Oxidation, Reduction & Redox Equations

## Track electrons in oxidation and reduction - Oxidation is electron loss and reduction is electron gain. The electrons transferred by one partner must be accepted by another in the overall redox process. - The reviewed iron/copper illustration shows Fe becoming Fe²⁺ while Cu²⁺ becomes Cu. Iron loses two electrons and copper(II) ions gain the same two electrons. - Oxygen gain or hydrogen loss can identify oxidation in suitable examples, but electron accounting and oxidation-state changes apply more generally. A reaction need not contain oxygen to be redox. ## Distinguish oxidation state from actual atomic charge - Oxidation state is a formal electron-accounting assignment based on an ionic approximation of bonds. It is not generally the measured partial charge of an atom in a covalent molecule. - For a heteronuclear bond in the simple assignment used here, give the bond electrons to the more electronegative atom. Equal-atom bonds are shared equally for oxidation-state bookkeeping. - The sum of the assigned oxidation states, including atom counts, must equal the overall species charge. This is distinct from assuming the covalent compound contains free ions. ## Assign uncombined elements and monatomic ions - An uncombined element has oxidation state zero, whether monatomic, molecular or an extended solid. This is not restricted to its standard-state allotrope. - For a monatomic ion, oxidation state equals its ionic charge: Fe³⁺ is +3 and S²⁻ is −2. - Distinguish a molecular subscript from an ionic charge. The two atoms in Cl₂ do not each have oxidation state +2, and ozone oxygen is still an uncombined element. ## Use atom counts in a neutral-compound balance - For a neutral formula, the atom-count-weighted sum of oxidation states is zero. Multiply each assigned value by the number of atoms of that element. - For SO₃ with oxygen assigned −2, let sulfur be x. Then x+3(−2)=0, giving x=+6. - Write the sign explicitly. A subscript is an atom count; it is not an oxidation-state multiplier that can be ignored. ## Use the overall charge of a polyatomic ion - For a polyatomic ion, the weighted oxidation-state sum equals the ion’s total charge, rather than always zero. - For SO₄²⁻ with oxygen −2, x+4(−2)=−2, giving sulfur +6. - The−2 ion charge belongs to the whole sulfate ion. It is not the oxidation state of sulfur and is not a subscript giving two extra oxygen atoms. ## Recognise the hydrogen rule and metal-hydride exception - Hydrogen commonly has oxidation state +1 in compounds with more electronegative elements, such as water. - In a metal hydride such as NaH, sodium is +1 and hydrogen is −1. Use the species and bonding context rather than an exception-free slogan. - Uncombined H₂ has oxidation state zero. A rule for hydrogen in compounds must not be applied to the elemental molecule. ## Handle oxygen and fluorine exceptions - Oxygen is commonly −2, but it is −1 in hydrogen peroxide. In OF₂, fluorine is −1 and oxygen is +2. - Use electronegativity and the overall charge balance to select the appropriate assignment. Fluorine in its compounds is assigned −1 in the examples covered here. - Do not use the peroxide assignment for every oxygen-containing compound, or assign−2 to oxygen in OF₂ despite the more electronegative fluorine. ## Interpret Roman numerals in compound names - A Roman numeral in a Stock name identifies an element’s oxidation state. Iron(II) and iron(III) distinguish+2 from +3. - For neutral FeCl₂, two chloride assignments−1 require iron +2. For FeCl₃, three require iron +3. - The numeral is not the number of atoms or the charge of the whole neutral compound. Common names such as sodium chloride do not always need a numeral. ## Identify oxidising and reducing agents - An oxidising agent accepts electrons and is itself reduced. A reducing agent donates electrons and is itself oxidised. - In Mg+Fe²⁺→Mg²⁺+Fe, magnesium donates two electrons and Fe²⁺ accepts them. The agent name describes what it does to the other partner. - Name the reactant species precisely. Fe²⁺, rather than the neutral product Fe, is the oxidising agent in this written reaction. ## Explain why one substance can play different redox roles - A substance’s redox role depends on the reaction partners and conditions. Hydrogen peroxide contains oxygen at −1, which can become−2 or0 in different reactions. - In supplied reaction A peroxide oxygen becomes water oxygen at −2, so peroxide is reduced and acts as oxidising agent. - In reaction B peroxide oxygen becomes O₂ at 0, so peroxide is oxidised and acts as reducing agent. These equations illustrate roles; they are not mixing instructions. ## Distinguish a formula average from individual sites - The total sulfur oxidation-state contribution in S₄O₆²⁻ is +10 when each oxygen is −2. Dividing by four gives a formula-average value+2.5. - That average does not establish that every sulfur atom has the same local environment or assignment. A structure is needed to distinguish individual sites. - A fractional formula average is not evidence that half an electron has been transferred in a single elementary event. Use the average only for the aggregate accounting it represents. ## Classify a redox reaction from oxidation-state changes - For 2Na+Cl₂→2NaCl, sodium changes from 0 to +1 and is oxidised. Chlorine changes from 0 to −1 and is reduced. - Each sodium atom loses one electron; the two chlorine atoms together accept two. Both atoms and transferred electrons must balance. - This