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Balancing Redox Reactions

Balancing Redox Reactions. I Conservation of Charge In the reaction Mg +2 + Al  Al +3 + Mg, we have a problem This is seen in the half reactions Mg +2 + 2e-  Mg Al  Al +3 + 3e-

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Balancing Redox Reactions

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  1. Balancing Redox Reactions I Conservation of Charge In the reaction Mg+2 + Al  Al+3 + Mg, we have a problem This is seen in the half reactions Mg+2 + 2e-  Mg Al  Al+3 + 3e- This is not balanced. We must have the same number of electrons lost and gained in a redox reaction. Chemical reactions must have conservation of mass and conservation of charge II Balancing Redox Reactions A. By simple half reactions If there is the same type of charge on both sides, we can use the half reactions to balance the charges. Gains 2 e- Loses 3 e-

  2. Write out the half reactions • Multiply each reaction by a number to make the electrons in both reactions the same. • Add the two reactions together to get the final reaction • Example • Mg+2 + 2e-  Mg • Al  Al+3 + 3e- • 3 Mg+2 + 2 Al  2 Al+3 + 3 Mg • This is now balanced for both charge and mass. • Example in notes 3 Mg+2 + 6e-  3 Mg 2 Al  2 Al+3 + 6e- Multiply by 3 Multiply by 2

  3. B. Ion + Electron Method (Acidic) • Most redox occur in solution. • Use this method for more complex redox reactions • (I- + IO3- I2) • Write both half reactions • Balance all atoms except H and O • Balance O atoms by adding H2O to the opposite side • Balance H atoms by adding H+ to the opposite side • Add the proper number of electrons to one side to make both sides of each half reaction have the same charge. • Multiply each half reaction to get the same number of electrons (Just like in other method), then add the half reactions.

  4. Example Cr2O7-2(aq) + Cl-(aq) Cr+3(aq) + Cl2(g) Cr2O7-2  Cr+3 Cr2O7-2  2 Cr+3 14 H+ + Cr2O7-2 + 6 e-  3 Cr+3 + 7 H2O Cl- Cl2 2 Cl-  Cl2 14 H+ + + 7 H2O + 2 e- Multiply by 3 14 H+ + Cr2O7-2 + 6 e-  2 Cr+3 + 7 H2O 6 Cl-  3Cl2 + 6 e- 14 H+ + Cr2O7-2 + 6 Cl- 2 Cr+3 + 7 H2O + 3 Cl2 Do example in notes!

  5. C. Ion – Electron Method (Basic solution) If the reaction occur in a basic solution, perform all the same steps, but add one more. 7. Add enough OH- to EACH side to cancel out the added H+ (This turns the extra H+ on one side into more H2O) Example CN- + MnO4- CNO- + MnO2 CN-  CNO- CN- + H2O  CNO- + 2 H+ + 2 OH- + 2 OH- CN- + 2 OH- + H2O  CNO- + 2 H2O CN- + 2 OH-  CNO- + H2O + 2 e-

  6. MnO4- MnO2 MnO4-  MnO2 + 2 H2O 4 H+ + MnO4-  MnO2 + 2 H2O + 4 OH- + 4 OH- 4 H2O + MnO4-  MnO2 + 2 H2O + 4 OH- 2 H2O + MnO4- + 3 e-  MnO2 + 4 OH- 2 Now multiply each reaction to get the same electrons 2 H2O + MnO4- + 3 e-  MnO2 + 4 OH- CN- + 2 OH-  CNO- + H2O + 2 e- 4 H2O + 2 MnO4- + 6 e- 2 MnO2 + 8 OH- 3 CN- + 6 OH-  3 CNO- + 3 H2O + 6 e- X 2 X 3 2 H2O + 3 CN- + 2 MnO4- 3 CNO- + 2 MnO2 + 2 OH-

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