1. Topics 9 & 19 Electrochemistry
IB CHEMISTRY
Topics 9 & 19 Electrochemistry
Jeff Venables Northwestern High School

2. Oxidation-Reduction Reactions
Zn added to HCl yields the spontaneous reaction
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g).
The oxidation number of Zn has increased from 0 to 2+.
The oxidation number of H has reduced from 1+ to 0.
Zn is oxidized to Zn2+ while H+ is reduced to H2.
H+ causes Zn to be oxidized and is the oxidizing agent.
Zn causes H+ to be reduced and is the reducing agent.
Note that the reducing agent is oxidized and the oxidizing agent is reduced.

4. Determining Oxidation Numbers
The oxidation numbers of all atoms in a molecule or ion add up to the total charge (0 for a molecule, the ion’s charge for an ion)
Elements by themselves are the charge shown
O is almost always -2 (exception H2O2)
F is almost always -1
Group 1 elements are +1
Group 2 elements are +2

5. Determining Oxidation Numbers
Examples: Determine the oxidation numbers of each atom in the following:
NaC2H3O2
Na2SO4
C2O42-
SO32-

6. Balancing Oxidation-Reduction Reactions
Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.
Conservation of charge: electrons are not lost in a chemical reaction.
Half Reactions
Half-reactions are a convenient way of separating oxidation and reduction reactions.

7. Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)
Half Reactions
The half-reactions for
Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)
are
Sn2+(aq)  Sn4+(aq) +2e-
2Fe3+(aq) + 2e-  2Fe2+(aq)
Oxidation: electrons are products.
Reduction: electrons are reactants.
Loss of Gain of
Electrons is Electrons is
Oxidation Reduction

8. Balancing Equations by the Method of Half Reactions
Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple).
MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2.
The equivalence point is given by the presence of a pale pink color.
If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

9. What is the balanced chemical equation?
1. Write down the two half reactions.
2. Balance each half reaction:
a. First with elements other than H and O.
b. Then balance O by adding water.
c. Then balance H by adding H+.
d. Finish by balancing charge by adding electrons.
Multiply each half reaction to make the number of electrons equal.
Add the reactions and simplify.
Check!

10. The two incomplete half reactions are MnO4-(aq)  Mn2+(aq)
For KMnO4 + Na2C2O4:
The two incomplete half reactions are
MnO4-(aq)  Mn2+(aq)
C2O42-(aq)  2CO2(g)
Adding water and H+ yields
8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O
There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left:
5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O

11. 10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O
In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons:
C2O42-(aq)  2CO2(g) + 2e-
To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives:
10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O
5C2O42-(aq)  10CO2(g) + 10e-

12. 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)
Adding gives:
16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)
Which is balanced!

13. Examples – Balance the following oxidation-reduction reactions:
Cr (s) + NO3- (aq)  Cr3+ (aq) + NO (g)
PO33- (aq) + MnO4- (aq)  PO43- (aq) + MnO2 (s)