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22:[["$","$L31",null,{"feature":"flashcards"}],"$L32",["$","$L33",null,{"href":"/igcse/chemistry/edexcel/electrolysis/flashcards","label":"Electrolysis","subjectCode":"igcse_chemistry","surface":"flashcards"}],["$","$L34",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L35",null,{"decks":[{"id":"bf3ad98b-bbf3-4de1-a3dc-4cc1250265f0","title":"Electrolysis Flashcards","topicId":"4a465fc7-0527-5ab2-81c8-93302fb111b9","topicSlug":"electrolysis","cards":[{"id":"65cccb18-409d-48e6-a0b6-e3562f5c03b3","cardType":"basic","points":2,"front":"What two conditions are required for a substance to conduct electricity?","back":"

Contains charged particles
The particles are free to move

","mcqOptions":[{"text":"

Contains charged particles
The particles are free to move

","correct":true,"explanation":"A substance conducts when it contains charged particles, such as ions or electrons, and those particles can move through the substance to carry electric charge."},{"text":"

Contains neutral particles
The particles are free to move

","correct":false,"explanation":"Neutral particles do not carry electric charge, even if they can move. The correct conditions are charged particles and particles that are free to move."},{"text":"

Contains charged particles
The particles are fixed in place

","correct":false,"explanation":"Fixed charged particles cannot travel through the substance, so they cannot carry a current. Conduction requires charged particles that are free to move."},{"text":"

Contains uncharged atoms
The particles are fixed in a lattice

","correct":false,"explanation":"Uncharged atoms do not carry charge, and particles fixed in a lattice cannot move through the substance. Both mobile charge and charged particles are required."}]},{"id":"ee2262db-5d3e-4919-8ba4-ea8a96964c33","cardType":"cloze","points":null,"front":"Covalent compounds do not conduct electricity because they are made of [ ] and have no mobile charged particles.","back":"Covalent compounds do not conduct electricity because they are made of neutral molecules and have no mobile charged particles.","mcqOptions":[{"text":"neutral molecules","correct":true,"explanation":"Covalent compounds consist of neutral molecules, so they have no overall charge to carry electricity. Their electrons are also held within bonds rather than moving freely."},{"text":"charged molecules","correct":false,"explanation":"Covalent compounds are made of neutral molecules, not charged molecules. Charged particles could carry electricity only if they were also able to move."},{"text":"free ions","correct":false,"explanation":"Free ions are characteristic of ionic substances in molten or aqueous states. Covalent compounds are usually made of neutral molecules, which do not carry charge."},{"text":"mobile electrons","correct":false,"explanation":"Mobile electrons can conduct electricity in metals, but covalent compounds are made of neutral molecules with electrons held in covalent bonds."}]},{"id":"aea0ca21-0e52-43c7-9f96-a77ee23a1df3","cardType":"cloze","points":null,"front":"Covalent compounds do not conduct electricity because they are made of neutral molecules and have no [ ].","back":"Covalent compounds do not conduct electricity because they are made of neutral molecules and have no mobile charged particles.","mcqOptions":[{"text":"mobile charged particles","correct":true,"explanation":"Covalent compounds contain neutral molecules and lack mobile charged particles. Without ions or electrons that can move through the substance, an electric current cannot flow."},{"text":"stationary charged particles","correct":false,"explanation":"Stationary charged particles cannot carry a current because they cannot move through the substance. The missing feature is mobile charged particles, not merely charge."},{"text":"neutral molecules","correct":false,"explanation":"Neutral molecules describe what covalent compounds are made of, but the second missing point is that they lack mobile charged particles."},{"text":"free-moving atoms","correct":false,"explanation":"Individual free-moving atoms are not the relevant charge carriers in covalent compounds. The correct phrase is mobile charged particles, such as ions or electrons."}]},{"id":"93badd6f-fd4a-47fc-9fc1-120e00905c62","cardType":"basic","points":2,"front":"Why do ionic compounds only conduct electricity when molten or aqueous?","back":"The ions are free to move; in solids, they are fixed in a lattice.","mcqOptions":[{"text":"The ions are free to move; in solids, they are fixed in a lattice.","correct":true,"explanation":"In molten or aqueous ionic compounds, the ions can move and carry charge. In a solid, the ions are held in fixed positions in a lattice."},{"text":"The electrons are free to move; in solids, they are fixed in a lattice.","correct":false,"explanation":"Ionic conduction is caused by moving ions, not free electrons. In solid ionic compounds, the ions are fixed in a lattice and cannot carry charge."},{"text":"The ions are fixed in place; in solids, they are free to move.","correct":false,"explanation":"This reverses the states: ions are free to move when molten or aqueous, but fixed in a lattice when solid, so solid ionic compounds do not conduct."},{"text":"The molecules are free to move; in solids, they are fixed in a lattice.","correct":false,"explanation":"Ionic compounds contain ions rather than molecules as their charge carriers. Their ions conduct when mobile and stop conducting when fixed in the solid lattice."}]},{"id":"62de90cd-ddf4-4b33-822b-d768646cdba0","cardType":"basic","points":1,"front":"Define electrolysis.","back":"The decomposition of an ionic compound using an electric current.","mcqOptions":[{"text":"The decomposition of an ionic compound using an electric current.","correct":true,"explanation":"Electrolysis uses an electric current to decompose an ionic compound into simpler substances, with ions moving to electrodes where chemical reactions occur."},{"text":"The formation of an ionic compound using an electric current.","correct":false,"explanation":"Electrolysis is decomposition, not formation. An electric current breaks an ionic compound into simpler substances at the electrodes."},{"text":"The decomposition of a covalent compound using an electric current.","correct":false,"explanation":"The standard definition refers to decomposing an ionic compound, whose mobile ions carry the current. Covalent compounds generally do not provide suitable ions for electrolysis."},{"text":"The decomposition of an ionic compound using an electric field.","correct":false,"explanation":"An electric field may influence charged particles, but the definition of electrolysis specifically states that an electric current decomposes the ionic compound."}]},{"id":"7cac258c-9bb0-4ffa-8c71-b0b92dabe855","cardType":"basic","points":2,"front":"To which electrode do cations and anions move?","back":"

