[{"data":1,"prerenderedAt":233},["ShallowReactive",2],{"topic-redox-reactions-and-redox-balancing":3},{"topic":4,"prev":220,"next":225},{"id":5,"slug":6,"title":7,"shortTitle":8,"lesson":5,"lastAlignmentAudit":9,"category":10,"description":11,"metaDescription":12,"objectives":13,"conceptSections":32,"workedExample":122,"oerSources":134,"selfStudyQuestions":136,"relatedTopicSlugs":161,"relatedElements":165,"relatedReferences":171,"keyEquations":174,"howTo":187,"conceptProbe":197,"practiceTeaser":201,"gatedContent":209,"seoKeywords":212,"objectiveCount":219},7,"redox-reactions-and-redox-balancing","Redox Reactions and Redox Balancing","Redox & Balancing","2026-07-15","reactions","Assign oxidation numbers, recognize redox reactions, identify oxidizing and reducing agents, write half-reactions, and balance redox equations by both the oxidation-number and half-reaction methods.","Master redox chemistry: assign oxidation numbers, identify redox reactions and agents, write half-reactions, and balance redox equations in acidic and basic solution.",[14,17,20,23,26,29],{"id":15,"text":16},"7.1","Assign oxidation numbers to atoms in compounds and polyatomic ions",{"id":18,"text":19},"7.2","Determine whether a reaction is redox or non-redox from oxidation-number changes",{"id":21,"text":22},"7.3","Identify the oxidized species, reduced species, oxidizing agent, and reducing agent in a redox reaction",{"id":24,"text":25},"7.4","Write balanced oxidation and reduction half-reactions for a redox process",{"id":27,"text":28},"7.5","Balance simple redox equations using the oxidation-number method",{"id":30,"text":31},"7.6","Balance redox equations in acidic and basic solution using the half-reaction method",[33,43,51,59,67,76,85,94,98,102,110,114,118],{"heading":34,"content":35,"relatedObjectives":36,"deepDive":39},"What Is an Oxidation Number?","\u003Cp>An \u003Cstrong>oxidation number\u003C/strong> (or oxidation state) is the hypothetical charge an atom would carry if every bond in the compound were completely ionic. It is a bookkeeping tool that lets you track electron distribution in molecules and ions, even covalent ones. Oxidation numbers are written with the sign before the digit (e.g., +2, −3) to distinguish them from ionic charges, which place the sign after the digit (e.g., 2+, 3−).\u003C/p>\u003Cp>Although oxidation numbers do not represent actual charges on atoms in covalent compounds, they provide a powerful way to classify reactions. Whenever one or more atoms change oxidation number during a reaction, that reaction is a \u003Cstrong>redox reaction\u003C/strong>.\u003C/p>",[37,38],35,36,[40],{"label":41,"body":42},"A useful fiction","\u003Cp>Oxidation numbers are openly hypothetical: no chlorine atom in HCl actually carries a full −1 charge, because the bond is shared, not transferred. The convention simply awards every shared electron pair to the more electron-greedy atom and tallies the result, like roommates assigning every shared expense to whoever wanted it more. The totals are a fiction, but a \u003Cem>consistent\u003C/em> fiction, and that is enough: when the tallies change between the two sides of an equation, real electrons really moved. Chemistry keeps oxidation numbers for the same reason accountants keep ledgers: not because the categories are physical, but because they catch every transfer.\u003C/p>",{"heading":44,"content":45,"relatedObjectives":46,"deepDive":47},"Rules for Assigning Oxidation Numbers","\u003Cp>Apply the following rules in order of priority to assign \u003Cstrong>oxidation numbers\u003C/strong> to every atom in a formula:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Free elements\u003C/strong> → oxidation number = 0 (e.g., Na, O\u003Csub>2\u003C/sub>, Fe, Cl\u003Csub>2\u003C/sub>).\u003C/li>\u003Cli>\u003Cstrong>Monatomic ions\u003C/strong> → oxidation number = ion charge (Na\u003Csup>+\u003C/sup> = +1, Cl\u003Csup>−\u003C/sup> = −1).\u003C/li>\u003Cli>\u003Cstrong>Fluorine\u003C/strong> is always \u003Cstrong>−1\u003C/strong>.\u003C/li>\u003Cli>\u003Cstrong>Hydrogen\u003C/strong> is usually \u003Cstrong>+1\u003C/strong> (except −1 in metal hydrides such as NaH or LiH, where H is bonded to a metal).\u003C/li>\u003Cli>\u003Cstrong>Oxygen\u003C/strong> is usually \u003Cstrong>−2\u003C/strong>, with three notable exceptions: −1 in peroxides (H\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub>, Na\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub>); approximately −½ in superoxides (KO\u003Csub>2\u003C/sub>, NaO\u003Csub>2\u003C/sub>); and positive when bonded to fluorine (e.g., +2 in OF\u003Csub>2\u003C/sub>) because F always wins the priority contest.