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Oxidation state
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=== Balancing redox === Oxidation states can be useful for balancing chemical equations for oxidation-reduction (or [[redox]]) reactions, because the changes in the oxidized atoms have to be balanced by the changes in the reduced atoms. For example, in the reaction of [[acetaldehyde]] with [[Tollens' reagent]] to form [[acetic acid]] (shown below), the [[carbonyl]] carbon atom changes its oxidation state from +1 to +3 (loses two electrons). This oxidation is balanced by reducing two {{chem2|Ag+}} cations to {{chem2|Ag^{0} }} (gaining two electrons in total). :[[File:Redox eqn 1.svg|600px]] An inorganic example is the Bettendorf reaction using [[tin dichloride]] ({{chem2|SnCl2}}) to prove the presence of [[arsenite]] ions in a concentrated [[hydrochloric acid|HCl]] extract. When arsenic(III) is present, a brown coloration appears forming a dark precipitate of [[arsenic]], according to the following simplified reaction: :{{chem2|2 As^{3+} + 3 Sn^{2+} -> 2 As^{0} + 3 Sn^{4+} }} Here three [[tin]] atoms are oxidized from oxidation state +2 to +4, yielding six electrons that reduce two arsenic atoms from oxidation state +3 to 0. The simple one-line balancing goes as follows: the two redox couples are written down as they react; :{{chem2|As^{3+} + Sn^{2+} <-> As^{0} + Sn^{4+} }} One tin is oxidized from oxidation state +2 to +4, a two-electron step, hence 2 is written in front of the two arsenic partners. One arsenic is reduced from +3 to 0, a three-electron step, hence 3 goes in front of the two tin partners. An alternative three-line procedure is to write separately the [[half-reaction]]s for oxidation and reduction, each balanced with electrons, and then to sum them up such that the electrons cross out. In general, these redox balances (the one-line balance or each half-reaction) need to be checked for the ionic and electron charge sums on both sides of the equation being indeed equal. If they are not equal, suitable ions are added to balance the charges and the non-redox elemental balance.
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