Cations move to the cathode (negative)
Anions move to the anode (positive)

","mcqOptions":[{"text":"

Cations move to the cathode (negative)
Anions move to the anode (positive)

","correct":true,"explanation":"Cations are positively charged and are attracted to the negative cathode. Anions are negatively charged and are attracted to the positive anode."},{"text":"

Cations move to the anode (positive)
Anions move to the cathode (negative)

","correct":false,"explanation":"This reverses the attractions. Positive cations move to the negative cathode, while negative anions move to the positive anode."},{"text":"

Cations move to the cathode (positive)
Anions move to the anode (negative)

","correct":false,"explanation":"The electrode names are matched with the wrong charges here. In electrolysis, the cathode is negative and attracts cations; the anode is positive and attracts anions."},{"text":"

Cations move to the anode (negative)
Anions move to the cathode (positive)

","correct":false,"explanation":"Both the destinations and electrode charges are reversed. Cations move to the negative cathode, whereas anions move to the positive anode."}]},{"id":"df62f361-bd30-4908-9098-2e0841397a32","cardType":"cloze","points":null,"front":"In electrolysis, reduction occurs at the [ ] and oxidation occurs at the anode.","back":"In electrolysis, reduction occurs at the cathode and oxidation occurs at the anode.","mcqOptions":[{"text":"cathode","correct":true,"explanation":"Reduction is gain of electrons, and cations gain electrons at the cathode during electrolysis. The cathode is therefore the electrode where reduction occurs."},{"text":"anode","correct":false,"explanation":"The anode is where oxidation occurs, meaning loss of electrons. Reduction occurs at the cathode, so confusing the two electrodes reverses the redox processes."},{"text":"electrolyte","correct":false,"explanation":"The electrolyte contains mobile ions and conducts the current, but it is not an electrode. Reduction occurs at the cathode, where particles gain electrons."},{"text":"positive electrode","correct":false,"explanation":"In electrolysis, the positive electrode is the anode, where oxidation occurs. Reduction takes place at the negative electrode, called the cathode."}]},{"id":"4ea33139-a557-48c0-9730-9623d8590ab6","cardType":"cloze","points":null,"front":"In electrolysis, reduction occurs at the cathode and oxidation occurs at the [ ].","back":"In electrolysis, reduction occurs at the cathode and oxidation occurs at the anode.","mcqOptions":[{"text":"anode","correct":true,"explanation":"Oxidation is loss of electrons, and anions lose electrons at the anode during electrolysis. Therefore oxidation occurs at the anode, while reduction occurs at the cathode."},{"text":"cathode","correct":false,"explanation":"The cathode is where reduction occurs because particles gain electrons there. Oxidation, involving electron loss, occurs at the anode."},{"text":"electrolyte","correct":false,"explanation":"The electrolyte provides mobile ions for conduction but is not an electrode where oxidation is defined to occur. Oxidation takes place at the anode."},{"text":"negative electrode","correct":false,"explanation":"The negative electrode is the cathode in electrolysis, and reduction occurs there. Oxidation occurs at the positive electrode, the anode."}]},{"id":"55a68d0b-71fc-408f-8059-72c80f6999ff","cardType":"basic","points":2,"front":"What are the observed products for the electrolysis of molten lead(II) bromide ($PbBr_2$)?","back":"