\u003C/li>\u003Cli>The sum of oxidation numbers in a \u003Cstrong>neutral molecule = 0\u003C/strong>; in a \u003Cstrong>polyatomic ion = the ion’s charge\u003C/strong>.\u003C/li>\u003C/ol>\u003Cp>To find an unknown value, assign all the known oxidation numbers first and solve the algebraic equation. For example, in Na\u003Csub>2\u003C/sub>SO\u003Csub>4\u003C/sub>: Na = +1 each, O = −2 each, so 2(+1) + S + 4(−2) = 0, giving S = +6.\u003C/p>\u003Cp>\u003Cstrong>Average (fractional) oxidation numbers\u003C/strong> appear when the algebraic sum rule, applied to a compound or ion, distributes over multiple atoms of the same element to yield a non-integer per-atom value. Examples: O = −½ in NaO\u003Csub>2\u003C/sub> (superoxide), since (+1) + 2x = 0 → x = −½; N = +1 in N\u003Csub>2\u003C/sub>O, since 2x + (−2) = 0 → x = +1; Fe = +8/3 in Fe\u003Csub>3\u003C/sub>O\u003Csub>4\u003C/sub> (a mixed Fe\u003Csup>2+\u003C/sup>/Fe\u003Csup>3+\u003C/sup> oxide whose per-Fe average is +8/3). Fractional values are valid outputs of the algebraic sum rule; the underlying physical structure may have crystallographically distinct sites with different per-atom oxidation states (Fe\u003Csub>3\u003C/sub>O\u003Csub>4\u003C/sub> contains 1 Fe\u003Csup>2+\u003C/sup> + 2 Fe\u003Csup>3+\u003C/sup> per formula unit), but the bookkeeping convention reports the average and that average is the answer LO 7.1 expects.\u003C/p>\u003Cp>\u003Cstrong>Averages are allowed.\u003C/strong> An oxidation number does not have to be a whole number. When several atoms of the same element sit in one species, the rules give an \u003Cem>average\u003C/em> across them: in Fe₃O₄, iron averages +8/3 (≈ +2.67); in thiosulfate, S₂O₃²⁻, sulfur averages +2; in tetrathionate, S₄O₆²⁻, sulfur averages +2.5. The average is taken across all atoms of that element in the formula unit. It is a bookkeeping figure for the formula unit as a whole — not an oxidation state belonging to any one atom.\u003C/p>\u003Cp>\u003Cstrong>One caution for later balancing.\u003C/strong> The average is what you report when the question asks you to assign oxidation numbers. It is not always what the half-reaction method needs. In S₂O₃²⁻ the two sulfurs are not actually identical — the terminal S sits at −1 while the central S sits at +5 — and in S₄O₆²⁻ the two bridging sulfurs sit at 0. When you balance S₂O₃²⁻ → S₄O₆²⁻ you need that per-atom picture to count electrons correctly (each oxidised terminal S goes −1 → 0, one electron per atom). Use the average for assignment questions, and the per-atom (structural) picture only when an electron count demands it.\u003C/p>",[37],[48],{"label":49,"body":50},"Test yourself: three assignments with a trap each","\u003Cp>Assign the oxidation number of the marked atom, then check below: Mn in KMnO\u003Csub>4\u003C/sub>, O in H\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub>, H in NaH.\u003C/p>\u003Cp>\u003Cstrong>Answers:\u003C/strong> In KMnO\u003Csub>4\u003C/sub>, K is +1 and each O is −2, so Mn must make the sum zero: \u003Cspan class=\"nowrap\">+1 + Mn + 4(−2) = 0\u003C/span> gives Mn = \u003Cstrong>+7\u003C/strong>. In H\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub>, oxygen is in a peroxide, one of its named exceptions: each O is \u003Cstrong>−1\u003C/strong> (check: \u003Cspan class=\"nowrap\">2(+1) + 2(−1) = 0\u003C/span>). In NaH, hydrogen is bonded to a metal, so it takes its hydride exception: H is \u003Cstrong>−1\u003C/strong> and Na is +1. The rules work in priority order precisely so these traps resolve themselves: the earlier rule (K = +1, Na = +1) pins the certain atoms, and the flexible atom absorbs what remains.\u003C/p>",{"heading":52,"content":53,"relatedObjectives":54,"deepDive":55},"Is the Reaction Redox? Decide by Oxidation-Number Changes","\u003Cp>A reaction is \u003Cstrong>redox\u003C/strong> when at least one atom’s oxidation number changes from the reactant side to the product side. Reactions in which \u003Cem>no\u003C/em> atom’s oxidation number changes are \u003Cstrong>non-redox\u003C/strong> — classic examples are simple double-replacement precipitations and acid-base neutralizations.\u003C/p>\u003Cp>The decision procedure is mechanical:\u003C/p>\u003Col>\u003Cli>Assign oxidation numbers to \u003Cstrong>every\u003C/strong> atom on both sides of the equation.\u003C/li>\u003Cli>List the atoms whose oxidation numbers differ between the two sides.\u003C/li>\u003Cli>If at least one atom’s number increases (and at least one other’s decreases), the reaction is redox. If no atom changes, it is non-redox.\u003C/li>\u003C/ol>\u003Cp>\u003Cstrong>Worked example.