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Balancing acidic redox eq

Using half reaction method balancing acidic redox equations.

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Balancing acidic redox eq

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  1. Balancing acidic redox equation By RA UNIS 2020

  2. ½ reactions following “SOHE” • Divide the equations into two ½ reactions. • S = Species ; balance elements other than oxygen and hydrogen first. • O = oxygen ; balance oxygen atoms using H2O. • H = hydrogen ; balance hydrogen atoms using H+. • E = electrons ; balance charges with electrons.

  3. Step 1 Cr2O72- + Mn2+ Cr3+ + MnO4- ½ reactions and balancing species: • Cr2O72-  2Cr3+ • Mn2+  MnO4-

  4. Step 2:balance Oxygen with H2O • Cr2O72-  2Cr3+ + 7H2O • 4H2O +Mn2+  MnO4-

  5. Step 3: balance hydrogen with H+ • 14H++ Cr2O72-  2Cr3+ + 7H2O • 4H2O +Mn2+  MnO4- + 8H+

  6. Step 4: balance the charges with electrons • 6e-+14H+ + Cr2O72-  2Cr3+ + 7H2O • 4H2O +Mn2+  MnO4- + 8H+ + 5e- Next electrons gained in the reduction reaction need to equal electrons lost in the oxidation reaction.

  7. Multiply first equation by 5 and the second equation by 6. Then you have 30 electrons gained and 30 electrons lost and we can cancel them out. • 5x (6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O) • 6x (4H2O +Mn2+  MnO4- + 8H+ + 5e-) ==> 30e-+ 70H+ + 5Cr2O72- 10 Cr3+ + 35H2O 24 H2O + 6Mn2+  6MnO4- + 48H+ + 30e-

  8. Add the two half equations to have a final redox reaction. 30e-+ 70H+ + 5Cr2O72- 10 Cr3+ + 35H2O 24 H2O + 6Mn2+  6MnO4- + 48H+ + 30e- ==> 70H+ + 5Cr2O72-+24 H2O + 6Mn2+ 10 Cr3+ + 35H2O +6MnO4- + 48H+

  9. Final step is cleaning up: Subtract excess H+ and H2O • 2270H+ + 5Cr2O72- +24 H2O + 6Mn2+ 10 Cr3+ + 1135H2O +6MnO4- + 48H+ 22H+ + 5Cr2O72-+ 6Mn2+ 10 Cr3+ + 11H2O + 6MnO4-

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