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Concept 01 / 04Redox Reactions

Oxidation and Reduction: Electron Transfer and Oxidation Numbers

Classical and electronic definitions of oxidation and reduction, oxidising and reducing agents, rules for assigning oxidation numbers, fractional and average oxidation states, oxidation states from structures (peroxide linkages), and the range of oxidation states of an element.

Figure 1
Classical definitions. Oxidation: addition of oxygen or an electronegative element, or removal of hydrogen or an electropositive element. Reduction: the reverse. For example, 2Mg+O2→2MgO\mathrm{2Mg + O_{2} \rightarrow 2MgO} oxidises magnesium; CuO+H2→Cu+H2O\mathrm{CuO + H_{2} \rightarrow Cu + H_{2}O} reduces copper oxide. Electronic definitions. Oxidation is loss of electrons; reduction is gain of electrons. They always occur together (redox). In Zn+Cu2+→Zn2++Cu\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}, zinc loses 2e−2e^{-} (oxidised) and Cu2+\mathrm{Cu^{2+}} gains them (reduced). Oxidising agent (oxidant): accepts electrons and is itself reduced. Reducing agent (reductant): donates electrons and is itself oxidised. Oxidation number (ON). The charge an atom would carry if all its bonds were completely ionic, with shared electrons assigned to the more electronegative atom. Rules: 1. Free elements: 00 (including O2\mathrm{O_{2}}, P4\mathrm{P_{4}}, S8\mathrm{S_{8}}). 2. Monatomic ions: equal to the charge. 3. Fluorine: always −1-1 in compounds. 4. Oxygen: −2-2, except peroxides (−1-1, as in H2O2\mathrm{H_{2}O_{2}}, Na2O2\mathrm{Na_{2}O_{2}}), superoxides (−12-\frac{1}{2}, as in KO2\mathrm{KO_{2}}), OF2\mathrm{OF_{2}} (+2+2) and O2F2\mathrm{O_{2}F_{2}} (+1+1). 5. Hydrogen: +1+1, except in metal hydrides (−1-1, as in NaH, CaH2\mathrm{CaH_{2}}). 6. Alkali metals +1+1; alkaline earth metals +2+2. 7. The sum of ONs equals 00 for a neutral species or the charge for an ion. Average and fractional ONs. Rule 7 gives an average when identical atoms are in different environments: Fe3O4\mathrm{Fe_{3}O_{4}} (+83+\frac{8}{3}, really one Fe2+\mathrm{Fe^{2+}} and two Fe3+\mathrm{Fe^{3+}}); S4O62−\mathrm{S_{4}O_{6}^{2-}} (+2.5+2.5, really two S at +5+5 and two bridging S at 00); C3O2\mathrm{C_{3}O_{2}} (+43+\frac{4}{3}); Br3O8\mathrm{Br_{3}O_{8}} (+163+\frac{16}{3}). ON from structure. When a molecule contains peroxide (O−O\mathrm{O-O}) linkages, those oxygens are −1-1, not −2-2. Examples: CrO5\mathrm{CrO_{5}} (butterfly structure with two peroxide groups) has Cr at +6+6, not +10+10; H2SO5\mathrm{H_{2}SO_{5}} (peroxomonosulfuric acid) has S at +6+6, not +8+8; H2S2O8\mathrm{H_{2}S_{2}O_{8}} has S at +6+6. In organic chemistry, count each bond to a more electronegative atom as +1+1 and to a less electronegative atom as −1-1 for that carbon. Range of ONs. The highest ON of a main group element is usually its group valence electrons (N: +5+5), the lowest is that number minus 88 (N: −3-3) (see figure). A species with the element in its highest state can act only as an oxidant (HNO3\mathrm{HNO_{3}}, KMnO4\mathrm{KMnO_{4}}); in its lowest state only as a reductant (NH3\mathrm{NH_{3}}, H2S\mathrm{H_{2}S}); in between, either (SO2\mathrm{SO_{2}}, H2O2\mathrm{H_{2}O_{2}}, HNO2\mathrm{HNO_{2}}). Key Takeaways: - Oxidation: loss of electrons, ON increases; reduction: gain of electrons, ON decreases. - The oxidant is reduced; the reductant is oxidised. - F is always −1-1; O is −2-2 except in peroxides, superoxides and fluorides. - Use structures when peroxide bonds are present (CrO5\mathrm{CrO_{5}}: +6+6). - Pitfall: a fractional ON is an average; no individual atom has a fractional oxidation state. Brain-Teaser 1: Find the oxidation number of Cr in K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}}, Mn in KMnO4\mathrm{KMnO_{4}}, S in Na2S2O3\mathrm{Na_{2}S_{2}O_{3}}, Fe in K4[Fe(CN)6]\mathrm{K_{4}[Fe(CN)_{6}]}. Solution: 2(+1)+2x+7(−2)=02(+1) + 2x + 7(-2) = 0 gives Cr +6+6. +1+x−8=0+1 + x - 8 = 0 gives Mn +7+7. 2+2x−6=02 + 2x - 6 = 0 gives average S +2+2. 4+x+6(−1)=04 + x + 6(-1) = 0 (cyanide is −1-1) gives Fe +2+2. Brain-Teaser 2: Why is the oxidation state of Cr in CrO5\mathrm{CrO_{5}} taken as +6+6 and not +10+10? Solution: CrO5\mathrm{CrO_{5}} has one Cr=O oxygen and four oxygens in two peroxide (O−O\mathrm{O-O}) groups. So x+(−2)+4(−1)=0x + (-2) + 4(-1) = 0, giving x=+6x = +6. The value +10+10 would exceed chromium's six valence electrons (3d54s13d^{5}4s^{1}), which is impossible. Brain-Teaser 3: Which of these can act both as oxidant and reductant: H2S\mathrm{H_{2}S}, SO2\mathrm{SO_{2}}, H2SO4\mathrm{H_{2}SO_{4}}, H2O2\mathrm{H_{2}O_{2}}? Solution: SO2\mathrm{SO_{2}} (S at +4+4, between −2-2 and +6+6) and H2O2\mathrm{H_{2}O_{2}} (O at −1-1, between −2-2 and 00). H2S\mathrm{H_{2}S} (S at −2-2, minimum) is only a reductant; H2SO4\mathrm{H_{2}SO_{4}} (S at +6+6, maximum) only an oxidant.

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