\u003C/strong> AgNO\u003Csub>3\u003C/sub>(aq) + NaCl(aq) → AgCl(s) + NaNO\u003Csub>3\u003C/sub>(aq). Ag = +1 on both sides; Na = +1 on both sides; N stays +5; O stays −2; Cl stays −1. No atom changes — \u003Cem>not\u003C/em> redox (it is a precipitation). Compare with 2 Na(s) + Cl\u003Csub>2\u003C/sub>(g) → 2 NaCl(s): Na goes 0 → +1 (oxidized) and Cl goes 0 → −1 (reduced) — this \u003Cem>is\u003C/em> redox.\u003C/p>\u003Cp>This redox/non-redox classification overlaps but is \u003Cem>not\u003C/em> the same as the four-category aqueous-reaction classification from Topic 6 (precipitation, acid-base, gas-forming, redox). Topic 6 asks you to recognize an aqueous reaction’s family from the reactant identities at a glance; this LO asks you to verify rigorously, atom by atom, whether oxidation numbers actually change.\u003C/p>\u003Cp>\u003Cstrong>Don’t mistake bond rearrangement for redox.\u003C/strong> Some reactions look like redox at first glance because covalent bonds break and re-form, but no atom changes oxidation number. A classic trap is hydrolysis of a covalent halide: PCl\u003Csub>5\u003C/sub> + 4 H\u003Csub>2\u003C/sub>O → H\u003Csub>3\u003C/sub>PO\u003Csub>4\u003C/sub> + 5 HCl. Track every element: P stays +5 (in PCl\u003Csub>5\u003C/sub> from x + 5(−1) = 0; in H\u003Csub>3\u003C/sub>PO\u003Csub>4\u003C/sub> from 3(+1) + x + 4(−2) = 0), Cl stays −1, H stays +1, O stays −2. No element shifts — it is hydrolysis, not redox. Ox-number bookkeeping is the only reliable test; visual cues (bond breaking, color change, gas evolution) can mislead.\u003C/p>",[38],[56],{"label":57,"body":58},"Why this matters: you are running redox right now","\u003Cp>Cellular respiration is the controlled oxidation of glucose: C\u003Csub>6\u003C/sub>H\u003Csub>12\u003C/sub>O\u003Csub>6\u003C/sub> + 6 O\u003Csub>2\u003C/sub> → 6 CO\u003Csub>2\u003C/sub> + 6 H\u003Csub>2\u003C/sub>O, the very same overall equation as burning the sugar in a flame. Your cells run it through dozens of enzyme-managed steps so the energy arrives in usable installments instead of one burst of heat, but the oxidation-number bookkeeping is identical: carbon’s oxidation number rises, oxygen’s falls. Rusting cars, draining batteries, bleaching stains, photosynthesizing leaves, and metabolizing lunch are one reaction family, and the oxidation-number check is how you recognize every member.\u003C/p>",{"heading":60,"content":61,"relatedObjectives":62,"deepDive":63},"Identifying Oxidation and Reduction","\u003Cp>Once you know a reaction is redox, identify the two complementary processes:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Oxidation\u003C/strong> — an \u003Cem>increase\u003C/em> in oxidation number (loss of electrons).\u003C/li>\u003Cli>\u003Cstrong>Reduction\u003C/strong> — a \u003Cem>decrease\u003C/em> in oxidation number (gain of electrons).\u003C/li>\u003C/ul>\u003Cp>A widely used mnemonic is \u003Cstrong>OIL RIG\u003C/strong>: Oxidation Is Loss, Reduction Is Gain (of electrons). The two processes always occur together — you cannot have oxidation without an accompanying reduction.\u003C/p>\u003Cp>In the reaction Fe\u003Csub>2\u003C/sub>O\u003Csub>3\u003C/sub> + 3 CO → 2 Fe + 3 CO\u003Csub>2\u003C/sub>, iron goes from +3 to 0 (reduced) and carbon goes from +2 to +4 (oxidized). Reactions that show \u003Cem>no\u003C/em> change in oxidation numbers — such as double-replacement or simple acid-base neutralisations — are not redox.\u003C/p>",[38],[64],{"label":65,"body":66},"Why the names sound backwards","\u003Cp>Both terms are older than the electron. “Oxidation” originally meant combining with oxygen, the most common electron thief chemists knew; when the electron was discovered, the definition generalized to any electron loss, with or without oxygen involved. “Reduction” comes from metallurgy: smelting “reduces” an ore to its metal, and the product literally weighs less than the ore because the oxygen leaves. The metal atoms gain electrons in the process, so “reduction” became the name for electron gain.\u003C/p>\u003Cp>So the vocabulary is a fossil record: two industrial-age words, re-founded on the electron, still carrying their original scars. OIL RIG exists because the history does not explain itself.\u003C/p>",{"heading":68,"content":69,"relatedObjectives":70,"deepDive":72},"Oxidizing and Reducing Agents","\u003Cp>Every redox reaction has an \u003Cstrong>oxidizing agent\u003C/strong> and a \u003Cstrong>reducing agent\u003C/strong>. The names describe what each species does to the \u003Cem>other\u003C/em> reactant:\u003C/p>\u003Cul>\u003Cli>The \u003Cstrong>oxidizing agent (oxidant)\u003C/strong> is \u003Cem>itself reduced\u003C/em> — it accepts electrons, causing another species to be oxidized.