Cathode: grey/silvery lead metal
Anode: brown bromine vapour

","mcqOptions":[{"text":"

Cathode: grey/silvery lead metal
Anode: brown bromine vapour

","correct":true,"explanation":"Lead(II) ions gain electrons at the cathode to form grey/silvery lead metal, while bromide ions lose electrons at the anode to form brown bromine vapour."},{"text":"

Cathode: brown bromine vapour
Anode: grey/silvery lead metal

","correct":false,"explanation":"The products have been placed at the wrong electrodes. Positive lead(II) ions are reduced to lead at the cathode, while negative bromide ions are oxidised to bromine at the anode."},{"text":"

Cathode: grey/silvery lead metal
Anode: yellow-green chlorine gas

","correct":false,"explanation":"Chlorine is formed from chloride ions, not bromide ions. Molten lead(II) bromide contains bromide ions, which are oxidised at the anode to produce brown bromine vapour."},{"text":"

Cathode: hydrogen gas
Anode: brown bromine vapour

","correct":false,"explanation":"Hydrogen forms at the cathode in some aqueous solutions, but molten lead(II) bromide contains no water. Lead(II) ions are reduced instead, producing grey/silvery lead metal."}]},{"id":"0a862132-4f7f-422a-b971-efc94cab8915","cardType":"basic","points":2,"front":"Write the half-equation for the formation of bromine at the anode.","back":"$$2Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + 2e^-$","mcqOptions":[{"text":"$$2Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + 2e^-$","correct":true,"explanation":"Two bromide ions each lose one electron at the anode, forming one bromine molecule and two electrons. The atoms and total charge are balanced on both sides."},{"text":"$$Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + e^-$","correct":false,"explanation":"A bromine molecule contains two bromine atoms, so two bromide ions are needed. The correct half-equation is $2Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + 2e^-$."},{"text":"$$2Br^-\\text{(l)} + 2e^- \\rightarrow Br_2\\text{(g)}$","correct":false,"explanation":"Electrons have been placed on the wrong side, representing reduction rather than oxidation. Bromide ions lose electrons at the anode: $2Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + 2e^-$."},{"text":"$$Br_2\\text{(l)} + 2e^- \\rightarrow 2Br^-\\text{(l)}$","correct":false,"explanation":"This is the reverse reduction of bromine to bromide ions, not bromide oxidation. At the anode, bromide ions lose electrons to form bromine: $2Br^-\\text{(l)} \\rightarrow Br_2\\text{(g)} + 2e^-$."}]},{"id":"b5b0e63b-37c3-4307-88d7-6e23057e5669","cardType":"basic","points":1,"front":"How do you identify oxidation from a half-equation?","back":"Electrons are on the right-hand side (electrons are lost).","mcqOptions":[{"text":"Electrons are on the right-hand side (electrons are lost).","correct":true,"explanation":"Oxidation is the loss of electrons. Therefore, in an oxidation half-equation, electrons appear among the products on the right-hand side."},{"text":"Electrons are on the left-hand side (electrons are gained).","correct":false,"explanation":"Electrons on the left-hand side show reduction, because the species gains electrons. Oxidation is electron loss, so electrons must appear on the right-hand side."},{"text":"Electrons are on the right-hand side (electrons are gained).","correct":false,"explanation":"The position is correct, but the process has been misidentified. Electrons on the right-hand side are products, showing that they have been lost during oxidation."},{"text":"Electrons are on the left-hand side (electrons are lost).","correct":false,"explanation":"This combines the position for reduction with the description of oxidation. Electron loss places electrons on the right-hand side, whereas electron gain places them on the left."