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Balancing Redox Equations in Acidic Conditions

Balancing Redox Equations in Acidic Conditions. Take the reaction between potassium permanganate （ KMnO 4 ) and sodium sulfite (NaSO 3 ) 7 steps are required to balance the full ionic equation from 2 separate half equations ( oxidation & reduction ). Step 1.

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Balancing Redox Equations in Acidic Conditions

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  1. Balancing Redox Equations in Acidic Conditions • Take the reaction between potassium permanganate （KMnO4) and sodium sulfite (NaSO3) • 7 steps are required to balance the full ionic equation from 2 separate half equations (oxidation & reduction)

  2. Step 1 • Write 2 half-equations for the reaction. • MnO4-→Mn2+ • SO32-→SO42-

  3. Step 2 • Balance oxygen atoms using H2O • MnO4- → Mn2+ + 4H2O • SO32- + H2O → SO42-

  4. Step 3 • Balance H with H+ • 8H+ + MnO4-→ Mn3+ + 4H2O • SO32- + H2O → SO42- + 2H+

  5. Step 4 • Balance the charges with electrons • 8H+ + MnO4- + 5e-→ Mn2+ 4H2O • SO32- + H2O → SO42- + 2H+ + 2e-

  6. Step 5 • Multiply the 2 half-equations by whole numbers (the lowest common multiple of the 2 stoichiometric coefficients in front of the electrons) so that electrons gained in the reduction reaction equals electrons given out by the oxidation reaction • In this case the reduction reaction needs to be multiplied by 2 while the oxidationreaction needs to be multiplied by 5 to get a common number of electrons 10 • 16H+ + 2MnO4- + 10e-→ 2Mn2+ 8H2O • 5SO32- + 5H2O → 5SO42- + 10H+ + 10e-

  7. Step 6 • Add the 2 half-reactions • 16H+ + 2MnO4- + 10e- + 5SO32- + 5H2O→ 2Mn2+ 8H2O + 5SO42- + 10H+ + 10e- • The electrons cancel on the both sides to give: • 16H+ + 2MnO4- + 5SO32- + 5H2O → 2Mn2+ 8H2O + 5SO42- + 10H+

  8. Step 7 • Subtract H+ & H2O which occur on both sides of the equation • The consumption of 16H+ & the production of 10H+ is equal to a net consumption of 6H+ • The consumption of 5 H2O molecules & the production of 8 H2O molecules is equal to the net production of 3 H2O molecules • 6H+ + 2MnO4- + 5SO32-→ 2Mn2+ 3H2O + 5SO42-

  9. Points to note • The H2O molecules always appear on the right hand-side (RHS) of the reduction (8H+ + MnO4- + 5e- → Mn2+ 4H2O) • reaction but on the left-hand side (LHS) of the oxidation (SO32- + H2O → SO42- + 2H++ 2e-reaction • The H+ ions always appear on the opposite side of H2O molecules in both of the half-equations and the net ionic equation. • 6H+ + 2MnO4- + 5SO32- → 2Mn2+ 3H2O + 5SO42-

  10. Balancing redox equations for neutral or alkaline conditions • The reaction taking place is between potassium permanganate and sodium sulfite to form manganese dioxide (MnO2) • The method used for balancing equations in acidic conditions is used; then 1 OH- is added for every H+ in the equation

  11. Step 1 • Write the 2 half-reactions • MnO4-→ MnO2 • SO32-SO42-

  12. Steps 2-7 • Follow the same steps as steps 2-4 for the reaction in acidic media to get the following 2 half-equations: • 4H+ + MnO4- + 3e-→ MnO2 +2H2O (x2) • SO32- + H2O → SO42- + 2H+ + 2e- (x3) • Multiplying the reduction reaction by 2, the oxidation by 3, adding together and simplifying gives: • 2H+ + 2MnO4- + 3SO32-→ 2MnO2 + H2O + 3SO42-

  13. Step 8 • Add OH- to convert any H+ to H2O. Any OH- added to 1 side of the equation must also be added to the other side. • 2H+ + 2OH- 2MnO4- + 3SO32-→ 2MnO2 + H2O + 3SO42- + 2OH- • Which simplifies to: • 2H2O + 2MnO4- + 3SO32-→ 2MnO2 + H2O + 3SO42- +2 OH- • Which simplifies further to: • H2O + 2MnO4- + 3SO32-→ 2MnO2 + 3SO42- +2 OH-

  14. Balancing Redox equations in strongly alkaline conditions • Take the reaction between potassium permanganate & sodium sulfite in strongly alkaline media • MnO4-→ MnO42- • SO32-SO42- • Following the same steps of balancing the equation in acidic media, then adding OH- for every H+ results in: • 2MnO4- + SO32- + 2OH-→ 2MnO42- + SO42- + H2O

  15. Redox Titrations • Similar to acid-base titrations • Acid-base titration: transfer of 1 or more hydrogen ions (protons) from the acid to the base • Redox Titration: transfer of one/more electrons from a reducing agent to an oxidizing agent

  16. Oxidizing agents for redox titrations • Acidified manganate (VII) ions (permanagate) • 8H+ + MnO4- + 5e-→ Mn2+ 4H2O • MnO4- is purple but Mn2+ is almost colourless

  17. Oxidizing agent for redox titrations 2 • Acidified dichromate (VI) ions • 14H+ + Cr2O72- + 6e- 2Cr3+ + 7H2O • Cr2O72-are orange in colour • Cr3+ is green • Can be used as primary standards (a reagent which is very pure, & can be used to prepare a solution of known concentration)

  18. Some more oxidizing agents • Iron (III) ions/salts • Fe3+ + e- Fe2+ • Iodine: • I2 + 2e-  2I- • I2 is red brown while I- is colourless • Acidified hydrogen peroxide • H2O2 + 2H+ + 2e-  2H2O

  19. Reducing agents for redox titrations • Iron(II) salts/ions • Fe2+  Fe3+ + e- • Hydrogen peroxide if a more powerful oxidizing agent e.g. dichromate (VI) or manganate (VII) is present • H202  2H+ + O2 + 2e- • Iodide ions: 2I-  I2 + 2e- • Sodium thiosulfate (VI): 2S2O32-  S4O62- + 2e-

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