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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/edexcel/oxidation-states-and-redox-equations/flashcards","label":"Oxidation states and redox equations","subjectCode":"a_level_chemistry","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"fb0f3852-9fc9-4837-ae6f-35566caaf281","title":"Oxidation states and redox equations Flashcards","topicId":"7b5b39ad-4453-565b-a2b1-384b797b3b9c","topicSlug":"oxidation-states-and-redox-equations","cards":[{"id":"11f3381e-fdf5-43a9-b0b7-da03e2316150","cardType":"basic","points":1,"front":"How is oxidation defined in terms of electron transfer?","back":"The loss of electrons.","mcqOptions":[{"text":"The loss of electrons.","correct":true,"explanation":"Oxidation occurs when a species loses one or more electrons, causing its oxidation state to increase. The mnemonic OIL RIG links oxidation with electron loss."},{"text":"The gain of electrons.","correct":false,"explanation":"Gaining electrons defines reduction, not oxidation, and causes oxidation state to decrease. Oxidation is the loss of electrons, which raises oxidation state."},{"text":"The gain of protons.","correct":false,"explanation":"Proton transfer describes acid-base reactions rather than oxidation in electron-transfer terms. Oxidation is specifically the loss of electrons from a species."},{"text":"The loss of neutrons.","correct":false,"explanation":"Neutron loss is a nuclear process and does not define a redox change. Oxidation is the loss of electrons, normally producing an increase in oxidation state."}]},{"id":"41d10fb3-7701-4fae-b762-1629c46dcc7e","cardType":"basic","points":1,"front":"How is reduction defined in terms of oxidation state?","back":"A decrease in oxidation state.","mcqOptions":[{"text":"A decrease in oxidation state.","correct":true,"explanation":"Reduction involves an atom or ion gaining electrons, making its oxidation state less positive or more negative. It is therefore identified by a decrease in oxidation state."},{"text":"An increase in oxidation state.","correct":false,"explanation":"An increase in oxidation state indicates oxidation because electrons have been lost. Reduction involves electron gain and therefore a decrease in oxidation state."},{"text":"No change in oxidation state.","correct":false,"explanation":"An unchanged oxidation state means that the species has undergone no redox change. Reduction requires electron gain, shown by a decrease in oxidation state."},{"text":"A decrease in electrical charge.","correct":false,"explanation":"Formal charge and oxidation state are different bookkeeping concepts, so a charge decrease does not by itself define reduction. Reduction is a decrease in oxidation state."}]},{"id":"c656e101-125f-440b-9f11-581928c41c37","cardType":"cloze","points":null,"front":"In chemical notation, charges are written with the sign [ ] the number (e.g. $2+$), while oxidation states are written with the sign before the number (e.g. $+2$).","back":"In chemical notation, charges are written with the sign after the number (e.g. $2+$), while oxidation states are written with the sign before the number (e.g. $+2$).","mcqOptions":[{"text":"after","correct":true,"explanation":"An ionic charge is conventionally written with its magnitude first and sign afterwards, as in $2+$. An oxidation state reverses this order, as in $+2$."},{"text":"before","correct":false,"explanation":"Placing the sign first follows the convention for oxidation states, such as $+2$, rather than ionic charges. For a charge, the sign comes after the number, as in $2+$."},{"text":"above","correct":false,"explanation":"Superscript placement puts the complete charge above the formula, but it does not describe the order within that charge. The sign follows the number, as in $2+$."},{"text":"below","correct":false,"explanation":"Subscripts usually indicate numbers of atoms, not the order of an ionic charge. Within a charge, the numerical magnitude is followed by its sign, as in $2+$."