrons (are reduced) to form potassium metal.
  Reaction: K⁺ + e⁻ → K

• Positive electrode (anode): Negatively charged ions (anions) are attracted here. The anions are I⁻ ions. They lose electrons (are oxidized) to form iodine molecules.
  Reaction: 2I⁻ → I₂ + 2e⁻

Therefore, the products are potassium at the negative electrode and iodine at the positive electrode, which matches option A.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

The correct answer is D. ions in the molten lead bromide and electrons in the wires.

Explanation:

• In the molten lead bromide (electrolyte): The current is carried by ions (Pb²⁺ and Br⁻ ions) moving toward the electrodes. The positive lead ions move to the negative electrode (cathode), and the negative bromide ions move to the positive electrode (anode).

• In the wires (external circuit): The current is carried by electrons moving through the metal wires from one electrode to the other.

So, the correct pairing is ions in the molten lead bromide and electrons in the wires, which matches option D.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

The correct answer is C. the ions in solid lead bromide are not free to move around.

Explanation:

• Lead bromide (PbBr₂) is an ionic compound. In its solid state, the ions (Pb²⁺ and Br⁻) are held in a rigid lattice structure by strong electrostatic forces. Because they are fixed in place, they cannot move freely to carry an electric current.

• For electrolysis to occur and the bulb to light, the electrolyte must contain mobile charged particles. In molten (liquid) lead bromide, the ions gain enough energy to overcome the lattice forces and move freely, allowing the circuit to be completed. In solid form, this cannot happen.

Why the other options are incorrect:

• A. Solid lead bromide is ionic, not covalently bonded.

• B. The issue is not about tight packing for electrons—ionic solids conduct electricity via ions, not electrons (unlike metals).

• D. The current in an ionic compound is carried by ions, not atoms. Even if atoms were free to move, they are uncharged and wouldn't carry a current.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

The correct answer is C. Pb²⁺ + 2e⁻ → Pb (reduction).

Explanation:

• At the negative electrode (cathode) in electrolysis, positively charged ions (cations) gain electrons. This gain of electrons is called reduction.

• In molten lead bromide, the cations are Pb²⁺ ions. Each Pb²⁺ ion gains 2 electrons to form a neutral lead atom:
  Pb²⁺ + 2e⁻ → Pb

• Therefore, the correct half-equation is Pb²⁺ + 2e⁻ → Pb, and the type of reaction is reduction (since reduction = gain of electrons).

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

The correct answer is B. 2Br⁻ → Br₂ + 2e⁻ (oxidation).

Explanation:

• At the positive electrode (anode) in electrolysis, negatively charged ions (anions) lose electrons. This loss of electrons is called oxidation (OIL RIG: Oxidation Is Loss, Reduction Is Gain).

• In molten lead bromide, the anions are Br⁻ ions. Two Br⁻ ions each lose 1 electron (total of 2 electrons) to form a neutral bromine molecule:
  2Br⁻ → Br₂ + 2e⁻

• Therefore, the correct half-equation is 2Br⁻ → Br₂ + 2e⁻, and the type of reaction is oxidation.

Why the other options are incorrect:

• A: The equation 2Br⁻ + 2e⁻ → Br₂ shows electrons on the reactant side (gain of electrons), which would be reduction, but at the positive electrode, anions lose electrons (oxidation). This equation is also unbalanced in terms of charge.

• C: Although Br⁻ → Br + e⁻ is an oxidation half-equation, it is not balanced in terms of atoms—bromine exists as diatomic molecules (Br₂), not single Br atoms, in its elemental form. The correct balanced equation must show Br₂.

• D: The equation Br⁻ + e⁻ → Br shows gain of electrons (reduction) and is also not balanced for the diatomic nature of bromine.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

The correct answer is C.

Explanation:

To identify which solution could be sulphuric acid (H₂SO₄), we need to consider the products formed during electrolysis:

• Sulphuric acid is an acid that ionizes in water to produce H⁺ ions and SO₄²⁻ ions.

• At the cathode (negative electrode), H⁺ ions are preferentially reduced (over water or other cations) to produce hydrogen gas:
  2H⁺ + 2e⁻ → H₂

• At the anode (positive electrode), since sulphate ions (SO₄²⁻) are very difficult to oxidize, water is oxidized instead, producing oxygen gas:
  2H₂O → O₂ + 4H⁺ + 4e⁻

Therefore, the products for sulphuric acid electrolysis are hydrogen at the cathode and oxygen at the anode.

Looking at the table:

• Option C shows hydrogen at the cathode and oxygen at the anode, which matches exactly.

Why the other options are incorrect:

• A (copper and chlorine) – suggests a solution containing Cu²⁺ and Cl⁻ ions (e.g., copper chloride).

• B (copper and oxygen) – suggests a solution containing Cu²⁺ ions but with an inert anode where water is oxidized (e.g., copper sulphate).

• D (hydrogen and chlorine) – suggests a solution containing H⁺ and Cl⁻ ions (e.g., hydrochloric acid).

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

The correct answer is B.

