1. Electrochemistry

2. homework
Chapter 18
Electrochemistry
AP Practice test

3. Oxidation- Reduction Reactions
Redox or oxidation-reduction reactions are reactions that involve a transfer of electrons.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
(think of the charge, OIL RIG)
So in the reaction
4 K + O2 → 4 K O2-
Potassium get oxidized, oxygen get reduced

4. Oxidation States
Oxidation state is the theoretical charge on all atoms if all bonds were ionic.
The sum of the oxidation states must be equal to the charge of the ion or molecule.

5. Using oxidation states
In the reaction…
2 Na +2 H2O →2 NaOH + H2
Note the changes
Sodium went from 0 to 1
2 of the hydrogen atoms went from +1 to 0 (the other two were unchanged)

6. Identification of Redox Components.
Specify which of the following equations represents oxidation-reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced.
CH4(g) + H2O(g)  CO(g) + 3H2(g)
2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2Ag(s)
H+(aq) + 2CrO42-(aq)  Cr2O72-(aq) + H2O(l)

7. Half reactions Ce4+ + Sn2+ → Ce3+ + Sn4+ Half reactions
Ce4+ + e- → Ce3+
Sn2+ → 2e- + Sn4+
Electrons lost must equal electrons gained!
2 Ce4+ +2 e- →2 Ce3+
Merge the two half reactions
2 Ce4+ + Sn2+ → 2 Ce3+ + Sn4+

8. Redox reactions in acidic solutions
It will be noted in the problem
Balance all elements except hydrogen and oxygen.
Balance oxygen by adding H2O (which is always prevalent in an acidic solution)
Balance hydrogen by adding H+
Then balance the charge adding electrons and proceed normally.

9. Example In an acidic solution Cr2O7 2- + Cl- → Cr3+ + Cl2
Half reactions
Cr2O7 2- → Cr3+
Cl- → Cl2

10. Reduction side Cr2O7 2- → Cr3+ Cr2O7 2- → 2 Cr3+
Cr2O7 2- → 2 Cr H2O
Cr2O H+→ 2 Cr H2O
Cr2O H++ 6 e- →2Cr3++7 H2O

11. Oxidation side Cl- → Cl2 2 Cl- → Cl2 2 Cl- → Cl2 + 2 e-
I have to equal 6 e- so multiply by 3
6 Cl- → 3 Cl2 + 6 e-

12. Combine my half reactions
Cr2O H++ 6 e- → 2 Cr H2O
6 Cl- → 3 Cl2 + 6 e-
And you get
Cr2O H++6Cl-→2Cr3++3 Cl2+7H2O
The electrons cancel out .

13. Example In an acidic solution MnO4- + H2O2 → Mn2+ + O2 Half reactions
MnO4- → Mn2+
H2O2 → O2

14. Top Equation MnO4- → Mn2+ MnO4- → Mn2+ + 4 H2O
MnO H+→ Mn H2O
MnO H++ 5 e-→ Mn H2O

15. Bottom Equation H2O2 → O2 H2O2 → O2 + 2 H+ H2O2 → O2 + 2 H+ + 2 e-
I need to equal 5 e- so…
That won’t work…
2MnO H++ 10 e-→ 2 Mn H2O
5 H2O2 → 5 O H e-

16. Add them together 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O
5 H2O2 → 5 O H e-
And you get
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O
Notice the H+ canceled out as well.

17. Balancing Redox Equations in a basic solution
Look for the words basic or alkaline
Follow all rules for an acidic solution.
After you have completed the acidic reaction add OH- to each side to neutralize any H+.
Combine OH- and H+ to make H2O.
Cancel out any extra waters from both sides of the equation.

18. Example We will use the same equation as before In a basic solution
MnO4- + H2O2 → Mn2+ + O2
2 MnO H++ 5 H2O2
→ 2 Mn O2 + 8 H2O

19. Basic solution Since this is a basic solution we can’t have excess H+.
We will add OH- to each side to neutralize all H+
2 MnO H++ 5 H2O2 + 6OH-
→2 Mn2+ +5 O2 +8 H2O + 6OH-
We added 6 OH- because there were 6H+

20. Cont.
H+ + OH- → H2O
Combine the hydroxide and hydrogen on the reactant side to make water
2 MnO H2O + 5 H2O2
→ 2 Mn2++ 5 O2+ 8 H2O + 6OH-
Cancel out waters on both sides
2 MnO H2O2
→2 Mn O2 +2 H2O +6OH-

