1. CHAPTER 15
Electro-chemistry
15.4 Electrochemical Cells

2. The lemon battery Now we know it is an oxidation reaction
A chemical reaction between magnesium and the citric acid generates electrons
Now we know it is an oxidation reaction
Electron flow
Magnesium
Citric acid

3. The lemon battery Now we know it is an oxidation reaction
A chemical reaction between magnesium and the citric acid generates electrons
Now we know it is an oxidation reaction
This means that somewhere else, a reduction reaction must also take place to absorb the electrons
LED
Electron flow
Copper
Citric acid

4. Electron flow Electron flow
LED
Oxidation and reduction reactions are separated
Electron flow
Electron flow
Copper
Magnesium
Citric acid
electrochemical cell: a device in which redox reactions take place.

5. Three main components of an electrochemical cell:
1) Two electrodes
2) The electrolyte
3) A conducting path that connects the electrodes externally
electrochemical cell: a device in which redox reactions take place.

6. An electrochemical cell
Three main components of an electrochemical cell:
1) Two electrodes
The anode electrode is where the oxidation occurs
The cathode electrode is where the reduction occurs

7. An electrochemical cell
Three main components of an electrochemical cell:
1) Two electrodes
2) The electrolyte
Electrodes are immersed in a conductive solution. This solution, or electrolyte, contains free ions.

8. An electrochemical cell
Three main components of an electrochemical cell:
1) Two electrodes
2) The electrolyte
3) A conducting path that connects the electrodes externally
There must be a connection between oxidation and reduction
Electrons must be able to flow from the anode to the cathode

9. There are two types of electrochemical cells:
Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell)
Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte
(or galvanic cells)

10. There are two types of electrochemical cells:
Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell)
Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte
(or galvanic cells)
In both cases, the cells are constructed with two half-cells:
Anode where oxidation happens
Cathode where reduction happens

11. A voltaic cell
The flow of electrical current is completed inside the battery
Negative ions move from the cathode to the anode
Positive ions move from the anode to the cathode to maintain electroneutrality
inside the battery

12. Chemistry in a voltaic cell
Consider the following setup:
Mg loses electrons more easily than Cu so electrons would flow from the Mg side
e– flow
e– flow
Oxidation (anode):
Mg(s) → Mg2+(aq) + 2e–
Reduction (cathode):
Cu2+(aq) + 2e– → Cu(s)
Mg(s)
Cu(s)

13. Chemistry in a voltaic cell
Consider the following setup:
e– flow
e– flow
Why does the flow of electrons only happen for a short time?
Mg(s)
Cu(s)
LED no longer lights up

14. Chemistry in a voltaic cell
Consider the following setup:
Oxidation (anode):
Mg(s) → Mg2+(aq) + 2e–
e– flow
e– flow
Mg(s)
Cu(s)
Mg2+ ions have nowhere to go
Mg2+ and e– are attracted to each other
The reaction stops as soon as it starts

15. Chemistry in a voltaic cell
Oxidation (anode):
Mg(s) → Mg2+(aq) + 2e–
Reduction (cathode)
Cu2+(aq) + 2e– → Cu(s)
e– flow
e– flow
If e– flows away, there is an excess of positive ions (Mg2+)
If e– flows in, Cu2+ is consumed, and there is an excess of negative ions (SO42–)
Mg(s)
Cu(s)

16. Chemistry in a voltaic cell
The solution: provide a path for positive ions to move from the anode half-cell to the cathode half-cell
e– flow
e– flow
If e– flows away, there is an excess of positive ions (Mg2+)
If e– flows in, Cu2+ is consumed, and there is an excess of negative ions (SO42–)
Mg(s)
Cu(s)

17. salt bridge: an electrical connection between the oxidation and the reduction half-cells of an electrochemical cell.

18. Notation

19. Electromotive force
Electrons that flow from the anode to the cathode are “pushed” by the electromotive force
electromotive force (emf): the difference in the electrical potential between the anode and the cathode of an electrochemical cell.

20. Faraday constant = 96,500 C/mole
Electromotive force
Based on the emf, we can calculate the amount of energy that the cell can provide:
cell emf
(the driving force of the cell reaction)
electrical energy output
moles of electrons
Faraday constant = 96,500 C/mole
electromotive force (emf): the difference in the electrical potential between the anode and the cathode of an electrochemical cell.