}]},{"id":"b3b9bf47-9a61-4ecc-b993-597849ee5a3a","cardType":"cloze","points":null,"front":"In chemical notation, charges are written with the sign after the number (e.g. $2+$), while oxidation states are written with the sign [ ] the number (e.g. $+2$).","back":"In chemical notation, charges are written with the sign after the number (e.g. $2+$), while oxidation states are written with the sign before the number (e.g. $+2$).","mcqOptions":[{"text":"before","correct":true,"explanation":"Oxidation states place the sign first and then the numerical value, as in $+2$ or $-1$. Ionic charges conventionally use the opposite order."},{"text":"after","correct":false,"explanation":"Writing the sign after the number gives ionic charge notation, such as $2+$, rather than oxidation-state notation. Oxidation states place the sign before the number."},{"text":"above","correct":false,"explanation":"An oxidation state may be displayed above an element symbol, but the blank concerns order within the value. Its sign comes before the number, as in $+2$."},{"text":"below","correct":false,"explanation":"Placing information below an element usually indicates a subscript rather than an oxidation state. In oxidation-state notation, the sign is written before the number."}]},{"id":"5cffb16c-dcbd-472f-a30e-64f484e0f5e5","cardType":"basic","points":1,"front":"What is the oxidation state of any uncombined element (e.g. $Mg$, $Cl_2$)?","back":"Zero.","mcqOptions":[{"text":"Zero.","correct":true,"explanation":"Atoms in an uncombined element have neither gained nor lost electrons relative to that element’s standard reference form. Each atom is therefore assigned oxidation state zero."},{"text":"Positive one.","correct":false,"explanation":"Confusing elemental state with a common ionic charge leads to this value. Every uncombined element, including a diatomic molecule such as $Cl_2$, has oxidation state zero."},{"text":"Negative one.","correct":false,"explanation":"A negative value may be common for some elements in compounds, but an uncombined element has not been assigned electrons from another element. Its oxidation state is zero."},{"text":"Its group number.","correct":false,"explanation":"An element’s group number can help predict some compound oxidation states, but it does not apply to the uncombined substance. All uncombined elements have oxidation state zero."}]},{"id":"ee08b293-f95e-43af-9f1d-63ab99654b22","cardType":"basic","points":1,"front":"What must the sum of oxidation states equal in a polyatomic ion?","back":"The overall charge of the ion.","mcqOptions":[{"text":"The overall charge of the ion.","correct":true,"explanation":"Adding the oxidation states of every atom in a polyatomic ion must reproduce its net ionic charge. For example, the values in sulfate sum to $-2$."},{"text":"The neutral value of the ion.","correct":false,"explanation":"A sum of zero applies to an electrically neutral compound, not generally to a polyatomic ion. For an ion, the oxidation-state sum equals its overall charge."},{"text":"The average oxidation state of the ion.","correct":false,"explanation":"Oxidation states are summed rather than averaged when checking an ion’s formula. Their total must equal the ion’s overall charge, accounting for every atom present."},{"text":"The central atom charge of the ion.","correct":false,"explanation":"The central atom’s oxidation state is only one contribution to the total. The sum for all atoms must equal the polyatomic ion’s overall charge."}]},{"id":"4354e8bd-e852-43c0-973a-2346b8181a94","cardType":"cloze","points":null,"front":"Oxygen usually has an oxidation state of $-2$, except in peroxides where it is $\\text{[...]}$ and in $OF_2$ where it is $+2$.","back":"Oxygen usually has an oxidation state of $-2$, except in peroxides where it is $\\boxed{-1}$ and in $OF_2$ where it is $+2$.","mcqOptions":[{"text":"$$-1$","correct":true,"explanation":"Peroxides contain an O-O bond and the peroxide ion has charge $-2$ shared between two equivalent oxygen atoms. Each oxygen therefore has oxidation state $-1$."