}]},{"id":"052da57e-e49a-47d3-8fba-11a108d31946","cardType":"basic","points":1,"front":"In aqueous electrolysis, what determines if a metal forms at the cathode?","back":"The metal forms if it is less reactive than hydrogen.","mcqOptions":[{"text":"The metal forms if it is less reactive than hydrogen.","correct":true,"explanation":"At the cathode, either metal ions or hydrogen ions can be reduced. A metal less reactive than hydrogen is preferentially discharged, so the metal forms."},{"text":"The metal forms if it is more reactive than hydrogen.","correct":false,"explanation":"More reactive metals are harder to reduce than hydrogen ions, so hydrogen forms instead at the cathode. The metal is discharged when it is less reactive than hydrogen."},{"text":"The metal forms if it is more reactive than oxygen.","correct":false,"explanation":"Cathode competition is between metal ions and hydrogen ions, not oxygen. The relevant rule is that a metal forms when it is less reactive than hydrogen."},{"text":"The metal forms if it is less reactive than oxygen.","correct":false,"explanation":"Oxygen is produced at the anode and does not determine cathode discharge. The metal forms at the cathode when it is less reactive than hydrogen."}]},{"id":"ad75ee32-a7ff-41ff-80bc-f86a8e9ebb6b","cardType":"cloze","points":null,"front":"During aqueous electrolysis, if the metal is more reactive than hydrogen, [ ] forms at the cathode.","back":"During aqueous electrolysis, if the metal is more reactive than hydrogen, hydrogen gas forms at the cathode.","mcqOptions":[{"text":"hydrogen gas","correct":true,"explanation":"When the metal is more reactive than hydrogen, its ions are not discharged at the cathode. Hydrogen ions from water are reduced instead, producing hydrogen gas."},{"text":"oxygen gas","correct":false,"explanation":"Oxygen is usually formed at the anode when no halide ions are present, not at the cathode. At this cathode, hydrogen gas forms because the metal is more reactive than hydrogen."},{"text":"the metal","correct":false,"explanation":"A metal more reactive than hydrogen is not discharged at the cathode in aqueous electrolysis. Water provides hydrogen ions, which are reduced to form hydrogen gas instead."},{"text":"a halogen gas","correct":false,"explanation":"Halogen formation depends on halide ions being oxidised at the anode. At the cathode, a metal more reactive than hydrogen leads to hydrogen gas formation."}]},{"id":"f66de047-a348-4a66-87c6-e340d773bdcf","cardType":"cloze","points":null,"front":"During aqueous electrolysis, if the metal is more reactive than hydrogen, hydrogen gas forms at the [ ].","back":"During aqueous electrolysis, if the metal is more reactive than hydrogen, hydrogen gas forms at the cathode.","mcqOptions":[{"text":"cathode","correct":true,"explanation":"The cathode is the negative electrode in electrolysis and is where reduction occurs. Hydrogen ions gain electrons there, so hydrogen gas forms at the cathode."},{"text":"anode","correct":false,"explanation":"The anode is where oxidation occurs and where halide ions or hydroxide ions may form products. Hydrogen gas is produced by reduction at the cathode."},{"text":"electrolyte","correct":false,"explanation":"The electrolyte is the liquid or molten substance containing mobile ions, not an electrode. Hydrogen gas forms at the cathode, where hydrogen ions gain electrons."},{"text":"positive electrode","correct":false,"explanation":"In electrolysis, the positive electrode is the anode, where oxidation takes place. Hydrogen gas forms at the negative electrode, called the cathode."}]},{"id":"ef1fa76b-caf1-43a2-bf13-11fbae7a8f1d","cardType":"basic","points":2,"front":"What is the rule for products at the anode in aqueous electrolysis?","back":"

If halide ions are present, the halogen forms
If no halide is present, oxygen forms

","mcqOptions":[{"text":"

If halide ions are present, the halogen forms
If no halide is present, oxygen forms

","correct":true,"explanation":"At the anode, halide ions are preferentially oxidised to halogens when present. Without halide ions, hydroxide ions from water are oxidised and oxygen forms."},{"text":"

If halide ions are present, the oxygen forms
If no halide is present, halogen forms