\u003C/li>\u003Cli>The \u003Cstrong>reducing agent (reductant)\u003C/strong> is \u003Cem>itself oxidized\u003C/em> — it donates electrons, causing another species to be reduced.\u003C/li>\u003C/ul>\u003Cp>This naming convention can seem backwards at first. A useful tip: identify the agent by what it \u003Cem>does to its partner\u003C/em>, not by what happens to itself.\u003C/p>\u003Cp>\u003Cstrong>Name the whole species, not just the changing element.\u003C/strong> In CH\u003Csub>4\u003C/sub> + 2 O\u003Csub>2\u003C/sub> → CO\u003Csub>2\u003C/sub> + 2 H\u003Csub>2\u003C/sub>O the reducing agent is CH\u003Csub>4\u003C/sub> and the oxidizing agent is O\u003Csub>2\u003C/sub>; in MnO\u003Csub>4\u003C/sub>\u003Csup>−\u003C/sup> + Fe\u003Csup>2+\u003C/sup> redox the oxidizing agent is the permanganate ion MnO\u003Csub>4\u003C/sub>\u003Csup>−\u003C/sup> and the reducing agent is Fe\u003Csup>2+\u003C/sup>.\u003C/p>\u003Cp>Several common reaction subclasses are redox processes:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Combustion\u003C/strong> — a fuel (reductant) reacts with O\u003Csub>2\u003C/sub> (oxidant), producing heat and often flame.\u003C/li>\u003Cli>\u003Cstrong>Single-displacement\u003C/strong> — a more reactive metal (reductant) replaces a less reactive metal ion from solution, e.g., Zn(s) + Cu\u003Csup>2+\u003C/sup>(aq) → Zn\u003Csup>2+\u003C/sup>(aq) + Cu(s).\u003C/li>\u003Cli>\u003Cstrong>Disproportionation\u003C/strong> — the \u003Cem>same\u003C/em> element is simultaneously oxidized and reduced, e.g., oxygen in H\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub> goes from −1 to both 0 (in O\u003Csub>2\u003C/sub>) and −2 (in H\u003Csub>2\u003C/sub>O). In a disproportionation the species is \u003Cem>both\u003C/em> oxidizing agent and reducing agent (it acts on itself).\u003C/li>\u003Cli>\u003Cstrong>Comproportionation\u003C/strong> — the inverse of disproportionation: the same element appears in \u003Cem>two different\u003C/em> oxidation states on the reactant side and merges to a single intermediate oxidation state on the product side. Example: NH\u003Csub>4\u003C/sub>NO\u003Csub>3\u003C/sub> → N\u003Csub>2\u003C/sub>O + 2 H\u003Csub>2\u003C/sub>O. Here N starts at −3 (in NH\u003Csub>4\u003C/sub>\u003Csup>+\u003C/sup>) and +5 (in NO\u003Csub>3\u003C/sub>\u003Csup>−\u003C/sup>) and ends at +1 (in N\u003Csub>2\u003C/sub>O). Per formula unit one N loses 4 e\u003Csup>−\u003C/sup> and one N gains 4 e\u003Csup>−\u003C/sup> — electrons balance internally.\u003C/li>\u003C/ul>",[71],38,[73],{"label":74,"body":75},"Common mistake: forgetting the agent acts on someone else","\u003Cp>An oxidizing agent is itself reduced; a reducing agent is itself oxidized. The inversion stops feeling backwards once you read “agent” the way you would for a travel agent: someone who arranges the trip for \u003Cem>another\u003C/em> person. The oxidizing agent arranges the other reactant’s oxidation, paying for it by accepting the electrons itself.\u003C/p>\u003Cp>The vocabulary is everywhere in daily life: bleach works because hypochlorite is a strong oxidizing agent attacking stain molecules, and many “antioxidants” in food act as reducing agents that sacrifice their own electrons to reactive oxidizers before your cells have to.\u003C/p>",{"heading":77,"content":78,"relatedObjectives":79,"deepDive":81},"Balancing Redox by the Oxidation-Number Method","\u003Cp>Simple redox equations can be balanced using the \u003Cstrong>change-in-oxidation-number method\u003C/strong>:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Assign oxidation numbers\u003C/strong> to every atom on both sides of the unbalanced equation.\u003C/li>\u003Cli>\u003Cstrong>Identify\u003C/strong> which atoms are oxidized (increase) and which are reduced (decrease).\u003C/li>\u003Cli>\u003Cstrong>Calculate the electron change\u003C/strong> per atom for each process.\u003C/li>\u003Cli>\u003Cstrong>Equalise electron transfer\u003C/strong>: multiply the oxidized and reduced species by coefficients so that total electrons lost = total electrons gained.\u003C/li>\u003Cli>\u003Cstrong>Balance remaining atoms\u003C/strong> (non-redox elements, then H and O) by inspection.\u003C/li>\u003Cli>\u003Cstrong>Verify\u003C/strong> that both atoms and charges balance on each side.\u003C/li>\u003C/ol>\u003Cp>This method works well for reactions that do not occur in aqueous solution or that involve straightforward stoichiometry. For more complex redox equations — especially those in acidic or basic solution — the half-reaction method covered later in this topic is generally more systematic.\u003C/p>",[80],37,[82],{"label":83,"body":84},"Try it: balance iron burning in chlorine","\u003Cp>Balance Fe + Cl\u003Csub>2\u003C/sub> → FeCl\u003Csub>3\u003C/sub> by the electron-counting method, then check below.