},{"text":"$$-2$","correct":false,"explanation":"Using oxygen’s usual value overlooks the peroxide exception created by the O-O bond. In peroxides, the total $-2$ is shared between two oxygens, giving each $-1$."},{"text":"$$+1$","correct":false,"explanation":"Reversing the sign incorrectly treats oxygen as the less electronegative element. Oxygen in a peroxide has oxidation state $-1$, obtained from the peroxide group’s total charge of $-2$."},{"text":"$$0$","correct":false,"explanation":"Zero is reserved for oxygen in its uncombined elemental form, such as $O_2$. In a peroxide compound, each oxygen is assigned oxidation state $-1$."}]},{"id":"dc76789d-d5b3-4df8-8982-fb910e9e3758","cardType":"cloze","points":null,"front":"Oxygen usually has an oxidation state of $-2$, except in peroxides where it is $-1$ and in $OF_2$ where it is $\\text{[...]}$.","back":"Oxygen usually has an oxidation state of $-2$, except in peroxides where it is $-1$ and in $OF_2$ where it is $\\boxed{+2}$.","mcqOptions":[{"text":"$$+2$","correct":true,"explanation":"Fluorine always has oxidation state $-1$ in compounds. Its total in $OF_2$ is therefore $-2$, so oxygen must be $+2$ to make the molecule neutral."},{"text":"$$-2$","correct":false,"explanation":"Applying oxygen’s usual value ignores that fluorine is more electronegative and must be $-1$. Two fluorines contribute $-2$, requiring oxygen to be $+2$."},{"text":"$$+1$","correct":false,"explanation":"This halves fluorine’s combined contribution instead of including both fluorine atoms. Since two fluorines total $-2$, oxygen must have oxidation state $+2$ in neutral $OF_2$."},{"text":"$$0$","correct":false,"explanation":"Zero would apply to uncombined elemental oxygen, not oxygen bonded to fluorine. The two fluorines contribute $-2$ altogether, so oxygen is assigned $+2$."}]},{"id":"0603c7f4-ded9-4ce2-a1e3-173b8ccfe9df","cardType":"basic","points":1,"front":"When does hydrogen have an oxidation state of $-1$?","back":"In metal hydrides (e.g. $NaH$).","mcqOptions":[{"text":"In metal hydrides (e.g. $NaH$).","correct":true,"explanation":"Highly electropositive metals transfer electron density to hydrogen, forming hydride ions. Hydrogen therefore has an oxidation state of $-1$ in metal hydrides such as $NaH$."},{"text":"In covalent hydrides (e.g. $HCl$).","correct":false,"explanation":"In $HCl$, chlorine is more electronegative than hydrogen, so hydrogen has an oxidation state of $+1$. Hydrogen is $-1$ in metal hydrides such as $NaH$."},{"text":"In elemental hydrogen (e.g. $H_2$).","correct":false,"explanation":"Any element in its uncombined elemental form has an oxidation state of $0$, so hydrogen is $0$ in $H_2$. It is $-1$ in metal hydrides."},{"text":"In aqueous acids (e.g. $HNO_3$).","correct":false,"explanation":"Hydrogen normally has an oxidation state of $+1$ in acids such as $HNO_3$. The exceptional oxidation state of $-1$ occurs in metal hydrides such as $NaH$."}]},{"id":"debe2cb0-f07b-460d-888d-42c7b8c59b15","cardType":"basic","points":1,"front":"What is the oxidation state of chlorine in the chlorate(I) ion ($ClO^-$)?","back":"$+1$.","mcqOptions":[{"text":"$+1$.","correct":true,"explanation":"Oxygen is $-2$, and the ion has an overall charge of $-1$. Therefore $x+(-2)=-1$, giving chlorine an oxidation state of $+1$."},{"text":"$-1$.","correct":false,"explanation":"This mistakes the ion’s overall charge for chlorine’s oxidation state. Since oxygen is $-2$, chlorine must be $+1$ for the total to equal $-1$."},{"text":"$+3$.","correct":false,"explanation":"This value would not produce the ion’s $-1$ charge: $+3-2=+1$. Solving $x-2=-1$ shows that chlorine has oxidation state $+1$."},{"text":"$+5$.","correct":false,"explanation":"This confuses chlorate(I) with a chlorine oxyanion containing more oxygen. In $ClO^-$, one oxygen contributes $-2$, so chlorine must be $+1$."}]},{"id":"05523de2-8a08-4562-9fc1-86c64bc788a8","cardType":"basic","points":2,"front":"Define an oxidising agent in terms of electron transfer and its own oxidation state change.","back":"

