Balancing Redox Reactions Occurring in Acidic Solution by the Half-Reaction Method 1) Assign oxidation numbers to those species that change their oxidation states. 2) Write the unbalanced half reactions, for each half reaction do the following: a) Balance all atoms except O and H b) Balance oxygen atoms by adding H2O to one side of the half reaction c) Balance hydrogen atoms by adding H+ to one side of the half reaction d) Balance the charge by adding electrons to one side of the half reaction 3) If needed, multiply one or both balanced half-reactions by integers to make the number of electrons lost in the oxidation half reaction equal to the number of electrons gained in the reduction half reaction. 4) Add the half-reactions, the electrons should cancel out. If H+ , OH– , or H2O appear on both sides of the final equation, cancel out the duplications. 5) Check that the elements and the charges on both sides of the equation are balanced. Balancing Redox Reactions Occurring in Basic Solution by the Half-Reaction Method 1) Balance the oxidation and reduction half reactions as if it occurs in acidic solution (Steps 1–5) 2) To both sides of the equation add the number of OH– ions equal to the number of H+ ions in the equation. 3) Combine the OH– and H+ ions that appear on the same side of a reaction to form H2O molecules. 4) If H2O appears on both sides of the final equation, cancel out the duplications. 5) Check that the elements and the charges on both sides of the equation are balanced. Examples: 1. Balance the redox reaction, Br2 + SO2 Br– + SO42– in acidic solution using the half reaction method. 2. Balance Al(s) + NO2– NH3 + AlO2– in basic solution using the half reaction method. Answer: The answer to question 1 is worked out in detail on the next page. Try to answer question 2 yourself. Check your Answer: OH– + 2Al + H2O + NO2– 2 AlO2– + NH3Example 1: Balance the redox reaction, Br2 + SO2 Br– + SO42– in acidic solution using the half reaction method. 1) Assign oxidation numbers to those species that change their oxidation states. Br2 + SO2 Br– + SO42– 0 +4 –1 +6 In this reaction the Br changes oxidation state from 0 in Br2 to –1 in Br– and S changes from +4 in SO2 to +6 in SO42–. Thus, Br2 gains electrons (each Br goes from a charge of 0 + e– –1) (Br2 is reduced) and the S in SO2 loses electrons (+4 +6 + 2e– ), SO2 is oxidized. 2) Write the unbalanced half reactions, for each half reaction do the following: a) Balance all atoms except O and H Br2 2 Br– SO2 SO42– b) Balance oxygen atoms by adding H2O to one side of the half reaction Br2 2 Br– 2H2O + SO2 SO42– c) Balance hydrogen atoms by adding H+ to one side of each half reaction Br2 2 Br– 2H2O + SO2 SO42– + 4H+ d) Balance the charge by adding electrons to one side of each half reaction 2 e– + Br2 2 Br– 2H2O + SO2 SO42– + 4H+ + 2 e– 3) If needed, multiply one or both balanced half-reactions by integers to make the number of electrons lost in the oxidation half reaction equal to the number of electrons gained in the reduction half reaction. Done. 2 electrons are gained in the reduction reaction which equals the 2 electrons lost in the oxidation reaction. 4) Add the half-reactions, the electrons should cancel out. If H+ , OH– , or H2O appear on both sides of the final equation, cancel out the duplications. 2 e– + Br2 2 Br– 2H2O + SO2 SO42– + 4H+ + 2 e– _______________________________________________________________________________________ 2H2O + SO2 + Br2 2 Br– + SO42– + 4H+ 5) Check to make sure the equation is balanced. Reactants: 4 H, 1 S, 4 O, 2 Br, charge = 0 Products: 4 H, 1 S, 4 O, 2 Br, charge = 2(-1) + (-2) + 4(+1) =
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