1. Redox Reactions and Electrochemistry
Unit 6
Redox Reactions and
Electrochemistry

2. Faraday’s Law
Michael Faraday found the amount of chemical change that occurs during electrolysis is directly proportional to the amount of electricity that passes through an electrolytic cell.
Basically, it’s an experimental measurement of electric current and time which can be used to calculate the amount of chemical change

3. To deposit 1 mole of a metal, we must look at the number of electrons in a balanced half-reaction
ex: Cu e-  Cu(s)
This example needs 2 moles of electrons for every 1 mole of Cu(s)
Faraday stated:
1 Faraday (F) = 1 mole of e-
We will also need the formula:
q = It
where q = charge (Coulombs), I = current (Amps) and t = time (seconds)

4. Experimentally, it was found that 1 Faraday carries a charge of 96500C.
Therefore, 1F = 1 mole e- = 96500C
Steps:
Convert minutes to seconds
Find Coulombs (q = It)
Find Faradays (F = q/96500)
Write a balanced half-reaction and determine moles of all species
Calculate grams of solid (m = (n)(MM))
**You may need to do these steps in reverse**

5. ex: How many grams of copper are deposited on the cathode if an electric current of 2.00A is run through a solution of CuSO4 for 20 minutes?
20 min x 60 sec/min = 1200s
q = It = (2)(1200) = 2400C
F = 2400C/96500 = F
= moles e-
Cu e-  Cu(s)
mol
mol
m = (n)(MM) = ( mol)(63.55g/mol)
= 0.791g of Cu(s) deposited

6. The Activity Series
All half reactions can be placed in an ordered series according to their affinity for electrons
Recall, atomic radii get smaller from left to right and the ionization energy increases from left to right.
In other words, the pull from the nucleus for the electrons gets stronger as we move from left to right across a period.

7. This corresponds to the reactivity of ions and molecules in half reactions.
Li+ has the least affinity for it’s electron. That is, it’s electron is easily lost
F2 has the greatest affinity for it’s electrons. That is, it’s electrons are not easily lost.
Therefore, through experimental data, a Half-Reaction Activity Series was developed.

8. Recall that a loss of electrons is an oxidation and a gain of electrons is a reduction (LEO GER)
Therefore, if an ion/molecule has a low affinity for it’s electron it is an oxidizing reaction (it is willing to get rid of the electron(s)) and vice versa
You will also need to know about oxidizing and reducing agents.
Reducing Agent is an ion/molecule that is involved in an oxidation reaction (it causes the other half-reaction to be a reduction)
The red agents are on the left side of the half-reaction
The strongest red agents (SRA) are the molecules that have the least affinity for their electrons.

9. Oxidizing Agent is an ion/molecule that is involved in a reduction reaction (it causes the other half-reaction to be an oxidation)
The ox agents are on the right side of the half-reaction
The strongest ox agents (SOA) are the molecules that have the greatest affinity for their electrons.
The oxidizing reactions must be flipped when they are in the presence of a stronger reducing reaction.

10. ex: A zinc metal strip is placed into a 1
ex: A zinc metal strip is placed into a 1.0M copper (II) nitrate solution. Predict the two half reactions.
We know that nitrate is a spectator ion so we must look up copper and zinc on our table
Cu e-  Cu(s) Zn e-  Zn(s)
According to our tables, Cu2+ has the greater affinity for its electrons, so it is the reducing reaction.
So, flip the zinc reaction to make it an oxidation and then add the two reactions.
Cu e-  Cu(s)
Zn(s)  Zn e-
Cu2+ + Zn(s)  Cu(s) + Zn2+

11. The previous example could have been written as Zn(s)│Zn2+││Cu2+│Cu(s)
product
reactant
product
reactant
Ox Reaction
Red Reaction
ex: Write the net reaction for:
Al(s) │ Al3+ │ │Ag+ │Ag(s)
Ag e-  Ag(s) 3
3Ag+ + Al(s)  3Ag(s) + Al3+
Al(s)  Al e-

12. Spontaneous Reactions
Spontaneous reactions occur when there is a positive change in electric potential between the two half-reactions.
The potentials listed on your SRPS are for the reduction reaction. If you’ve flipped a reaction (from how it is on the sheet), you must also flip the sign (pos. or neg) too.
There are two types of problems that will require you to calculate spontaneity:

13. reactant│product││reactant│product ex: Pb │ Pb2+ │ │Ni2+ │Ni
Chemical 1││Chemical 2
Recall that it goes:
reactant│product││reactant│product
ex: Pb │ Pb2+ │ │Ni2+ │Ni
Flip the sign from the chart b/c the rxn was flipped
Pb  Pb2+ + 2e-
+0.13V
-0.26V
Ni2+ + 2e-  Ni
-0.13V
NOT SPONTANEOUS!