Gains electrons
Is reduced (oxidation state decreases)

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

Gains electrons
Is reduced (oxidation state decreases)

","correct":true,"explanation":"An oxidising agent accepts electrons from another species, causing that species to be oxidised. By gaining electrons, the oxidising agent is itself reduced and its oxidation state decreases."},{"text":"

Loses electrons
Is oxidised (oxidation state increases)

","correct":false,"explanation":"These are the changes undergone by a reducing agent, not an oxidising agent. An oxidising agent gains electrons and is reduced, so its oxidation state decreases."},{"text":"

Gains electrons
Is oxidised (oxidation state increases)

","correct":false,"explanation":"Gaining electrons corresponds to reduction, not oxidation, and normally decreases oxidation state. An oxidising agent gains electrons and is therefore reduced during the reaction."},{"text":"

Loses electrons
Is reduced (oxidation state decreases)

","correct":false,"explanation":"Losing electrons is oxidation and causes oxidation state to increase, so these statements are inconsistent. An oxidising agent instead gains electrons and is reduced."}]},{"id":"38a89d2d-e05a-4e3f-a17b-6c5ee275f06e","cardType":"basic","points":2,"front":"Define a reducing agent in terms of electron transfer and its own oxidation state change.","back":"

Loses electrons
Is oxidised (oxidation state increases)

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

Loses electrons
Is oxidised (oxidation state increases)

","correct":true,"explanation":"A reducing agent donates electrons to another species, causing that species to be reduced. The reducing agent therefore loses electrons, is oxidised, and increases its oxidation state."},{"text":"

Gains electrons
Is reduced (oxidation state decreases)

","correct":false,"explanation":"These are the changes undergone by an oxidising agent, not a reducing agent. A reducing agent loses electrons and is oxidised, so its oxidation state increases."},{"text":"

Loses electrons
Is reduced (oxidation state decreases)

","correct":false,"explanation":"Losing electrons constitutes oxidation rather than reduction and causes oxidation state to rise. A reducing agent loses electrons and is therefore oxidised during the reaction."},{"text":"

Gains electrons
Is oxidised (oxidation state increases)

