Chemistry · Ch 6 — Redox Reactions
Redox Reaction in Terms of Oxidation Number
Redox Reaction in Terms of Oxidation Number
With the idea of oxidation number in hand, oxidation, reduction, oxidising agent and reducing agent can all be redefined in one general way that covers both ionic and covalent redox reactions. Oxidation is an increase in the oxidation number of an element in a given substance; reduction is a decrease in the oxidation number of an element in a given substance. An oxidising agent is a substance that brings about an increase in some other element's oxidation number, while its own key element's oxidation number itself decreases (it is reduced). A reducing agent is a substance that brings about a decrease in some other element's oxidation number, while its own key element's oxidation number itself increases (it is oxidised). These are exactly the electron-transfer definitions of Section 6.1.2 restated in oxidation-number language, since a rise in oxidation number corresponds to a loss of electrons and a fall corresponds to a gain. To identify whether a given reaction is redox, and to name its oxidant and reductant, the working method is always the same: assign oxidation numbers to every atom in the reactants and products, using the rules of Section 6.2.1, and see which elements' oxidation numbers actually change. If at least one element's oxidation number rises while another's falls, the reaction is redox; the element whose oxidation number rises has been oxidised (so the substance carrying it is the reducing agent), and the element whose oxidation number falls has been reduced (so the substance carrying it is the oxidising agent). Applying this to 3H3AsO3(aq) + BrO3⁻(aq) → Br⁻(aq) + 3H3AsO4(aq): arsenic rises …
Worked out. Worked example finding the oxidation number of the central atom in three compounds using rules 2, 3 and 6. (a) KMnO4: with K = +1 and O = -2, the neutral-molecule rule gives (+1) + ON(Mn) + 4(-2) = 0, so ON(Mn) = +7. (b) K2Cr2O7: with K = +1 and O = -2, 2(+1) + 2 x ON(Cr) + 7(-2) = 0, so 2 x ON(Cr) = +12 and ON(Cr) = +6. (c) Ca3(PO4)2: with Ca = +2 (alkaline-earth ion) and O = -2, 3(+2) + 2 x ON(P) + 8(-2) = 0, so 2 x ON(P) = +10 an …
Worked out. Worked example finding the oxidation number of the element other than oxygen or hydrogen in eight ions/molecules, using O = -2 as the fixed reference. i. SO3²⁻: -2 = ON(S) + 3(-2), so ON(S) = -2 + 6 = +4. ii. BrO3⁻: -1 = ON(Br) + 3(-2), so ON(Br) = -1 + 6 = +5. By the same method the rest work out to: iii. Cl in ClO4⁻ = +7; iv. N in NH4⁺ = -3; v. N in NO3⁻ = +5; vi. N in NO2⁻ = +3; vii. S in SO3 = +6; viii. N in N2O5 = +5. Together these show nitrogen alone spanning oxidation states from -3 up to +5 depending on the …
Worked out. A supplementary note explaining that some elements show a FRACTIONAL oxidation number in certain compounds -- examples given are C3O2 (carbon = +4/3), Br3O8 (bromine = +16/3), Na2S4O6, sodium tetrathionate (sulfur = +2.5) and C8H18, octane (carbon = -9/4). The note explains this fraction is only ever an AVERAGE: different atoms of the same element within one species can genuinely sit at different individual oxidation states, and the fraction is just their mean. This is illustrated with the tetrathionate ion, S4O6²⁻: its four sulfur atoms are not identical -- two central S atoms are joined S-S with oxidation number 0 each, while the two outer S atoms (each bonded to three O) carry oxidation number +5 each -- so the average, (2 x 0 + 2 x 5) / 4 = 10/4 = 2.5, is th …
Worked out. Worked example on 3H3AsO3(aq) + BrO3⁻(aq) → Br⁻(aq) + 3H3AsO4(aq). Assigning oxidation numbers to As and Br before and after: in H3AsO3, ON(As) = +3 (from 3(+1) + ON(As) + 3(-2) = 0); in H3AsO4, ON(As) = +5 (from 3(+1) + ON(As) + 4(-2) = 0). In BrO3⁻, ON(Br) = +5 (from ON(Br) + 3(-2) = -1); in Br⁻, ON(Br) = -1. Since As increases from +3 to +5 (a loss of 2 electrons per As, oxidation) while Br decreases from +5 to -1 (a gain of 6 electrons per Br, reduction), the reaction is confirmed redox. As is the reducing agent (oxidised via H3AsO3), and Br, via BrO3⁻ …