\u003C/p>\u003Cp>\u003Cstrong>Answer:\u003C/strong> Iron goes from 0 to +3 (loses 3 electrons per atom); each chlorine goes from 0 to −1 (gains 1 electron per atom, 2 electrons per Cl\u003Csub>2\u003C/sub> molecule). Total electrons lost must equal total gained: the least common multiple of 3 and 2 is 6, so use 2 Fe (giving 6 e\u003Csup>−\u003C/sup>) and 3 Cl\u003Csub>2\u003C/sub> (taking 6 e\u003Csup>−\u003C/sup>): \u003Cstrong>2 Fe + 3 Cl\u003Csub>2\u003C/sub> → 2 FeCl\u003Csub>3\u003C/sub>\u003C/strong>. Verify: 2 Fe and 6 Cl on each side. The electron ledger chose the coefficients for you; inspection would have found them eventually, but the ledger never guesses.\u003C/p>",{"heading":86,"content":87,"relatedObjectives":88,"deepDive":90},"Why Use the Half-Reaction Method?","\u003Cp>Many aqueous redox reactions — especially those involving polyatomic ions, water, or H\u003Csup>+\u003C/sup>/OH\u003Csup>−\u003C/sup> — are too complex for inspection or for the oxidation-number method alone. The \u003Cstrong>half-reaction method\u003C/strong> provides a systematic, step-by-step procedure that works for any aqueous redox equation.\u003C/p>\u003Cp>The method exploits a fundamental truth about redox chemistry: oxidation and reduction always occur together, and the total electrons lost must equal the total electrons gained. By separating the two processes, you can balance each independently and then combine them, guaranteeing both atom and charge balance in the final equation.\u003C/p>",[89],47,[91],{"label":92,"body":93},"Why this matters: half-reactions become real in a battery","\u003Cp>Splitting a redox equation into half-reactions can feel like an artificial paper exercise. It is not: a battery is a device that physically enforces the split. The oxidation half runs at one electrode, the reduction half at the other, and the electrons that the equation shows being transferred are forced to travel through an external wire to get from one half to the other. That electron traffic through the wire \u003Cem>is\u003C/em> the electric current powering your phone.\u003C/p>\u003Cp>So each half-reaction you write here is rehearsal for electrochemistry, where the two halves get separate rooms and the transfer becomes measurable, usable energy.\u003C/p>",{"heading":95,"content":96,"relatedObjectives":97},"Identifying Half-Reactions","\u003Cp>The first step is to \u003Cstrong>split the overall reaction into two half-reactions\u003C/strong> — one showing oxidation (electron loss) and one showing reduction (electron gain). Each half-reaction contains one reactant and one product that share the same element.\u003C/p>\u003Cp>To decide which species is oxidized and which is reduced, assign oxidation numbers. The species whose oxidation number \u003Cem>increases\u003C/em> is being oxidized; the species whose oxidation number \u003Cem>decreases\u003C/em> is being reduced.\u003C/p>\u003Cp>Example: In the reaction between Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> and Fe\u003Csup>2+\u003C/sup>, chromium goes from +6 to +3 (reduced) and iron goes from +2 to +3 (oxidized). The half-reactions are:\u003C/p>\u003Cul>\u003Cli>Oxidation: Fe\u003Csup>2+\u003C/sup> → Fe\u003Csup>3+\u003C/sup>\u003C/li>\u003Cli>Reduction: Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → Cr\u003Csup>3+\u003C/sup>\u003C/li>\u003C/ul>",[89],{"heading":99,"content":100,"relatedObjectives":101},"Balancing Atoms in Each Half-Reaction","\u003Cp>Once you have the two half-reactions, balance the atoms in a specific order:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Balance all atoms except O and H first.\u003C/strong> For the chromium half-reaction, there are 2 Cr on the left, so place a coefficient of 2 on Cr\u003Csup>3+\u003C/sup>: Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → 2 Cr\u003Csup>3+\u003C/sup>.\u003C/li>\u003Cli>\u003Cstrong>Balance oxygen by adding H\u003Csub>2\u003C/sub>O.\u003C/strong> There are 7 oxygen atoms on the left and none on the right, so add 7 H\u003Csub>2\u003C/sub>O to the right: Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → 2 Cr\u003Csup>3+\u003C/sup> + 7 H\u003Csub>2\u003C/sub>O.\u003C/li>\u003Cli>\u003Cstrong>Balance hydrogen by adding H\u003Csup>+\u003C/sup>.\u003C/strong> The right side now has 14 H atoms (from 7 H\u003Csub>2\u003C/sub>O), so add 14 H\u003Csup>+\u003C/sup> to the left: 14 H\u003Csup>+\u003C/sup> + Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → 2 Cr\u003Csup>3+\u003C/sup> + 7 H\u003Csub>2\u003C/sub>O.\u003C/li>\u003C/ol>\u003Cp>This order — non-O/H atoms, then oxygen, then hydrogen — ensures you never introduce an imbalance that you have to undo later.