example also shows why oxidation-state increase means oxidation and decrease means reduction. Follow the actual signs rather than the position of an element in the equation. ## Recognise a reaction that is not redox - A balanced reaction can occur without oxidation-state changes. Acid–base neutralisation is a useful contrast to electron-transfer reactions. - In H⁺+OH⁻→H₂O, hydrogen remains +1 and oxygen remains −2. The ionic charges combine, but those oxidation states do not change. - Changes in total species charge or proton transfer alone do not prove electron transfer. Check the oxidation states before labelling a reaction redox. ## Write an oxidation half-equation - An oxidation half-equation places released electrons on the product side. Balance the atoms and then the total electrical charge. - Fe²⁺→Fe³⁺+e⁻ has one iron atom on each side and charge +2 on each side after the electron is included. - The coefficient counts electrons, not iron atoms. Adding an electron to the wrong side would increase the charge mismatch rather than repair it. ## Write a reduction half-equation - A reduction half-equation places accepted electrons on the reactant side. Cu²⁺+2e⁻→Cu balances one copper atom and zero net charge. - The oxidation state decreases from +2 to 0, corresponding to two electrons gained per copper ion. - Check both atom count and charge. Equal atom counts alone would not distinguish a correct electron coefficient from an incorrect one. ## Balance an oxygen-containing half-equation in acid - For the supplied MnO₄⁻→Mn²⁺ reduction in acid, first balance manganese, then oxygen with water and hydrogen with H⁺. - The balanced half-equation is MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O. Manganese falls from +7 to +2 and gains five electrons. - The medium and stated product matter. This acidified Mn²⁺ equation must not be replaced by a MnO₂ equation from different conditions. ## Account for two redox-active atoms in one ion - One Cr₂O₇²⁻ contains two chromium atoms, each at +6. Formation of two Cr³⁺ therefore requires six electrons in total. - In acid the half-equation is Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O. - Do not use a three-electron total for the whole dichromate ion merely because each individual chromium changes by three. Multiply by the atom count. ## Combine half-equations using equal electron totals - MnO₄⁻ in acid accepts five electrons per ion, while Fe²⁺→Fe³⁺ supplies one per iron ion. Multiply the iron half-equation by five. - After adding and cancelling electrons, MnO₄⁻+8H⁺+5Fe²⁺→Mn²⁺+4H₂O+5Fe³⁺. - Electrons cancel from the overall ionic equation. The final coefficient ratio follows from electron conservation, not from guessing equal amounts of the two ions. ## Remove spectator ions from an ionic equation - In Zn+CuSO₄→ZnSO₄+Cu under the supplied dissociation model, sulfate appears unchanged on both sides. - Cancel sulfate to obtain Zn+Cu²⁺→Zn²⁺+Cu. Zinc donates two electrons and copper(II) ions accept them. - Remove only species unchanged in the same form on both sides. An atom that changes charge or chemical form is not a spectator merely because its element appears twice. ## Use a balanced redox equation for amounts - The iodine–thiosulfate equation consumes two moles of thiosulfate per mole of I₂. The ratio comes from the balanced coefficients. - The supplied 0.0100 mol I₂ needs 0.0200 mol thiosulfate. Starting with 0.0300 mol leaves 0.0100 mol thiosulfate after iodine is consumed. - Distinguish I₂ molecules from individual iodine atoms or iodide ions. A mole of I₂ corresponds to two moles of iodine atoms and forms two moles of I⁻ in this reaction. ## Balance a half-equation in alkaline solution - For MnO₄⁻→MnO₂ in alkaline solution, manganese falls from +7 to +4 and accepts three electrons per atom. - The balanced alkaline half-equation is MnO₄⁻+2H₂O+3e⁻→MnO₂+4OH⁻. Atoms and charge match on both sides. - One method balances in acid first, then adds OH⁻ to both sides to neutralise H⁺ and cancels excess water. The final alkaline equation must not retain uncancelled H⁺. ## Recognise cold-alkali disproportionation - Disproportionation occurs when the same starting species is both oxidised and reduced in one overall reaction. - For the supplied cold dilute hydroxide products, chlorine at 0 becomes−1 in Cl⁻ and+1 in ClO⁻. The equation is Cl₂+2OH⁻→Cl⁻+ClO⁻+H₂O. - The starting chlorine species fills both redox roles. Different product oxidation states show the two directions; no separate external electron source is required in the overall equation. ## Balance hot-alkali disproportionation - For the supplied hot concentrated hydroxide products, chlorine becomes−1 in Cl⁻ and+5 in ClO₃⁻. - One chlorine atom rising from 0 to +5 loses five electrons overall, balanced by five chlorine atoms each falling from 0 to −1. - The simplest equation is 3Cl₂+6OH⁻→5Cl⁻+ClO₃⁻+3H₂O. Do not substitute the cold-alkali ClO⁻ product into the hot-condition calculation. ## Audit atoms, charge and electron cancellation - A valid half-equation conserves each element and total electrical charge. A valid overall redox equation also cancels electrons transferred between its partners. - For the acidified MnO₄⁻ half-equation, check one manganese, four oxygen, eight hydrogen and equal net charge, rather than checking only one feature. - An equation can have the correct manganese change but the wrong water, hydrogen-ion or electron coefficient. Record medium, species and the smallest integer ratio as part of the final answer.