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Writing ion electron half equations…. ….and balanced redox equations (……. polyatomic ions or molecules are involved….) …in acidic and neutral solutions.

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Presentation on theme: "Writing ion electron half equations…. ….and balanced redox equations (……. polyatomic ions or molecules are involved….) …in acidic and neutral solutions."— Presentation transcript:

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

2 MnO 4 - Cr 2 O 7 2- H 2 O 2 HSO 3 - NO 3 - SO 2 MnO 2 Redox rxns involving polyatomic ions or molecules. permanganate = manganate (VII) dichromate hydrogen peroxide hydrogen sulfite nitrate sulfur dioxide manganese dioxide = manganese (IV) oxide

3 An acidified potassium permanganate solution (purple) will turn colourless when iron (II) sulfate is added. Red: MnO 4 -  Mn 2+ Writing ion – electron half equations – example 1 + 4H 2 O+ 8H e - Ox:Fe 2+  Fe 3+ + e - (x5) Ox:5Fe 2+  5Fe e - Overall: MnO 4 -  Mn H 2 O+ 8H + +5Fe 2+ +5Fe 3+ The permanganate ion (MnO 4 - ) reacts with the Fe 2+ ion to form Mn 2+ and Fe 3+, a colourless (or v. pale orange) solution.

4 The Rules a 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. You don’t have to use them all… but make sure you do them in order

5 An acidified potassium dichromate solution (orange) will turn green when iron (II) sulfate is added. Red: Cr 2 O 7 2-  Cr 3+ Example 2 + 7H 2 O+ 14H e - (x6) 2 Overall:Cr 2 O 7 2-  2Cr 3+ +7H 2 O +6Fe H + + 6Fe 2+ The dichromate ion (Cr 2 O 7 2- ) reacts with the Fe 2+ ion to form Cr 3+ (green) and Fe 3+. Ox:Fe 2+  Fe 3+ + e - Ox:6Fe 2+  6Fe e -

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

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


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