","correct":false,"explanation":"The products have been reversed. Halide ions are oxidised to halogens at the anode, whereas oxygen forms when no halide ions are present."},{"text":"

If metal ions are present, the metal forms
If no metal is present, hydrogen forms

","correct":false,"explanation":"This describes neither the anode rule nor the competing anions. At the anode, halide ions produce halogens; if no halide is present, oxygen forms."},{"text":"

If halide ions are present, the hydrogen forms
If no halide is present, oxygen forms

","correct":false,"explanation":"Hydrogen is a cathode product formed when suitable metal ions are not discharged. At the anode, halide ions are oxidised to form the corresponding halogen."}]},{"id":"bc2b473b-3ba5-422a-abde-b3357825a538","cardType":"basic","points":2,"front":"Identify the cathode and anode products for sodium chloride solution ($NaCl(aq)$).","back":"

Cathode: hydrogen
Anode: chlorine

","mcqOptions":[{"text":"

Cathode: hydrogen
Anode: chlorine

","correct":true,"explanation":"In aqueous sodium chloride, sodium is more reactive than hydrogen, so hydrogen forms at the cathode. Chloride ions are oxidised at the anode to chlorine."},{"text":"

Cathode: sodium
Anode: chlorine

","correct":false,"explanation":"Sodium is more reactive than hydrogen and is not discharged from aqueous solution. Hydrogen forms at the cathode, while chloride ions still produce chlorine at the anode."},{"text":"

Cathode: hydrogen
Anode: oxygen

","correct":false,"explanation":"Although oxygen can form at the anode when no halide is present, sodium chloride contains chloride ions. Chloride is oxidised preferentially, producing chlorine."},{"text":"

Cathode: sodium
Anode: oxygen

","correct":false,"explanation":"Both products reflect common reversals or omissions. In aqueous sodium chloride, sodium is not discharged because it is too reactive, and chloride ions form chlorine at the anode."}]},{"id":"f3a380a6-5b25-4ff9-b052-6a0a2a1ec109","cardType":"basic","points":2,"front":"Why does the solution remain alkaline after the electrolysis of $NaCl(aq)$?","back":"$$Na^+$ and $OH^-$ ions are left in solution, forming sodium hydroxide.","mcqOptions":[{"text":"$$Na^+$ and $OH^-$ ions are left in solution, forming sodium hydroxide.","correct":true,"explanation":"Sodium ions are not discharged at the cathode, while water produces hydroxide ions. These ions remain together in solution as sodium hydroxide, so the solution stays alkaline."},{"text":"$$H^+$ and $Cl^-$ ions are left in solution, forming hydrochloric acid.","correct":false,"explanation":"This reverses the effect of electrolysis: hydrogen ions are removed at the cathode and chloride ions are discharged at the anode. The correct result is sodium hydroxide from $Na^+$ and $OH^-$ ions."},{"text":"$$Na^+$ and $Cl^-$ ions are left in solution, forming sodium chloride.","correct":false,"explanation":"Chloride ions are discharged at the anode, so they are not left in solution. Water supplies hydroxide ions, leaving $Na^+$ and $OH^-$ to form sodium hydroxide."},{"text":"$$H^+$ and $OH^-$ ions are left in solution, forming water.","correct":false,"explanation":"Hydrogen ions are discharged at the cathode rather than remaining in solution. The ions left are $Na^+$ and $OH^-$, which form sodium hydroxide and keep the solution alkaline."}]},{"id":"3681b2ef-6f7f-444f-953e-8f8c556f112d","cardType":"basic","points":2,"front":"What are the products of the electrolysis of dilute sulfuric acid?","back":"

Cathode: hydrogen
Anode: oxygen

","mcqOptions":[{"text":"

Cathode: hydrogen
Anode: oxygen

","correct":true,"explanation":"In dilute sulfuric acid, hydrogen ions are reduced at the cathode to form hydrogen, while hydroxide ions from water are oxidised at the anode to form oxygen."},{"text":"

Cathode: oxygen
Anode: hydrogen

","correct":false,"explanation":"The electrodes have been reversed. Reduction occurs at the cathode and produces hydrogen, whereas oxidation at the anode produces oxygen."},{"text":"

Cathode: sulfur
Anode: oxygen

","correct":false,"explanation":"Sulfur is not present as a discharged ion in dilute sulfuric acid. Hydrogen forms at the cathode, while oxygen forms at the anode from water or hydroxide ions."},{"text":"