\u003C/p>",[89],{"heading":103,"content":104,"relatedObjectives":105,"deepDive":106},"Balancing Charge with Electrons","\u003Cp>After atoms are balanced, the two sides of each half-reaction will usually have unequal total charges. You fix this by \u003Cstrong>adding electrons to the more positive side\u003C/strong>.\u003C/p>\u003Cp>For the iron half-reaction: Fe\u003Csup>2+\u003C/sup> → Fe\u003Csup>3+\u003C/sup>. The left side is 2+ and the right is 3+. Adding one electron to the right gives 3+ + 1− = 2+, matching the left: Fe\u003Csup>2+\u003C/sup> → Fe\u003Csup>3+\u003C/sup> + e\u003Csup>−\u003C/sup>.\u003C/p>\u003Cp>For the chromium half-reaction: 14 H\u003Csup>+\u003C/sup> + Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → 2 Cr\u003Csup>3+\u003C/sup> + 7 H\u003Csub>2\u003C/sub>O. Left side charge: 14(+1) + (2−) = +12. Right side charge: 2(+3) = +6. Adding 6 e\u003Csup>−\u003C/sup> to the left gives +12 + 6(−1) = +6, balancing the charge.\u003C/p>\u003Cp>\u003Cstrong>Key rule:\u003C/strong> electrons appear on the product side of oxidation half-reactions (electrons are lost) and on the reactant side of reduction half-reactions (electrons are gained).\u003C/p>",[89],[107],{"label":108,"body":109},"Common mistake: balancing charge to zero instead of to equal","\u003Cp>Charge balance means the two sides carry the \u003Cem>same\u003C/em> total charge, not that either side reaches zero. In Fe\u003Csup>2+\u003C/sup> → Fe\u003Csup>3+\u003C/sup> + e\u003Csup>−\u003C/sup>, both sides total 2+, and that is perfectly balanced; nothing about the half-reaction needs to vanish to neutrality.\u003C/p>\u003Cp>The related slip is adding electrons to the more negative side, which drives the sides further apart. One habit prevents both errors: compute each side’s total charge first, write the two numbers down, and add electrons to the more positive side until the numbers match. If you find yourself aiming at zero, you have switched problems without noticing.\u003C/p>",{"heading":111,"content":112,"relatedObjectives":113},"Equalizing and Combining Half-Reactions","\u003Cp>The two half-reactions will usually show different numbers of electrons. Since electrons are transferred, not created or destroyed, you must \u003Cstrong>multiply each half-reaction by an integer\u003C/strong> so the electron counts match.\u003C/p>\u003Cp>The iron half-reaction transfers 1 e\u003Csup>−\u003C/sup> and the chromium half-reaction transfers 6 e\u003Csup>−\u003C/sup>. Multiply the iron half-reaction by 6:\u003C/p>\u003Cul>\u003Cli>6 Fe\u003Csup>2+\u003C/sup> → 6 Fe\u003Csup>3+\u003C/sup> + 6 e\u003Csup>−\u003C/sup>\u003C/li>\u003Cli>6 e\u003Csup>−\u003C/sup> + 14 H\u003Csup>+\u003C/sup> + Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup> → 2 Cr\u003Csup>3+\u003C/sup> + 7 H\u003Csub>2\u003C/sub>O\u003C/li>\u003C/ul>\u003Cp>Now add the two half-reactions. The 6 electrons on the left cancel the 6 electrons on the right, giving the balanced equation:\u003C/p>\u003Cp>14 H\u003Csup>+\u003C/sup>(aq) + Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup>(aq) + 6 Fe\u003Csup>2+\u003C/sup>(aq) → 2 Cr\u003Csup>3+\u003C/sup>(aq) + 6 Fe\u003Csup>3+\u003C/sup>(aq) + 7 H\u003Csub>2\u003C/sub>O(l)\u003C/p>",[89],{"heading":115,"content":116,"relatedObjectives":117},"Adapting for Basic Solution","\u003Cp>The steps above work directly for \u003Cstrong>acidic solution\u003C/strong> (where H\u003Csup>+\u003C/sup> is available). For reactions in \u003Cstrong>basic solution\u003C/strong>, two equivalent approaches give the same balanced equation:\u003C/p>\u003Cp>\u003Cstrong>Approach 1: Balance acidic, then convert.\u003C/strong>\u003C/p>\u003Col>\u003Cli>Balance the equation as though it were in acidic solution (using H\u003Csup>+\u003C/sup> and H\u003Csub>2\u003C/sub>O).\u003C/li>\u003Cli>Count the H\u003Csup>+\u003C/sup> ions in the equation. Add that same number of OH\u003Csup>−\u003C/sup> ions to \u003Cem>both\u003C/em> sides.\u003C/li>\u003Cli>On the side where H\u003Csup>+\u003C/sup> and OH\u003Csup>−\u003C/sup> appear together, combine them into H\u003Csub>2\u003C/sub>O.\u003C/li>\u003Cli>Cancel any H\u003Csub>2\u003C/sub>O molecules that appear on both sides.\u003C/li>\u003C/ol>\u003Cp>This conversion works because adding equal amounts of OH\u003Csup>−\u003C/sup> to both sides does not change the equation’s balance — it simply replaces the acidic H\u003Csup>+\u003C/sup> with the basic-solution species H\u003Csub>2\u003C/sub>O and OH\u003Csup>−\u003C/sup>.