","correct":false,"explanation":"Gaining electrons causes reduction and a decrease in oxidation state, so these statements conflict. A reducing agent instead loses electrons and is oxidised."}]},{"id":"8d1b4f7a-8460-437e-8fd4-1a64d0e3bfb7","cardType":"basic","points":1,"front":"What is the oxidation state of sulfur in $SO_4^{2-}$?","back":"$+6$.","mcqOptions":[{"text":"$+6$.","correct":true,"explanation":"Four oxygen atoms contribute a total oxidation state of $-8$. Since the sulfate ion has charge $-2$, sulfur must be $+6$ because $+6-8=-2$."},{"text":"$+4$.","correct":false,"explanation":"This often results from overlooking the ion’s overall charge or miscounting oxygen. Four oxygens total $-8$, so sulfur must be $+6$ to give $-2$ overall."},{"text":"$+2$.","correct":false,"explanation":"This incorrectly treats the ion’s $2-$ charge as the magnitude of sulfur’s oxidation state. Balancing oxygen’s total of $-8$ requires sulfur to be $+6$."},{"text":"$-2$.","correct":false,"explanation":"This copies the sulfate ion’s overall charge onto sulfur. Oxidation states must sum to $-2$; with oxygen contributing $-8$, sulfur is $+6$."}]},{"id":"6d109c17-42a1-4903-b03d-d5c932cbda0a","cardType":"cloze","points":null,"front":"To balance a half-equation in acidic conditions, oxygen is balanced using $\\text{[...]}$ and hydrogen is balanced using $H^+$.","back":"To balance a half-equation in acidic conditions, oxygen is balanced using $\\boxed{H_2O}$ and hydrogen is balanced using $H^+$.","mcqOptions":[{"text":"$$H_2O$","correct":true,"explanation":"In acidic half-equations, water molecules are added to the side lacking oxygen atoms. Hydrogen atoms introduced this way can then be balanced separately using $H^+$ ions."},{"text":"$$O_2$","correct":false,"explanation":"Adding molecular oxygen is not the standard half-equation method and can alter the redox process being represented. In acidic conditions, oxygen atoms are balanced using $H_2O$."},{"text":"$$H^+$","correct":false,"explanation":"Hydrogen ions are used to balance hydrogen atoms, not oxygen atoms. Oxygen is first balanced by adding $H_2O$, after which $H^+$ balances hydrogen."},{"text":"$$OH^-$","correct":false,"explanation":"Hydroxide ions are associated with balancing equations in alkaline conditions, not the initial acidic method. In acidic conditions, oxygen is balanced by adding $H_2O$."}]},{"id":"0dc83557-ed4b-4829-9b4e-ef998e8626e1","cardType":"cloze","points":null,"front":"To balance a half-equation in acidic conditions, oxygen is balanced using $H_2O$ and hydrogen is balanced using $\\text{[...]}$.","back":"To balance a half-equation in acidic conditions, oxygen is balanced using $H_2O$ and hydrogen is balanced using $\\boxed{H^+}$.","mcqOptions":[{"text":"$$H^+$","correct":true,"explanation":"In acidic conditions, hydrogen atoms are balanced by adding $H^+$ ions to the side with fewer hydrogens. Charge is then balanced separately using electrons."},{"text":"$$OH^-$","correct":false,"explanation":"Hydroxide ions are used when converting or balancing half-equations under alkaline conditions. In acidic conditions, hydrogen atoms are balanced by adding $H^+$ ions."},{"text":"$$H_2O$","correct":false,"explanation":"Water is used to balance oxygen atoms and may introduce hydrogen atoms that still need balancing. In acidic conditions, those hydrogen atoms are balanced using $H^+$."},{"text":"$$H_2$","correct":false,"explanation":"Adding hydrogen gas would introduce a new reacting species rather than perform the standard balancing procedure. In acidic half-equations, hydrogen atoms are balanced using $H^+$ ions."}]},{"id":"07d243a1-8b17-457a-9251-4f20cf941e55","cardType":"basic","points":1,"front":"What is the final step in balancing any half-equation after atoms, oxygen, and hydrogen are balanced?","back":"Balance the overall charge by adding electrons ($e^-$).","mcqOptions":[{"text":"Balance the overall charge by adding electrons ($e^-$).","correct":true,"explanation":"Once all atoms are balanced, electrons are added to the more positive side until both sides have equal overall charge. This also shows whether oxidation or reduction occurs."},{"text":"Balance the overall charge by adding hydrogen ions ($H^+$).","correct":false,"explanation":"Hydrogen ions balance hydrogen atoms in acidic conditions, so adding them at the end would disrupt the atom balance. Overall charge is finally balanced using electrons, $e^-$."},{"text":"Balance the overall charge by adding hydroxide ions ($OH^-$).","correct":false,"explanation":"Hydroxide ions are involved in alkaline balancing and would alter both oxygen and hydrogen counts. After all atoms are balanced, overall charge must be balanced using electrons, $e^-$."},{"text":"Balance the number of atoms by adding electrons ($e^-$).","correct":false,"explanation":"Electrons carry charge but contain no atoms, so they cannot correct an atom imbalance. After the atoms are balanced, electrons are added to balance the overall charge."