14. Separate the equation into 2 half-reactions
2) Balanced Equations
Separate the equation into 2 half-reactions
Then find the two potentials and add to find spontaneity.
ex: 3Ca Cr  3Ca + 2Cr3+
**The multipliers don’t have an effect on the potential from the sheet.**
Rxn 1: 3Ca2+ + 6e-  3Ca
-2.87V
Rxn 2: 2Cr  2Cr3+ + 6e-
+0.74V
-2.13V
NOT SPONTANEOUS!

15. The Electrochemical Cell
This cell is also known as a voltaic cell or a galvanic cell
These cells involve spontaneous redox reactions
It produces a current as a result of the change from chemical energy to electrical energy
The cell is broken into 2 half-reactions
The 2 half-reactions are connected by a salt bridge

16. The salt bridge allows for ion transfer between the 2 half-reactions
The salt bridge allows for ion transfer between the 2 half-reactions. The flow is opposite to the flow of electrons.
There is a metal bar in each half-reaction solution. These metal bars are called electrodes.
One electrode is a cathode and the other is an anode.
The cathode reaction is always the reduction and the anode reaction is always the oxidation (LEOA and GERC)
The solution in each half-reaction is called the electrolytic solution because it allows for ion flow and therefore, it conducts electricity.

17. The electrodes are connected by external wires.
It is through the wires that the electrons flow. The flow is always anode to cathode.
The wire is hooked up to a voltmeter that measures the cell potential.
Remember, these cells are spontaneous so the cell potential must be positive or these cells won’t work.
So, when writing out the half-reactions, the higher reaction (on the SRPS) is the reduction reaction and the lower reaction (on the SRPS) is the oxidation reaction.

18. ex: Draw an electrochemical cell that contains ZnSO4 and CuSO4 solutions. What’s the overall cell potential?
Copper is higher = reduction (cathode) Cu2+ + 2e-  Cu(s)
Zinc is lower = oxidation (anode) Zn(s)  Zn2+ + 2e-
E0 = (+0.76V) + (+0.34V) = +1.10V

19. The previous example is known as the Daniell Cell
Eventually, you would see that the copper electrode would get bigger (as more copper solid is deposited) and the zinc electrode would get smaller (as more solid zinc turns into ionic zinc)
If ever you are given two solutions and you need to figure out which ions are not spectators, find the two closest to 0V.
Also, nitrate will only react in acidic solutions.

20. ex: Draw a galvanic cell containing K2Cr2O7 solution and Ni(NO3)2 solution. Determine the E0.
Cr2O7- and Ni2+ are closest to 0V
Look up K+, Cr2O7-1, and Ni2+
Anode: Ni(s)  Ni2+ + 2e-
Cathode: 14H+ + Cr2O e-  2Cr3+ + 7H2O
0.26V
1.23V
1.49V

21. The Electrolytic Cell
These cells convert electrical energy into chemical energy.
They involve an electric current (battery) that causes an otherwise non-spontaneous redox reaction to occur.
There are still two electrodes, an electrolytic solution and wires just like a galvanic cell. There is no salt bridge!
But, you must have a battery and only one beaker.

22. The negative terminal of the battery is hooked up to the cathode and the positive terminal is hooked up to the anode.
The electrons still flow from anode to cathode through the wires and battery.
In the electrolytic solution, the flow of ions conducts the current, completing the circuit.
For these cells, the higher reaction is oxidation and the lower reaction is reduction
However, the oxidation is still at the anode and the reduction is still at the cathode.

23. Many times for these cells, we will dissolve a salt in water.
Recall, if you’re trying to find out the two half-reactions, find the two closest to 0V.
Many times for these cells, we will dissolve a salt in water.
ex: Draw the electrolytic cell of KI in water.
So, we have K+, I- and HOH
I- and HOH are closest to 0V + e-
Carbon
Cathode I2
Anode H2
OH- I- K+

24. Electroplating
This is where a cheap inert material is covered with a precious metal.
In these examples, the two reactions are the reverses of one half-reaction.
ex: Take a cheap iron ring and electroplate it with gold.
The gold metal bar
Anode: Au(s)  Au3+ + 3e-
Cathode: Au3+ + 3e-  Au(s)
The ring

26. Practical Applications
There are two types of electrolytic cells:
A. Primary Cells
1) Dry Cell (acid form) are also known as Leclanche Cells.
- they are the common battery.
2) Dry Cell (alkaline form)
- are more effective than the acid dry cell because they can be miniaturized and provide more energy.
- also, acid corrosion of the container does not occur.

27. B. Secondary Cells
1) Lead Storage Cells
- used in cars.
- the battery is discharged as current is consumed. When the car’s engine is running, the battery becomes recharged.
- most of these batteries last between 3-5 years and die because of sludge build-up
2) Nickel-Cadmium Cell
- a rechargeable cell
-used in power tools, flashlights, calculators, etc…