\u003C/p>\u003Cp>\u003Cstrong>Approach 2: Direct basic-solution shortcut\u003C/strong> (produces the same balanced equation as Approach 1). Instead of routing through H\u003Csup>+\u003C/sup>, balance oxygen directly in basic medium using OH\u003Csup>−\u003C/sup> and H\u003Csub>2\u003C/sub>O:\u003C/p>\u003Cul>\u003Cli>To add \u003Cem>n\u003C/em> oxygen atoms to a deficient side, add \u003Cstrong>2\u003Cem>n\u003C/em> OH\u003Csup>−\u003C/sup>\u003C/strong> to that side AND \u003Cstrong>\u003Cem>n\u003C/em> H\u003Csub>2\u003C/sub>O\u003C/strong> to the opposite side. (Each pair of OH\u003Csup>−\u003C/sup> on one side balances with one H\u003Csub>2\u003C/sub>O on the other side, and contributes 1 O atom net to the original side.)\u003C/li>\u003Cli>Then balance charge by adding electrons to the more positive side, exactly as in the acidic procedure.\u003C/li>\u003C/ul>\u003Cp>Worked example: balance Mn\u003Csup>2+\u003C/sup> → MnO\u003Csub>2\u003C/sub> in basic solution (oxidation half). Need 2 O on the right. Add 4 OH\u003Csup>−\u003C/sup> to the left and 2 H\u003Csub>2\u003C/sub>O to the right: 4 OH\u003Csup>−\u003C/sup> + Mn\u003Csup>2+\u003C/sup> → MnO\u003Csub>2\u003C/sub> + 2 H\u003Csub>2\u003C/sub>O. Charge: left = 4(−1) + (+2) = −2; right = 0. Add 2 e\u003Csup>−\u003C/sup> to the right (less negative): 4 OH\u003Csup>−\u003C/sup> + Mn\u003Csup>2+\u003C/sup> → MnO\u003Csub>2\u003C/sub> + 2 H\u003Csub>2\u003C/sub>O + 2 e\u003Csup>−\u003C/sup>.\u003C/p>\u003Cp>Both approaches give the same balanced equation; pick whichever pattern you find easier to track.\u003C/p>",[89],{"heading":119,"content":120,"relatedObjectives":121},"Verification, Common Errors, and Quick Checks","\u003Cp>Always verify your final balanced equation by checking \u003Cem>both\u003C/em> atom balance and charge balance on each side. A correctly balanced redox equation satisfies:\u003C/p>\u003Cul>\u003Cli>Same number of each type of atom on both sides\u003C/li>\u003Cli>Same total charge on both sides\u003C/li>\u003Cli>All electrons cancelled (none remaining in the final equation)\u003C/li>\u003C/ul>\u003Cp>\u003Cstrong>Common mistakes to watch for:\u003C/strong>\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Forgetting subscripts:\u003C/strong> In Cr\u003Csub>2\u003C/sub>O\u003Csub>7\u003C/sub>\u003Csup>2−\u003C/sup>, there are \u003Cem>two\u003C/em> Cr atoms and \u003Cem>seven\u003C/em> O atoms. The total oxidation-number sum equals the ion charge (−2), not zero.\u003C/li>\u003Cli>\u003Cstrong>Confusing charge and oxidation number:\u003C/strong> Write +3 for an oxidation state but 3+ for an ionic charge. Mixing these up causes sign errors in algebra.\u003C/li>\u003Cli>\u003Cstrong>Adding electrons to the wrong side:\u003C/strong> electrons go to the more positive side.\u003C/li>\u003Cli>\u003Cstrong>Failing to multiply half-reactions\u003C/strong> so electron counts match before adding.\u003C/li>\u003Cli>\u003Cstrong>In basic solution, forgetting to add OH\u003Csup>−\u003C/sup> to \u003Cem>both\u003C/em> sides\u003C/strong> — adding to only one side destroys the balance.\u003C/li>\u003Cli>\u003Cstrong>Not cancelling H\u003Csub>2\u003C/sub>O\u003C/strong> molecules that appear on both sides after the basic-solution conversion.\u003C/li>\u003C/ul>\u003Cp>A quick reasonableness check: metals tend to be oxidized (they lose electrons and form cations), while electronegative non-metals such as O\u003Csub>2\u003C/sub> and halogens tend to be reduced.\u003C/p>",[89],{"title":123,"problem":124,"steps":125,"answer":133},"Balancing a Redox Reaction in Acidic Solution","Balance the reaction: Cr₂O₇²⁻(aq) + Fe²⁺(aq) → Cr³⁺(aq) + Fe³⁺(aq) in acidic solution.",[126,127,128,129,130,131,132],"Write two half-reactions: Oxidation: Fe²⁺ → Fe³⁺. Reduction: Cr₂O₇²⁻ → Cr³⁺.","Balance non-O/H atoms: The Cr half-reaction needs a coefficient of 2 on Cr³⁺: Cr₂O₇²⁻ → 2 Cr³⁺.","Balance O with H₂O: 7 O on the left, so add 7 H₂O to the right: Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O.","Balance H with H⁺: 14 H on the right, so add 14 H⁺ to the left: 14 H⁺ + Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O.","Balance charge with electrons: Left side charge = 14(+1) + (−2) = +12. Right side charge = 2(+3) = +6. Add 6 e⁻ to the left: 6 e⁻ + 14 H⁺ + Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O. For Fe: Fe²⁺ → Fe³⁺ + e⁻.","Equalize electrons: Multiply the Fe half-reaction by 6: 6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻.","Add half-reactions and cancel 6 e⁻: 14 H⁺ + Cr₂O₇²⁻ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O.","Balanced equation: 14 H⁺(aq) + Cr₂O₇²⁻(aq) + 6 Fe²⁺(aq) → 2 Cr³⁺(aq) + 6 Fe³⁺(aq) + 7 H₂O(l). Check: Cr 2=2 ✓, Fe 6=6 ✓, O 7=7 ✓, H 14=14 ✓, charge 24+=24+ ✓.",[135],"OpenStax Chemistry 2e, Ch 4.2: Classifying Chemical Reactions — Oxidation-Reduction