Explanation:

To identify which solution could be copper(II) nitrate (Cu(NO₃)₂), we need to consider the ions present and the products formed during electrolysis:

• Copper(II) nitrate dissociates in water into Cu²⁺ ions and NO₃⁻ ions.

• At the cathode (negative electrode), Cu²⁺ ions are less reactive than hydrogen (from water), so they are preferentially reduced to form copper metal:
  Cu²⁺ + 2e⁻ → Cu

• At the anode (positive electrode), nitrate ions (NO₃⁻) are very difficult to oxidize. Instead, water is oxidized, producing oxygen gas:
  2H₂O → O₂ + 4H⁺ + 4e⁻

Therefore, the products for copper(II) nitrate electrolysis are copper at the cathode and oxygen at the anode.

Looking at the table:

• Option B shows copper at the cathode and oxygen at the anode, which matches exactly.

Why the other options are incorrect:

• A (copper and chlorine) – suggests a solution containing Cu²⁺ and Cl⁻ ions (e.g., copper chloride).

• C (hydrogen and oxygen) – suggests a solution containing H⁺ ions but no metal ions that are less reactive than hydrogen (e.g., sulphuric acid or sodium nitrate).

• D (hydrogen and chlorine) – suggests a solution containing H⁺ and Cl⁻ ions (e.g., hydrochloric acid).

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 8:

The correct answer is B. chlorine, bleaches damp blue litmus paper to white.

Explanation:

For the electrolysis of a concentrated solution of sodium chloride (brine):

• At the anode (positive electrode), chloride ions (Cl⁻) are preferentially discharged (over hydroxide ions from water) because the solution is concentrated. They lose electrons to form chlorine gas (Cl₂).

• At the cathode (negative electrode), hydrogen ions (from water) are preferentially reduced over sodium ions, producing hydrogen gas (H₂).

Therefore, Gas X (at the anode) is chlorine.

Test for chlorine gas:

• Chlorine gas turns damp blue litmus paper white (bleaches it) due to its oxidizing properties. (It may briefly turn red first due to acidity, but the definitive test is bleaching.)

Why the other options are incorrect:

• A: Chlorine does turn damp blue litmus red briefly (because it forms hydrochloric acid), but the characteristic test is bleaching it to white, not just turning it red.

• C: Oxygen relights a glowing splint, but oxygen is not produced at the anode in concentrated NaCl electrolysis—chlorine is.

• D: Hydrogen ignites with a "pop" with a burning splint, but hydrogen is produced at the cathode (Gas Y), not the anode (Gas X).

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

The correct answer is B. 2H⁺ + 2e⁻ → H₂.

Explanation:

• Gas Y is produced at the cathode (negative electrode). In the electrolysis of concentrated sodium chloride solution, the cation discharged at the cathode is hydrogen ions (H⁺) from the water, not sodium ions (Na⁺), because hydrogen is less reactive than sodium.

• At the cathode, H⁺ ions gain electrons (reduction) to form hydrogen gas. Each H⁺ ion gains 1 electron to form a hydrogen atom, but hydrogen exists as a diatomic molecule (H₂), so two H⁺ ions gain a total of 2 electrons to produce one H₂ molecule:
  2H⁺ + 2e⁻ → H₂

Why the other options are incorrect:

• A. H⁺ + e⁻ → H is not balanced for the formation of hydrogen gas because hydrogen is diatomic (H₂), not single atoms (H). The correct equation must show H₂.

• C. 2H⁺ → H₂ + 2e⁻ shows loss of electrons (oxidation), but at the cathode, reduction (gain of electrons) occurs. This equation would be correct for the anode, not the cathode.

• D. Na⁺ + e⁻ → Na would be the reduction of sodium ions, but in aqueous solution, H⁺ ions are preferentially discharged over Na⁺ ions because sodium is too reactive. Sodium metal is only produced in the electrolysis of molten sodium chloride, not aqueous.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

The correct answer is A. aluminium at the cathode and 2O²⁻ → O₂ + 4e⁻ at the anode.

Explanation:

• At the cathode (negative electrode): Al³⁺ ions gain electrons (reduction) to form aluminium metal:
  Al³⁺ + 3e⁻ → Al
  So the product at the cathode is aluminium.

• At the anode (positive electrode): Oxide ions (O²⁻) lose electrons (oxidation) to form oxygen gas. Since oxygen is diatomic (O₂), two O²⁻ ions lose a total of 4 electrons to produce one O₂ molecule:
  2O²⁻ → O₂ + 4e⁻
  So the equation at the anode is 2O²⁻ → O₂ + 4e⁻.

Why the other options are incorrect:

• B: The equation O²⁻ → O + 2e⁻ is wrong because oxygen exists as O₂, not single O atoms, and the equation is not balanced for the diatomic nature of oxygen.

• C: This option incorrectly states the product at the cathode is oxygen (it is actually aluminium), and the equation shown (Al³⁺ + 3e⁻ → Al) is actually the cathode reaction, not the anode reaction.

• D: This option also incorrectly states the product at the cathode is oxygen, and the equation Al³⁺ → Al + 3e⁻ shows loss of electrons (oxidation), but at the cathode reduction occurs (gain of electrons). This equation is also incorrect because it implies aluminium ions lose electrons, which does not happen.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.