Writing more complex Redox Equations

Aim: To be able construct ionic half equations for reactions

involving complex ions.

Starter

• NaBr + H2SO4 HBr + NaHSO4

Redox? Yes or no?

2HBr + H2SO4 SO2 + Br2 + 2H2O

Half ionic equations?

Oxidation number?

Starter: NaI with Sulphuric acid

• NaI + H2SO4 HI + NaHSO4

• 8HI + H2SO4 H2S + 4I2 + 4H2O

Redox? – Strength of reducing agent?

Starter Disproportionation

• 3NaClO 2NaCl + NaClO3

• Ionic equation?

• Disproportionation?

Disproportionation 2

• 2NaOH + Cl2 NaOCl + NaCl + H2O

• Ionic equation?

• Disproportionation?

Disproportionation 3

• In hot alkali

• 6NaOH + 3Cl2 NaClO3 + 5NaCl + 3H2O

• Ionic equation?

• Disproportionation?

Redox reactions – Potassium Permanganate and Sodium thiosulphate.

• KMnO4

• Na2S2O3

• Oxidation number of Potassium?• Charge on the Manganate ion?• Oxidation number of Mn in MnO4

-?• Oxidation number of Na?• Charge on the thiosulphate ion?• Oxidation number of S in the thiosulphate ion –

S2O3-2?

Reaction between MnO4- and S2O3

-

wine into water!• MnO4

-

• Mn+4

• Mn+2

• S2O3- 2

• S4O6-2

• Observation : Purple to colourless.

• Redox?

Reduction of the Manganate ion MnO4

-

• MnO4- Mn+2

• Step 1: balance the oxygen by using water molecules.

• MnO4- Mn+2 + 4H20

• Step 2: Balance the hydrogen using H+1 ions.• MnO4

- + 8H+ Mn+2 + 4H20• Balance for Charge using electrons:• MnO4

- + 8H+ + 5e- Mn+2 + 4H20• Reduction or oxidation?

Oxidation of the Thiosulphate ion to Tetrathionate ion.

• 2S2O3- 2 S4O6

-2 + 2e-

• Sulphur change in oxidation number?

• S+2 S+2.5

• Reduction or oxidation? Why?

Overall redox equation

• MnO4- + 8H+ + 5e- Mn+2 + 4H20

• 2S2O3- 2 S4O6

-2 + 2e-

Ionic half equation for the reduction of the dichromate ion

• Cr2O72- 2Cr+3

• Step 1 balance Oxygen using water

• Cr2O72- 2Cr+3 + 7H2O

• Step 2 Balance Hydrogen using H+1ions

• Cr2O72- + 14H+1 2Cr+3 + 7H2O

• Step 3 Balance for Charge using electrons

• Cr2O72- + 14H+1 +6e- 2Cr+3 + 7H2O

Some examples

• Write ionic half equations for the following that take place in acidic solution.

• Iodate (v) ions IO3-1 to iodine.

• Manganese dioxide MnO2 to Mn+2.

• VO2+ ions to VO+2

• H2SO3 to Sulphuric acid

Answers

• 2IO3-1 +12H+ +10e- I2 + 6H2O

• MnO2 + 4H+ +2e-

Mn+2 + 4H2O

• VO2+ + 2H+ + 1e- VO2+ + H2O

• SO3-2 + H2O SO4

-2 + 2H+ + 2e-

Reaction between Iodine and Sodium Thiosulphate

• 2Na2S2O3 + I2 2NaI + Na2S4O6

• Half ionic equations to identify redox?• Iodine: • Thiosulphate: • How will you know when this reaction has

gone to completion?• How could you prove that the reaction is

consistent with the above equation using titration?

I2 + 2e-1 2I-1

2S2O3 -2 S4O6

2- + 2e-

A Redox Titration

0.01M sodium thiosulphate

10ml of 0.01M I2

•Titrate 10ml samples of 0.01M iodine solution with 0.01M Na2S2O3

•The end point for the titration is when the yellow iodine turns colourless.

•The end point can be made more accurate by adding a few drops of 1% starch when the iodine colour becomes very pale. A dark blue colour is produced which suddenly disappears at the end point.

Titration Results.

Burette readings

1 2 3 4

Burette reading final

Burette reading initial

Titre cm3

Interpreting the Results

• Mean Titre Sodium Thiosulphate =

• Calculate the number of moles of sodium thiosulphate that reacted with the iodine.

• Calculate the number of moles of Iodine used in the reaction.

• Prove that this is consistent with the equation for the reaction.

• 2Na2S2O3 + I2 2NaI + Na2S4O6

Redox questions

• Separate the following into their 2 half ionic equations

• 2FeCl2 + Cl2 2FeCl3

• 2KBr + Cl2 2KCl + Br2

• MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

Further redox questions – separate into half ionic equations

• MnO4- + 5Fe+2+ 8H+ Mn+2 +5Fe+3 +4H2O

• IO3- + 5I-1 + 6H+ 3I2 + 3H2O

Home work

• Questions 1 -12 Question sheet.• Revise for mock exam on January 3rd.• See revision guide.