Cathode: hydrogen
Anode: sulfur dioxide

","correct":false,"explanation":"Sulfur dioxide is not produced during electrolysis of dilute sulfuric acid. Water is decomposed overall, giving hydrogen at the cathode and oxygen at the anode."}]},{"id":"4920a5b6-b3fb-4982-b997-641de2648db0","cardType":"basic","points":2,"front":"What is the half-equation for the discharge of hydroxide ions at the anode?","back":"$$4OH^-\\text{(aq)} \\rightarrow O_2\\text{(g)} + 2H_2O\\text{(l)} + 4e^-$","mcqOptions":[{"text":"$$4OH^-\\text{(aq)} \\rightarrow O_2\\text{(g)} + 2H_2O\\text{(l)} + 4e^-$","correct":true,"explanation":"At the anode, hydroxide ions are oxidised, losing four electrons. The atoms balance to form oxygen gas and water, giving the correctly balanced half-equation."},{"text":"$$4OH^-\\text{(aq)} + 4e^- \\rightarrow O_2\t\\text{(g)} + 2H_2O\t\\text{(l)}$","correct":false,"explanation":"Electrons are on the wrong side, showing reduction rather than oxidation. At the anode hydroxide ions lose electrons, so the correct equation has $4e^-$ among the products."},{"text":"$$2OH^-\\text{(aq)} \\rightarrow O_2\t\\text{(g)} + H_2O\t\\text{(l)} + 2e^-$","correct":false,"explanation":"The coefficients do not conserve oxygen, hydrogen, and charge: two hydroxide ions cannot produce one oxygen molecule and one water molecule. Four hydroxide ions and four electrons are required."},{"text":"$$4OH^-\\text{(aq)} \\rightarrow 2O_2\t\\text{(g)} + 2H_2O\t\\text{(l)} + 4e^-$","correct":false,"explanation":"The oxygen coefficient has been doubled without changing the hydroxide coefficient, so oxygen atoms are not balanced. Four hydroxide ions produce one $O_2$ molecule, not two."}]},{"id":"b33cc822-ab49-47cc-bfd0-cad3c9a2f278","cardType":"basic","points":2,"front":"Predict the observations at the cathode and anode for copper(II) sulfate solution.","back":"

Cathode: pink-brown metal coating
Anode: bubbles of colourless gas

","mcqOptions":[{"text":"

Cathode: pink-brown metal coating
Anode: bubbles of colourless gas

","correct":true,"explanation":"Copper ions are reduced at the cathode, forming a pink-brown copper coating. Hydroxide ions or water are oxidised at the anode, producing bubbles of colourless oxygen gas."},{"text":"

Cathode: bubbles of colourless gas
Anode: pink-brown metal coating

","correct":false,"explanation":"The electrode products have been reversed. Copper ions gain electrons and coat the cathode, while oxygen gas is formed as bubbles at the anode."},{"text":"

Cathode: black metal coating
Anode: bubbles of colourless gas

","correct":false,"explanation":"Copper metal is pink-brown, not black; black deposits suggest a different substance or an observation error. Oxygen bubbles still form at the anode, but the cathode coating is copper."},{"text":"

