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Writing ion electron half equations….

No Brain Too Small…. Writing ion electron half equations…. ….and balanced redox equations. …in acidic and neutral solutions. (……. polyatomic ions or molecules are involved….). Redox rxns involving polyatomic ions or molecules. permanganate = manganate (VII). MnO 4 - Cr 2 O 7 2- H 2 O 2

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Writing ion electron half equations….

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  1. No Brain Too Small….. Writing ion electron half equations…. ….and balanced redox equations …in acidic and neutral solutions. (……. polyatomic ions or molecules are involved….)

  2. Redox rxns involving polyatomic ions or molecules. • permanganate = manganate (VII) • MnO4- • Cr2O72- • H2O2 • HSO3- • NO3- • SO2 • MnO2 • dichromate • hydrogen peroxide • hydrogen sulfite • nitrate • sulfur dioxide • manganese dioxide = manganese (IV) oxide

  3. Writing ion – electron half equations – example 1 • An acidified potassium permanganate solution (purple) will turn colourless when iron (II) sulfate is added. • The permanganate ion (MnO4-) reacts with the Fe2+ ion to form Mn2+ and Fe3+, a colourless (or v. pale orange)solution. Red: MnO4-  Mn2+ + 8H+ + 5 e- + 4H2O Ox: Fe2+ Fe3+ + e- (x5) Ox: 5Fe2+ 5Fe3+ + 5e- Overall: MnO4- Mn2+ + 8H+ +5Fe2+ + 4H2O +5Fe3+

  4. You don’t have to use them all… but make sure you do them in order The Rulesa few extra steps • Balance the atoms which aren’t O or H. • Add water molecules to the other side to balance any oxygen atoms. • Add H+ ions to the other side to balance the H’s in the water etc. • Balance the charges by adding electrons to the most positive side. • Cancel out terms that appear on both sides and combine.

  5. Example 2 An acidified potassium dichromate solution (orange) will turn green when iron (II) sulfate is added. The dichromate ion (Cr2O72-) reacts with the Fe2+ ion to form Cr3+ (green) and Fe3+. Red: Cr2O72- Cr3+ + 14H+ + 6 e- 2 + 7H2O Ox: Fe2+ Fe3+ + e- (x6) Ox: 6Fe2+ 6Fe3+ + 6e- Overall:Cr2O72-2Cr3+ + 14H+ +6Fe2+ +7H2O +6Fe3+

  6. Example 3 • An acidified hydrogen peroxide solution (colourless) will turn red-orange when potassium bromide solution is added. • The hydrogen peroxide (H2O2) reacts with the Br- ion to form water and Br2(red-orange). Red: H2O2 H2O + 2H+ + 2 e- 2 Ox: Br - Br2 2 + 2e- Overall: H2O2 + 2Br- 2H2O + Br2 + 2H+

  7. Example 4 An acidified potassium dichromate solution (orange) will turn green when sodium hydrogen sulfite is added. The dichromate ion (Cr2O72-) reacts with the hydrogen sulfite (HSO3-) ion to form Cr3+ (green) and SO42- (colourless). 5 4 Red: Cr2O72- Cr3+ + 14H+ + 6 e- 2 + 7H2O (x3) Ox: HSO3- SO42- + H2O + 3H+ + 2e- + 3H2O Ox: 3HSO3- 3SO42- + 9H+ + 6e- Overall:Cr2O72-2Cr3+ + 5H+ +3HSO3- +4H2O +3SO42-

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