Balancing Redox Equations Identifying if a redox reaction has occurred – In order for a reaction to be classified as a redox reaction there must be a change.

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Balancing Redox Equations Identifying if a redox reaction has occurred – In order for a reaction to be classified as a redox reaction there must be a change in the oxidation numbers of the Combination reactions involving at least 1 element in the reactants or products – if compounds combine to form a new compound it is not classified as a redox reaction NH 3 + HCl  NH 4 Cl (in this reaction the oxidation numbers of the elements do not change) Decomposition reactions involving at least 1 element in the products – if a compound decomposes to form a new compounds it is not classified as a redox reaction CaCO 3  CaO + CO 2 All single displacement reactions involve an element and are considered redox reactions Combustion reactions are redox reactions as they all have elemental O 2 in the reactants Double displacement reactions are not considered redox reactions

Balancing - 1st method using oxidation number changes: easy! best method for balancing reactions where the elements in the reactants and products occur in a single form on each side of the reaction – not for reactions involving acids and bases 1.write a skeleton reaction (no coefficients) Fe 2 O 3 + CO  Fe + CO 2 2.assign oxidation numbers to the elements in the equation Fe (+3)  Fe (0) reduction (-3 change) C (+2)  C (+4) oxidation (+2 change) O remains -2 on each side of the reaction 3.balance the oxidation and reduction charges to account for all of the electrons 2 x (-3 Fe ) = 3 x (+2 C ) 4.balance the equation by balancing the charges Fe 2 O 3 + 3CO  2Fe + 3CO 2

2nd method using 1/2 reactions a half reaction is a reaction that shows just the oxidation or reduction that takes place in the reaction 1.write a skeleton reaction (no coeficients) S (s) + HNO 3(aq)  SO 2(g) + NO (g) +H 2 O (l) (takes place in an acidic solution) S + H + + NO 3 -  SO 2 + NO + H 2 O 2.separate into 1/2 reactions S (0)  SO 2 (+4) oxidation NO 3 - (+5)  NO (+2) reduction – Note: O remains -2 and H + remains +1 on both sides of the equation and thus are not included in the 1/2 reactions

3.balance the 1/2 reactions by adding H 2 O then H + Oxidation: 2H 2 O + S  SO 2 (add the appropriate amount of water 1st) (balance with the appropriate amount of H + 2nd) 2H 2 O + S  SO 2 +4H + Reduction: NO 3 -  NO + 2H 2 O 4H + + NO 3 -  NO + 2H 2 O Add electrons to the sides of the 1/2 reactions to balance the charges 2H 2 O + S  SO 2 +4H + + 4e- 3e- + 4H + + NO 3 -  NO + 2H 2 O (note the neg. charge on the NO 3 - )

4. balance the charges on the 2 1/2 reactions 3 x (4e-) = 4 x (3e-) 3 x (2H 2 O + S  SO 2 +4H + + 4e- ) 4 x (3e- + 4H + + NO 3 - g NO + 2H 2 O) 6H 2 O + 3S  3SO 2 +12H e- 12e- + 16H + + 4NO 3 -  4NO + 8H 2 O 5. combine the 1/2 reactions 6H 2 O + 3S + 12e- + 16H + + 4NO 3 -  3SO 2 +12H e- + 4NO + 8H 2 O 6. subtract any items that are present on both sides of the equation and reduce to balance the overall equation 3S (s) + 4HNO 3 - (aq)  3SO 2(g) + 4NO (g) + 2H 2 O (l)