Chemistry Unit 3 Summary Notes Part 4

Topic 2.6 Redox

Balancing redox equations
Equations for redox reactions can be balanced by considering them as two half-reactions.

To balance a redox half–reaction follow the procedure:

1. Write down the formulae of the oxidised and reduced species
2. Balance the equation for the element being oxidised/reduced
3. Balance the equation for O by adding an appropriate number of water molecules
+
4. Balance the equation for H by adding an appropriate number of H ions
-
5. Balance the equation for charge by adding an appropriate number of electrons (e )
6. As a final check, ensure that the overall change in oxidation state of the atoms being
oxidised/reduced is equal to the number of electrons in your equation

Applying this to the half-reaction in which manganate(VII) is reduced to manganese(II) ions:
- 2+
1. MnO4 → Mn (manganese is reduced from +7 to +2)
2. The equation is already balanced for Mn (1 x Mn on each side)
- 2+
3. MnO4 → Mn + 4H2O (4 x O on each side)
- + 2+
4. MnO4 + 8H → Mn + 4H2O (8 x H on each side)
- + - 2+
5. MnO4 + 8H + 5e → Mn + 4H2O (overall charge of 2+ on each side)
-
6. (+7) (+2) (change of -5 corresponds to 5e on left)

Applying to the half-reaction in which ethanedioic acid is oxidised to carbon dioxide

1. (COOH)2 → CO2 (carbon is oxidised from +3 to +4)
2. (COOH)2 → 2CO2 (2 x C on each side)
3. The equation is already balanced for oxygen (4 x O on each side)
+
4. (COOH)2 → 2CO2 + 2H (2 x H on each side)
+ -
5. (COOH)2 → 2CO2 + 2H + 2e (no overall charge on each side)
-
6. 2 x (+3) 2 x (+4) (change of +2 corresponds to 2e on right)

To determine the balanced equation for the overall redox reaction, multiply the two pertinent half-equations
by suitable factors such that the number of electrons is the same in each, then add the two equations
together (thus cancelling the electrons).

So, combining the two half equations above to obtain the balanced equation for the reaction in which
manganate(VII) oxidises ethanedioic acid to carbon dioxide:
- + - 2+
MnO4 + 8H + 5e → Mn + 4H2O

+ -
(COOH)2 → 2CO2 + 2H + 2e 
- + - 2+

x2 2MnO4 + 16H + 10e → 2Mn + 8H2O 
+ -
x5 5(COOH)2 → 10CO2 + 10H + 10e 
- + - 2+ + -
+ 2MnO4 + 16H + 10e + 5(COOH)2 → 2Mn + 8H2O + 10CO2 + 10H + 10e
- + 2+
Cancel 2MnO4 + 6H + 5(COOH)2 → 2Mn + 8H2O + 10CO2
+7 +3 +2 +4 oxidation states


2 x (-5)
10 x (+1)

Note that the nett change in oxidation state is zero. This can be used as a final check that the equation is
balanced.




01/06/2009