Reactions (CC BY 4.0)",[137,139,142,145,147,149,151,153,155,157,159],{"question":138},"What is an oxidation number (oxidation state)?",{"question":140,"hint":141},"What are the rules for assigning oxidation numbers?","Start with the simplest rules: free elements, monatomic ions, then common assignments for O and H.",{"question":143,"hint":144},"How do you decide whether a reaction is redox or non-redox?","Look for at least one atom whose oxidation number changes.",{"question":146},"What is oxidation and what happens to the oxidation number during oxidation?",{"question":148},"What is reduction and what happens to the oxidation number during reduction?",{"question":150},"What is an oxidizing agent and what is a reducing agent?",{"question":152},"What is a disproportionation reaction?",{"question":154},"What is the oxidation-number method for balancing redox equations and when is it appropriate?",{"question":156},"What is a half-reaction and why is the half-reaction method useful?",{"question":158},"What must always be true about the total electrons lost and gained in a balanced redox equation?",{"question":160},"How does balancing in basic solution differ from balancing in acidic solution?",[162,163,164],"predicting-and-balancing-chemical-reactions","chemical-reactions-in-solution","electrochemistry",[166,167,168,169,170],"Fe","Mn","Cr","Cu","Zn",[172,173],"activity-series","standard-reduction-potentials",[175,179,183],{"label":176,"equation":177,"note":178},"Oxidation Number Sum (neutral)","sum of all oxidation numbers = 0","Applies to any neutral molecule or formula unit",{"label":180,"equation":181,"note":182},"Oxidation Number Sum (ion)","sum of all oxidation numbers = charge of ion","Applies to polyatomic ions",{"label":184,"equation":185,"note":186},"Electron Balance (overall)","total e- lost = total e- gained","True in every balanced redox equation",{"title":188,"steps":189},"How to Balance a Redox Equation in Acidic Solution Using Half-Reactions",[190,191,192,193,194,195,196],"Assign oxidation numbers to identify which element is oxidized and which is reduced.","Write two separate half-reactions: one for oxidation and one for reduction.","Balance all atoms except O and H in each half-reaction.","Balance oxygen atoms by adding H2O molecules to the side that needs oxygen.","Balance hydrogen atoms by adding H+ ions to the side that needs hydrogen.","Balance the charge by adding electrons (e-) to the more positive side of each half-reaction.","Multiply each half-reaction by an integer so that the total electrons lost equals the total electrons gained, then add the two half-reactions and cancel species that appear on both sides.",{"question":198,"answer":199,"type":200},"A student balances a redox equation in acidic solution and gets the correct answer, then is told the reaction actually occurs in basic solution. Can they convert their answer, or must they start over? Explain.","They can convert without starting over. To adapt an acidic-solution answer for basic solution, add OH- ions equal in number to the H+ ions to both sides of the equation. On the side where H+ and OH- appear together, combine them into H2O. Then cancel any water molecules that appear on both sides. This works because adding equal amounts of OH- to both sides does not change the overall balance — it simply replaces H+ with the basic-solution species H2O and OH-.","conceptual",[202,205],{"id":203,"problem":204,"type":200},"pt-7-1","Assign oxidation numbers to every atom in the following: (a) KMnO4, (b) Cr2O7 2-, (c) H2SO4, (d) Na2O2.",{"id":206,"problem":207,"type":208},"pt-7-2","Balance the following redox reaction in acidic solution using the half-reaction method: MnO4-(aq) + C2O4 2-(aq) -> Mn2+(aq) + CO2(g).","calculation",{"workedExampleCount":210,"hasWorksheets":211},12,true,[213,214,215,216,217,218],"redox reactions","oxidation numbers","oxidizing agent","reducing agent","half-reaction method","balancing redox equations",6,{"id":219,"slug":163,"lesson":219,"title":221,"shortTitle":222,"description":223,"category":10,"objectiveCount":5,"problemCount":224},"Chemical Reactions in Solution","Reactions in Solution","Understand reactions in aqueous solution: solubility rules, electrolytes, dissociation, ionic equations, precipitation, acid-base, and gas-forming reactions.",52,{"id":226,"slug":227,"lesson":226,"title":228,"shortTitle":229,"description":230,"category":231,"objectiveCount":5,"problemCount":232},8,"the-mole-and-chemical-formulas","The Mole and Chemical Formulas","Moles & Composition","Understand the mole concept, molar mass, percent composition, empirical formulas, and molecular formulas.","quantitative",46,1785108608053]