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Balancing Oxidation Reduction Equations

Balancing Oxidation Reduction Equations. Oxidation and Reduction. Oxidation – the atom loses electrons The charge becomes more positive Reduction – the atom gains electrons The charge becomes more negative Mass and charge must be conserved for all chemical equations.

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Balancing Oxidation Reduction Equations

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  1. Balancing Oxidation Reduction Equations

  2. Oxidation and Reduction • Oxidation – the atom loses electrons • The charge becomes more positive • Reduction – the atom gains electrons • The charge becomes more negative • Mass and charge must be conserved for all chemical equations.

  3. Steps for Balancing Redox reactions • Acidic Solutions Step 1: Write the half reactions for the oxidation and reduction. MnO4- + H2SO3 SO42- + Mn2+ + H2O (Mn goes from Mn7+ to Mn2+ so this is reduction) MnO4- Mn2+

  4. Steps for Balancing Redox reactions • S goes from S4+ to S6+, so this is oxidation H2SO3 SO42- Step 2: Balance the half rxns for atoms other than H and O. • Already balanced for the example. Step 3: Balance the O’s by adding H2O. MnO4-  Mn2+ + 4H2O H2O + H2SO3  SO42-

  5. Steps for Balancing Redox Reactions Step 4: Balance H atoms with H+ MnO4- + 8H+ Mn2+ + 4H2O H2O + H2SO3  SO42- + 4H+ Step 5: Balance the charge on each side by adding electrons. MnO4- + 8H+  Mn2+ + 4H2O total charge = +7  total charge = +2 • So, add 5e- to the reactants side to balance the charge.

  6. Steps for Balancing Redox Reactions 5 e- + MnO4- + 8H+  Mn2+ + 4H2O H2O + H2SO3  SO42- + 4H+ charge = 0 charge = 2+ So, H2O + H2SO3  SO42- + 4H+ + 2e- • Electrons must be added to opposite sides for each half reaction.

  7. Steps for Balancing Redox Reactions Step 6: Balance the electrons between both half reactions. (5 e- + MnO4- + 8H+  Mn2+ + 4H2O) x 2 10 e- + 2MnO4- + 16H+  2Mn2+ + 8H2O (H2O + H2SO3  SO42- + 4H+ + 2e-) x 5 5H2O + 5H2SO3  5SO42- + 20H+ + 10e- • Now the electrons will cancel out

  8. Steps for Balancing Redox Reactions Step 7: Cross cancel the electrons and other substances. 10 e- + 2MnO4- + 16H+  2Mn2+ + 8H2O 5H2O + 5H2SO3  5SO42- + 20H+ + 10e- Step 8: Add the half reactions together 5H2SO4 + 2MnO4-  2Mn2+ + 3H2O + 5SO42- + 4H+ 3 4

  9. Basic Solutions Follow the same steps as for acidic solutions, and add the following steps. Step 9: Add OH- to both sides to equal the H+ 5H2SO3 + 2MnO4-  2Mn2+ + 3H2O + 5SO42- + 4H+ 4 OH-  4OH-

  10. Basic Solutions Step 10: Combine the H+ and OH- to make water. 4 OH- + 5H2SO3 + 2MnO4-  2Mn2+ + 3H2O + 5SO42- + 4H2O Step 11: Cancel the H2O if they are on both sides. 4 OH- + 5H2SO3 + 2MnO4-  2Mn2+ + 7H2O + 5SO42-

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