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.
•
•
•
•
•
•
•
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-