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Balancing Redox Equations (continued):

Balancing Redox Equations (continued):. Balancing equations using oxidation numbers. Oxidation Numbers (O.N.):.

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Balancing Redox Equations (continued):

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  1. Balancing Redox Equations (continued): • Balancing equations using oxidation numbers. Oxidation Numbers (O.N.): • An oxidation number is the real or apparent charge on an atom or ion when all bonds are assumed to be ionic. (Always assume that electrons are held by the more electronegative element.) eg: If we assume that H2O is an ionic compound, then the apparent charge or O.N. of each of the elements is: +1 for H and -2 for O (H2O is composed of 2 H+and O2-)

  2. eg: What are the O.N.s of C and Cl in CCl4? • The chlorine is more electronegative than the carbon. Thus the O.N. of carbon is +4 and the O.N. of chlorine is -1. Determining O.N.s: The O.N. for any atom in its elemental state is zero. eg: Fe(s) O.N. = 0 Cl2(g) O.N. = 0 H2(g) O.N. = 0 The O.N. for many ions is the same as their typical charge. eg: alkali metals O.N. = +1 alkaline earth metals O.N. = +2 halogens (usually) O.N. = -1

  3. H O O H The O.N. for oxygen is -2, except in compounds where it is bonded to itself, then it is -1. eg: O2(s) O.N. = 0 H2O O.N. = -2 H2O2 O.N. = -1 a “peroxide” The O.N. for hydrogen is +1, except when it is combined with metals (metal hydrides) then it is -1. eg: H2(s) O.N. = 0 H2O O.N. = +1 NaH O.N. = -1

  4. The O.N.s are assigned so that the sum of O.N.s equal the net charge on the molecule or polyatomic ion. eg: SO42- • We know O.N. for O is -2,thus the O.N. for S here must be +6. 1(x) + 4(-2) = -2 x = +6 eg: KMnO4 • O.N. for K = +1 • O.N. for O = -2 1(+1) + 1(x) + 4(-2) = 0 x = +7 • O.N. for Mn = +7

  5. eg: How does the O.N. of N change in the following half-reaction? NO3- NO O.N.(N) = +5 O.N.(N) = +2 • In this half-reaction the N goes from O.N. = 5 to O.N. = 2, this means that 3e- were gained and this is a reduction process.

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