Transcript Document

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.
•
•
•
•
•
•
•
MnO4Cr2O72H 2O 2
HSO3NO3SO2
MnO2
•permanganate = manganate (VII)
•dichromate
•hydrogen peroxide
•hydrogen sulfite
•nitrate
•sulfur dioxide
•manganese dioxide = manganese (IV) oxide
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- + 8H+ + 5 e- 
Mn2+ + 4H2O
Ox:
Fe2+

Ox:
5Fe2+
 5Fe3+ + 5e-
Fe3+ + e-
(x5)
Overall: MnO4- + 8H+ +5Fe2+ Mn2+ + 4H2O +5Fe3+
The Rules
a few extra steps
You don’t have to use them
all… but make sure you do
them in order
• 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.
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- + 14H+ + 6 e-  2 Cr3+ + 7H2O
Ox:
Fe2+

Ox:
6Fe2+
 6Fe3+ + 6e-
Fe3+ + e-
(x6)
Overall:Cr2O72- + 14H+ + 6Fe2+ 2Cr3++7H2O +6Fe3+
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 + 2H+ + 2 e-
Ox:
2 Br
-


2H2O
Br2
+ 2e-
Overall: H2O2 + 2H+ + 2Br-  2H2O + Br2
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).
Red: Cr2O72-
5
4
+ 14H+ + 6 e-  2 Cr3+ + 7H2O
+ H 2O 
Ox:
HSO3-
Ox:
3HSO3- + 3H2O 
SO42- + 3H+ + 2e- (x3)
3SO42- + 9H+ + 6e-
Overall:Cr2O72- + 5H+ + 3HSO3- 2Cr3++4H2O +3SO42-
Use your Textbook to…
• Pg 76 – complete Question 5
Common Oxidants
• Oxygen:
O2 + 4e- → 2O2• Permanganate:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
• Dichromate:
Cr2O72- + 14H+ +6e- → 2Cr3+ + 7H2O
• Iron(III) ion: Fe3+ + e- → Fe2+
• Halogens:
X2 + 2e- → 2X• Hydrogen peroxide:
H2O2 + 2H+ + 2e- → 2H2O
• Iodate: IO3- + 6H+ + 6e- → I- + 3H2O
• Bromate:BrO3- + 6H+ + 6e- → Br- + 3H2O
Common Reductants
•
•
•
•
•
•
•
Hydrogen: H2 → 2H+ + 2eThiosulfate: 2S2O32- → S4O62- + 2eIron(II) ion: Fe2+ → Fe3+ + eHalides (Halogen ions): 2X- → X2 + 2eZinc: Zn → Zn2+ + 2eOxalate:
C2O42- → 2CO2 + 2eSulfur dioxide:
SO2 + 2H2O→ SO42- + 4H+ + 2e-
Today you Learnt..
• How to balance Redox equations
• That we can tell how a species changes by
looking at the colour change in the reaction
• Tomorrow:- What we can find out from the
changing colour
Summary table of Common Reactions
• Use page 72 in your text book to complete
the table
• When there is a colour change colour the
reactant and product the correct colour
• In box below each one write the balanced
half equation for that reaction
• Hw:Common oxidants hand out