Cathode: pink-brown metal coating
Anode: bubbles of green gas

","correct":false,"explanation":"The cathode observation is correct, but the anode gas is oxygen, which is colourless. Green gas would incorrectly suggest chlorine, which is not produced from copper(II) sulfate solution."}]},{"id":"929b9d61-598b-4dde-a85e-46b151b46f0f","cardType":"cloze","points":null,"front":"When electrolysing copper(II) sulfate with inert electrodes, the blue colour becomes [ ] because $Cu^{2+}$ ions are removed.","back":"When electrolysing copper(II) sulfate with inert electrodes, the blue colour becomes paler because $Cu^{2+}$ ions are removed.","mcqOptions":[{"text":"paler","correct":true,"explanation":"The blue colour is caused by hydrated copper(II) ions in solution. As $Cu^{2+}$ ions are discharged at the cathode, their concentration falls and the solution becomes paler."},{"text":"darker","correct":false,"explanation":"Removing $Cu^{2+}$ ions decreases the concentration of the blue species, so the colour becomes paler rather than darker. A darker colour would require more copper(II) ions."},{"text":"green","correct":false,"explanation":"Copper(II) sulfate solution is blue, and removing $Cu^{2+}$ ions reduces the colour intensity. Electrolysis with inert electrodes does not turn the solution green."},{"text":"colourless immediately","correct":false,"explanation":"The blue colour fades as copper(II) ions are removed, but it does not usually disappear immediately. The expected observation is that the solution becomes paler."}]},{"id":"e03e63cb-b102-437b-a01d-2462db9d0e87","cardType":"cloze","points":null,"front":"When electrolysing copper(II) sulfate with inert electrodes, the blue colour becomes paler because [ ] are removed.","back":"When electrolysing copper(II) sulfate with inert electrodes, the blue colour becomes paler because $Cu^{2+}$ ions are removed.","mcqOptions":[{"text":"$$Cu^{2+}$ ions","correct":true,"explanation":"Copper(II) ions give copper(II) sulfate solution its blue colour. They are attracted to the cathode, gain electrons, and form copper metal, so their concentration decreases."},{"text":"$$SO_4^{2-}$ ions","correct":false,"explanation":"Sulfate ions are not the ions removed at the cathode in this experiment. Copper(II) ions are discharged and deposited as copper, causing the blue colour to become paler."},{"text":"$$H^+$ ions","correct":false,"explanation":"Hydrogen ions are not responsible for the blue colour and are not the main ions discharged from copper(II) sulfate solution. The removed ions are $Cu^{2+}$ ions."},{"text":"$$OH^-$ ions","correct":false,"explanation":"Hydroxide ions may be involved in oxygen formation at the anode, but they do not cause the blue colour or explain its fading. Copper(II) ions are removed at the cathode."}]},{"id":"d2f27465-09a1-4563-8c26-31355baea2ef","cardType":"basic","points":1,"front":"How do you test for chlorine gas in an electrolysis experiment?","back":"It bleaches damp blue litmus paper.","mcqOptions":[{"text":"It bleaches damp blue litmus paper.","correct":true,"explanation":"Chlorine reacts with water on damp blue litmus paper to form acidic and bleaching substances. The paper turns red briefly and is then bleached white."},{"text":"It relights a glowing splint.","correct":false,"explanation":"Relighting a glowing splint tests for oxygen, not chlorine. Chlorine is identified because it bleaches damp blue litmus paper."},{"text":"It makes damp red litmus paper blue.","correct":false,"explanation":"Turning red litmus blue indicates an alkaline gas, which chlorine is not. Chlorine is tested using damp blue litmus paper, which it bleaches."},{"text":"It forms a white precipitate with limewater.","correct":false,"explanation":"A white precipitate with limewater is associated with carbon dioxide, not chlorine. Chlorine is identified by its bleaching action on damp blue litmus paper."}]},{"id":"8c4164ce-3095-450a-87e4-6fcc6b70bcad","cardType":"basic","points":1,"front":"In the electrolysis practical, why must electrodes not touch each other?","back":"To prevent a short-circuit (current would bypass the electrolyte).","mcqOptions":[{"text":"To prevent a short-circuit (current would bypass the electrolyte).","correct":true,"explanation":"If the electrodes touch, current can pass directly between them through the metal instead of travelling through the electrolyte. This creates a short-circuit and may damage the apparatus."},{"text":"To prevent the electrolyte from evaporating.","correct":false,"explanation":"Electrode contact does not cause the electrolyte to evaporate. It allows current to bypass the solution, creating a short-circuit, so the electrodes must remain separate."},{"text":"To prevent the electrodes from melting.","correct":false,"explanation":"The main danger is not simply that the electrodes melt. Touching provides a low-resistance path for current, causing a short-circuit and potentially overheating the circuit."},{"text":"To prevent the solution from becoming neutral.","correct":false,"explanation":"Electrode contact does not directly neutralise the solution. It creates a short-circuit because current bypasses the electrolyte, disrupting the electrolysis and potentially damaging the equipment."}]}]}],"examCountdown":{"examDate":"2027-05-19","paperLabel":"Paper 1"},"heading":"Electrolysis","title":"Electrolysis Flashcards","topbar":{"backHref":"/igcse/chemistry/edexcel/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"4a465fc7-0527-5ab2-81c8-93302fb111b9","walled":"$L37"}],"$L38","$L39"]