}]},{"id":"09b566d3-c88d-48d9-8dc8-8ce31bbcab9b","cardType":"basic","points":1,"front":"When combining two half-equations, what must happen to the electrons?","back":"They must cancel out completely (the number lost must equal the number gained).","mcqOptions":[{"text":"They must cancel out completely (the number lost must equal the number gained).","correct":true,"explanation":"When half-equations are combined, electrons produced in oxidation must equal electrons consumed in reduction. Multiplying half-equations may be necessary so the electron numbers cancel completely."},{"text":"They must remain on both sides (the number lost must be greater than the number gained).","correct":false,"explanation":"Leaving electrons in the final equation means charge and electron transfer have not been balanced. The correct method makes the number lost equal the number gained so electrons cancel completely."},{"text":"They must cancel out completely (the number lost must be less than the number gained).","correct":false,"explanation":"The numbers of electrons cannot be unequal, because that leaves a net electron imbalance. The half-equations must be scaled until the number lost equals the number gained."},{"text":"They must remain on one side (the number lost must equal the number gained).","correct":false,"explanation":"Even equal numbers of electrons must be removed from opposite sides rather than left in the overall equation. Combining half-equations requires complete cancellation of the electrons."}]},{"id":"bacdc2aa-f6ee-40f8-8f90-e6260441d744","cardType":"basic","points":2,"front":"How do you convert an acidic half-equation into an alkaline half-equation?","back":"Add $OH^-$ to both sides to react with all $H^+$ (forming $H_2O$), then cancel excess $H_2O$.","mcqOptions":[{"text":"Add $OH^-$ to both sides to react with all $H^+$ (forming $H_2O$), then cancel excess $H_2O$.","correct":true,"explanation":"Adding hydroxide ions to both sides neutralises every hydrogen ion by forming water. After combining water molecules, any excess $H_2O$ is cancelled to produce the alkaline half-equation."},{"text":"Add $H^+$ to both sides to react with all $OH^-$ (forming $H_2O$), then cancel excess $H_2O$.","correct":false,"explanation":"Adding $H^+$ neutralises hydroxide and is the wrong direction for converting an acidic equation into alkaline conditions. The correct method adds $OH^-$ to both sides to remove $H^+$."},{"text":"Add $OH^-$ to one side to react with all $H^+$ (forming $H_2O$), then cancel excess $H_2O$.","correct":false,"explanation":"Adding hydroxide to only one side changes the equation's charge balance. Hydroxide must be added to both sides, then water formed on each side is cancelled where possible."},{"text":"Add $H_2O$ to both sides to react with all $H^+$ (forming $OH^-$), then cancel excess $OH^-$.","correct":false,"explanation":"Water does not directly convert all hydrogen ions into hydroxide ions in this procedure. Add $OH^-$ to both sides so $H^+$ forms $H_2O$, then cancel excess water."}]},{"id":"dd3e49c3-e5f8-41e4-9258-993ccba509df","cardType":"basic","points":1,"front":"Define disproportionation.","back":"A reaction where the same element is both oxidised and reduced.","mcqOptions":[{"text":"A reaction where the same element is both oxidised and reduced.","correct":true,"explanation":"Disproportionation occurs when atoms of the same element undergo two simultaneous changes: some increase their oxidation state while others decrease theirs."},{"text":"A reaction where the same element is only oxidised.","correct":false,"explanation":"Oxidising the same element alone is not disproportionation. The defining feature is that the same element undergoes oxidation and reduction simultaneously in different products."},{"text":"A reaction where different elements are both oxidised and reduced.","correct":false,"explanation":"Disproportionation concerns one element appearing in two oxidation-state changes, not different elements. The same element must be both oxidised and reduced in the reaction."},{"text":"A reaction where the same element is only reduced.","correct":false,"explanation":"Reduction of one element alone is an ordinary redox change, not disproportionation. Disproportionation requires the same element to be both oxidised and reduced."}]},{"id":"9ef6029d-d7aa-43ca-bc22-b654f0300770","cardType":"basic","points":2,"front":"In the reaction $Cl_2 + 2OH^- \\to Cl^- + ClO^- + H_2O$, why is chlorine considered to have undergone disproportionation?","back":"Its oxidation state changes from $0$ to both $-1$ (reduction) and $+1$ (oxidation).","mcqOptions":[{"text":"Its oxidation state changes from $0$ to both $-1$ (reduction) and $+1$ (oxidation).","correct":true,"explanation":"Chlorine starts at oxidation state $0$ in $Cl_2$. It forms $Cl^-$ at $-1$ by reduction and $ClO^-$ at $+1$ by oxidation, so disproportionation occurs."},{"text":"Its oxidation state changes from $0$ to only $-1$ (reduction).","correct":false,"explanation":"This accounts only for chlorine forming $Cl^-$ and ignores chlorine in $ClO^-$. The correct answer includes both changes, from $0$ to $-1$ and from $0$ to $+1$."},{"text":"Its oxidation state changes from $0$ to only $+1$ (oxidation).","correct":false,"explanation":"This accounts only for chlorine forming $ClO^-$ and ignores chlorine in $Cl^-$. Chlorine undergoes both oxidation to $+1$ and reduction to $-1$."