21. Another example In a basic solution MnO4 − + SO32-→MnO4 2− + SO42-
Half reactions
MnO4 − → MnO4 2−
SO32-→ SO42-

22. Half reactions MnO4 − → MnO4 2− MnO4 - + e- → MnO4 2− SO32-→ SO42-
H2O + SO32-→ SO42-
H2O + SO32-→ SO H+
H2O + SO32-→ SO H+ +2e-
Double the top reaction

23. 2 MnO e- → 2 MnO4 2−
H2O + SO32-→ SO H+ +2e-
Combine them
2 MnO4 - + H2O + SO32-
→ 2 MnO4 2− +SO H+
Add OH-
2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2−+SO H++2 OH-

24. 2 MnO4 - + H2O + SO OH-
→ 2 MnO4 2− +SO H2O
finishing
2 MnO4 - + SO OH-
→ 2 MnO4 2− +SO42- + H2O

25. Major Redox Points to Remember.
Any redox reaction can be treated as the sum of the reduction and oxidation half-reactions.
Mass (atoms) and charge are conserved in each half-reaction.
Electrons lost in one half-reaction are gained in the other.
Even though the half-reactions are treated separately, electron loss and electron gain occur simultaneously.

26. Galvanic Cell

27. Galvanic cell
A galvanic cell is a device in which chemical energy is changed to electrical energy.
A galvanic cell uses a spontaneous redox reaction to produce an electric current that can be used to do work. The system does work on the surroundings.
A redox reaction involves the transfer of electrons from the reducing agent to the oxidizing agent.

28. Oxidation v reduction Oxidation. * Involves a loss of electrons.
 * Increase in oxidation number.
 * “To get more positive.”
 * Occurs at the anode of a galvanic cell.
Reduction.
* Involves a gain of electrons.
* Decrease in oxidation number.
* “To get more negative.”
* Occurs at the cathode of a galvanic cell.

29. Production of Current
Oxidation Reactions involve a transfer of electrons.
Electric current is a movement of electrons.
In order to produce a usable current, the electrons must be forced across a set path (circuit).
In order to accomplish this, an oxidizing agent and something to oxidize must be separated from a reducing agent with something to reduce.

30. Pictures
Oxidation reduction reaction in the same container will have electrons transferring, but we can’t harness them.
Separating the oxidation reaction from the reduction reaction, but connecting them by a wire would allow only electrons to flow.
oxidation
reduction
Oxidation
Reduction

31. Closer look
X → X+ + e-
X+ + e- → X
Oxidation
Reduction
We now have excess electrons being formed in the oxidizing solution and a need for electrons in the reducing solution with a path for them to flow through.
However, if electrons did flow through the wire it would cause a negative and positive solution to form.

32. That’s not possible Or at least it would require a lot of energy.
A negative solution would theoretically be formed by adding electrons, and a positive one by removing electrons.
The negative solution would then repel the electrons and stop them from flowing in, and a positive solution would attract the electrons pulling them back where they came from.
Making it so the charged solutions wouldn’t form.
In order for this to work, I would need a way for ions to flow back and forth but keeping the solutions mostly separated.

33. The Salt Bridge or the Porous Disk.
These devices allow ion flow to occur (circuit completion) without mixing the solutions. They are typically made of sodium sulfate or potassium nitrate

34. e-
Closer look e- e-
Salt Bridge
X → X+ + e-
X+ + e- → X
Oxidation
Reduction
Now electrons can flow across the wire from the oxidation reaction to the reduction reaction.
As the oxidation reaction becomes positive, it removes negative ions and adds positive ions to the salt bridge.
The reduction reaction does the reverse.

35. Closer look Zooming in on the oxidizing side
Salt Bridge
- ion
+ ion
Oxidation side e-
+ ion
- ion
- ion
+ ion
+ ion
- ion
Zooming in on the oxidizing side
This would make the salt bridge positive…

36. Closer look (Zooming in on the reducing side)
Salt Bridge
- ion
+ ion
- ion e-
+ ion
- ion
+ ion
- ion
Reducing side
+ ion
(Zooming in on the reducing side)
if the reverse wasn’t happening on this side.

37. Close up of salt bridge
+ ion
+ ion
+ ion
- ion
- ion
- ion
- ion
- ion
+ ion
+ ion
+ ion
- ion
The ions keep flowing in the salt bridge to keep everything neutral.
Electrons do also travel across the salt bridge.
This decreases the cell’s effectiveness.

38. Electrochemical cell This is the basic unit of a battery.
It is also called a galvanic cell, most commercial batteries have several galvanic cells linked together.
Batteries always have two terminals.
The terminal where oxidation occurs is called the anode.
The terminal where reduction occurs is called the cathode.

39. Galvanic Cell

40. Voltaic Cell: Cathode Reaction
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41. Voltaic Cell: Anode Reaction
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42. Electrochemical Half-Reactions in a Galvanic Cell
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