1. Lecture 5: Electrochemistry
Lecture 5 Topic Chapter 20
1. Redox agents & half-equations
Reducing & oxidizing agents
Solving redox by half-equation
Steps!
2. Voltaic cells are redox reactions
Separate “half-cells”
3. Batteries
Electromotive force
Batteries & Calculating Ecell
Fuel Cells
4. Corrosion & Electrolysis
Corrosion
Electrolysis

2. Redox agents & half-equations
Reduction agents are oxidized.
Oxidation agents are reduced.
Half-equations account for electrons & ‘co-factors’.
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Redox terminology Oxidizing agent (oxidant)
the substance that removes e- from the oxidized compound
The oxidizing agent is reduced during redox reactions.
Reducing agent (reductant)
the substance that donates e- to the compound being reduced
The reducing agent is oxidized during redox reactions.
Zn (solid) + 2H+ --> Zn+2 + H2 (gas)
oxidation numbers
reducing agent
oxidizing agent
Zn = oxidized Zn becomes + charged, thus lost e-
H = reduced H loses + charged, thus has gained e- p

4. Identifying redox agents
The rechargeable nickel-cadmium (nicad) battery uses this redox reaction to generate electricity:
Cd(s) + NiO2(s) + 2H2O(l) --> Cd(OH)2(s) + Ni(OH)2(s)
/ / /-2/ /-2/+1
a) Identify the substances being reduced & oxidized.
increasing ox# -> oxidized Cd decreasing ox# -> reduced Ni
b) Identify the oxidizing & reducing agents.
- thus = reducing agent
increasing ox# -> oxidized Cd decreasing ox# -> reduced Ni
- thus = oxidizing agent
Identify the oxidizing & reducing agents in this redox equation:
2H2O(l) + Al(s) + MnO4-(aq) --> Al(OH)4-(aq) + MnO2(s)
+1/ / /-2/ /-2
increasing ox# for Al -> Al is oxidized & is the reducing agent
decreasing ox# for Mn -> Mn is reduced & is the oxidizing agent p

5. Balancing redox half-equations
Usually balancing an equation means getting the same number of atoms of each element on each side of the equation.
But, for redox reactions: the number of e- lost and gained on each side must also be balanced.
To make this ‘easier’ we think of redox reactions as pairs of half-reactions written as half-equations.
Sn+2 + 2Fe+3 --> Sn+4 + 2Fe+2
oxidation: Sn+2 --> Sn+4 + 2e-
reduction: 2Fe+3 + 2e- --> 2Fe+2
If e- cancel out and the 2 half-reactions are added they sum to the overall equation.
Notice that the e- are added to opposite sides of the 2 half-equations. p

6. Steps for balancing half-equations
Why use this method? Isn’t it more complicated than simply assigning oxidation numbers?
Yes, half-equations appear complicated, but redox reactions are often complex & involve the solvent that the compounds are suspended in. Solvents can participate in the redox reaction itself, & need to accounted for.
Steps to follow for acidic solutions:
1. Rewrite the equation in half-reactions, one for each reactant
2. Balance each half-equation for mass starting with non-H and non-O elements.
3. Balance for O atoms by adding H2O where needed.
4. Balance for H atoms by adding H+1 ions where needed..
5. Balance for charge by adding e- to one side [Note that e- should be added to opposite sides of the two half-equations.]
6. If necessary, multiply half-equations by factors so that the same number of e- are added or removed in each half-equation.
Add the two half-equations, canceling out items that are the same on both sides. [Note that all e- should cancel.]
8. Double check to see that this final equation balances. p

7. Example: acidic aqueous solution
Consider the reaction between permanganate ion (MnO4-) and oxalate ion (C2O4-2) in an acidic solution.
Color changes & a gas is formed.
MnO4-1 + C2O > Mn+2 + CO2 (gas)
H+1 is involved.
MnO4-1  Mn+2
C2O4-2  CO2
MnO4-1  Mn+2 + 4H2O
C2O4-2  2CO2 -2
8H+1 + MnO4-1  Mn+2 + 4H2O +7 +2
5e- + 8H+1 + MnO4-1  Mn+2 + 4H2O
C2O4-2  2CO2 + 2e-
10e- + 16H+1 + 2MnO4-1  2Mn+2 + 8H2O X2
C2O4-2  10CO2 + 10e- X5
16H+1 + 2MnO CO2O4-2  2Mn+2 + 8H2O + 10CO2
So, we learn how much acid is required and how much water is produced. p

8. Another example
Complete & balance the following equation using half-reactions:
Cr2O7-2 + Cl-1  Cr+3 + Cl (in acid sol’n)
Cr2O7-2  Cr+3
Cl-1 --> Cl2
14H+1 + Cr2O7-2  2Cr+3 + 7HO2
2Cl-1  Cl2
+14 -2 +6 -2
6e- + 14H+1 + Cr2O7-2  2Cr+3 + 7H2O
(x3)
2Cl-1  Cl2 + 2e-
6e- + 14H+1 + Cr2O Cl-1  2Cr+3 + 7H2O + 3Cl2 + 6e-
Finally, check to be sure that atoms & charge are still balanced. p

9. One more! Complete & balance this equation using half-reactions:
Cu(s) + 4H+1 + 2NO3-1  Cu+2 + 2NO2(g) + 2H2O(l)
Cu(s)  Cu+2
4H+1 + 2NO3-1  2NO2(g) + 2H2O(l)
Cu(s)  Cu+2
4H+1 + 2NO3-1  2NO2(g) + 2H2O(l)
Cu(s)  Cu+2 + 2e-
2e- + 4H+1 + 2NO3-1  2NO2(g) + 2H2O(l)
2e- + 4H+1 + 2NO3-1 + Cu(s)  2NO2(g) + 2H2O(l) + Cu+2 + 2e- p

10. Last one! Complete & balance this equation using half-reactions:
2Mn+2 + 5NaBiO3 + 14H+1  2MnO Bi+3 + 5Na+1 + 7H2O(l)
2Mn+2 + 8H2O  2MnO H+1
5NaBiO3 + 30H+1  5Bi+3 + 5Na H2O
2Mn+2 + 8H2O  2MnO H e-
10e- + 5NaBiO3 + 30H+1  5Bi+3 + 5Na H2O
10e- + 5NaBiO3 + 30H+1 + 2Mn+2 + 8H2O  5Bi+3 + 5Na H2O + 2MnO H e- 14 7 p