},{"text":"Its oxidation state changes from $-1$ to both $0$ (oxidation) and $+1$ (oxidation).","correct":false,"explanation":"The starting oxidation state of chlorine is $0$ in $Cl_2$, not $-1$. The products contain chlorine at $-1$ and $+1$, representing reduction and oxidation respectively."}]},{"id":"c1f6d978-b00b-4fc5-a500-762e9e087496","cardType":"cloze","points":null,"front":"Fluorine always has an oxidation state of $\\text{[...]}$ in its compounds, while Aluminium is usually $+3$.","back":"Fluorine always has an oxidation state of $\\boxed{-1}$ in its compounds, while Aluminium is usually $+3$.","mcqOptions":[{"text":"$$-1$","correct":true,"explanation":"Fluorine is the most electronegative element and almost always gains one electron in compounds, giving it the oxidation state $-1$."},{"text":"$$+1$","correct":false,"explanation":"The $+1$ value reverses fluorine's usual electron-gaining behaviour. Fluorine is more electronegative than the elements it bonds with, so its oxidation state is $-1$."},{"text":"$$+3$","correct":false,"explanation":"The $+3$ value is associated with some commonly occurring oxidation states of elements such as aluminium, not fluorine. Fluorine has oxidation state $-1$ in its compounds."},{"text":"$$0$","correct":false,"explanation":"Oxidation state $0$ applies to fluorine in its elemental form, $F_2$, rather than in compounds. In compounds, fluorine is assigned $-1$."}]},{"id":"02426821-b7a9-4079-8521-400f4ce6e6c4","cardType":"cloze","points":null,"front":"Fluorine always has an oxidation state of $-1$ in its compounds, while Aluminium is usually $\\text{[...]}$.","back":"Fluorine always has an oxidation state of $-1$ in its compounds, while Aluminium is usually $\\boxed{+3}$.","mcqOptions":[{"text":"$$+3$","correct":true,"explanation":"Aluminium usually loses three electrons when forming compounds, so it has the oxidation state $+3$. This reflects its three valence electrons."},{"text":"$$-3$","correct":false,"explanation":"The negative sign suggests aluminium gains electrons, but aluminium is a metal that normally loses three electrons. Its usual oxidation state is therefore $+3$."},{"text":"$$+1$","correct":false,"explanation":"The $+1$ value would represent loss of only one electron. Aluminium typically loses all three of its outer electrons, giving the oxidation state $+3$."},{"text":"$$0$","correct":false,"explanation":"Oxidation state $0$ applies to an element in its uncombined form. Aluminium in compounds usually loses three electrons and has oxidation state $+3$."}]},{"id":"cf3b5bc7-a529-4a3d-bd77-8d97dfddc576","cardType":"basic","points":1,"front":"What is the oxidation state of nitrogen in $NO_3^-$?","back":"$+5$.","mcqOptions":[{"text":"$+5$.","correct":true,"explanation":"In $NO_3^-$, the three oxygens contribute $-6$ overall. Since the ion totals $-1$, nitrogen must be $+5$ because $+5-6=-1$."},{"text":"$+3$.","correct":false,"explanation":"Using $+3$ gives a total of $-3$ for $NO_3^-$, not the required $-1$. Three oxygens contribute $-6$, so nitrogen must be $+5$."},{"text":"$-3$.","correct":false,"explanation":"The negative value incorrectly treats nitrogen as gaining electrons relative to oxygen. With oxygen contributing $-6$ and the ion total being $-1$, nitrogen is $+5$."},{"text":"$+6$.","correct":false,"explanation":"The $+6$ value would make the ion total $0$, as $+6-6=0$. Because nitrate has charge $-1$, nitrogen must be $+5$."}]},{"id":"54486a68-cc63-43a3-b02c-bfa9f66a8808","cardType":"basic","points":1,"front":"What is a spectator ion?","back":"An ion that is present in solution but does not take part in the reaction or change oxidation state.","mcqOptions":[{"text":"An ion that is present in solution but does not take part in the reaction or change oxidation state.","correct":true,"explanation":"A spectator ion remains dissolved and unchanged while the reacting species undergo chemical change. It appears in the full ionic equation but cancels from the net ionic equation."},{"text":"An ion that is present in solution and is oxidised during the reaction or change oxidation state.","correct":false,"explanation":"An ion that is oxidised is an active redox species, not a spectator. A spectator ion remains present in solution without taking part or changing oxidation state."},{"text":"An ion that is present in solution and is reduced during the reaction or change oxidation state.","correct":false,"explanation":"Reduction means the ion participates in the redox process, so it is not a spectator. Spectator ions remain unchanged and do not change oxidation state."},{"text":"An ion that is absent from solution but does not take part in the reaction or change oxidation state.","correct":false,"explanation":"A spectator ion must actually be present in the solution, even though it does not react. Being absent means it cannot appear in the ionic equation at all."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Oxidation states and redox equations","title":"Oxidation states and redox equations Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"7b5b39ad-4453-565b-a2b1-384b797b3b9c","walled":"$L39"}],"$L3a","$L3b"]