21. Electromotive force
The Mg–Cu cell has a cell emf of Ecell = 2.71 V. The cell reaction is:
Calculate the total energy that can be obtained from this cell if 2.3 g of Mg is consumed in the reaction.
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

22. Electromotive force Half-reactions:
The Mg–Cu cell has a cell emf of Ecell = 2.71 V. The cell reaction is:
Calculate the total energy that can be obtained from this cell if 2.43 g of Mg is consumed in the reaction.
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Asked: Wcell, energy from this cell with 2.43 g of Mg
Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant
Relationships:
Half-reactions:
Mg(s) → Mg2+(aq) + 2e–
Cu2+(aq) + 2e– → Cu(s)

23. Electromotive force Half-reactions:
Asked: Wcell, energy from this cell with 2.43 g of Mg
Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant
Relationships:
Half-reactions:
Solve:
Mg(s) → Mg2+(aq) + 2e–
Cu2+(aq) + 2e– → Cu(s)
From the oxidation half-reaction, 1 mole Mg ~ 2 moles e–
So n = moles e–

24. Electromotive force Half-reactions:
Asked: Wcell, energy from this cell with 2.43 g of Mg
Given: The cell reaction, Ecell = 2.71 V, and the amount of reactant
Relationships:
Half-reactions:
Solve:
Answer: With 2.43 g of Mg, this cell will release 5,230 J of energy.
Mg(s) → Mg2+(aq) + 2e–
Cu2+(aq) + 2e– → Cu(s)
From the oxidation half-reaction, 1 mole Mg ~ 2 moles e–
So n = moles e–

25. Ecell = Ereduction + Eoxidation
The total cell voltage (Ecell) is a combination of the potential at each half-cell:
Ecell = reduction potential + oxidation potential
Ecell = Ereduction Eoxidation
For the oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e– Eoxidation = V
For the reduction half-reaction: Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V

26. Scientists decided to only keep track of Ereduction
The total cell voltage (Ecell) is a combination of the potential at each half-cell:
Ecell = reduction potential + oxidation potential
Ecell = Ereduction Eoxidation
For the oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e– Eoxidation = V
For the reduction half-reaction: Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V
Scientists decided to only keep track of Ereduction

27. Standard reduction potentials
Mg2+(aq) + 2e– → Mg(s) Ereduction = –2.37 V
Standard conditions:
Temperature = 25oC
Pressure = 1 atm (for gases)
Concentration = 1 M (for solutions)
The cell voltage under standard conditions is called the standard reduction potential, Eocell

28. Standard reduction potentials
Ecell = Ereduction + Eoxidation
We can’t measure the potential of single electrodes
We need a reference half-cell!

29. Standard reduction potentials
Ecell = Ereduction + Eoxidation
We can’t measure the potential of single electrodes
We need a reference half-cell!
Use this reaction as the reference where Ereduction = 0 V

30. Standard reduction potentials
The potential of all other cells is measured with respect to the reference cell
Reference half-cell

31. Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).

32. Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Given: The cell reaction, standard reduction potentials
Relationships: Eocell = Eoreduction + Eooxidation
Look up Table 15.4

33. Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V
Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Given: The cell reaction, standard reduction potentials
Relationships: Eocell = Eoreduction + Eooxidation
Zn(s) → Zn2+(aq) + 2e– Eooxidation = V
Cu2+(aq) + 2e– → Cu(s) Eoreduction = V
reverse reaction
From the Eoreduction table

34. Zn(s) → Zn2+(aq) + 2e– Eooxidation = +0.76 V
Using the standard reduction potentials, calculate the cell voltage (Eocell) of the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
Asked: Eocell of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Given: The cell reaction, standard reduction potentials
Relationships: Eocell = Eoreduction + Eooxidation
Zn(s) → Zn2+(aq) + 2e– Eooxidation = V
Cu2+(aq) + 2e– → Cu(s) Eoreduction = V
Solve: Eocell = Eoreduction + Eooxidation
Eocell = 0.34 V V = 1.10 V
Answer: The cell voltage is 1.10 V

35. Reaction spontaneity Eocell = Eoreduction + Eooxidation
The sign of Eocell also tells you whether the reaction is spontaneous or nonspontaneous
Eocell > 0 Spontaneous reaction
Eocell < 0 Nonspontaneous reaction

36. Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions.

37. The half-cells and their half-cell potentials:
Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions.
Asked: The spontaneity of the reaction under standard conditions
Given: The reaction and the table of standard reduction potentials
Relationships: Eocell = Eoreduction + Eooxidation
The half-cells and their half-cell potentials:
Zn(s) → Zn2+(aq) + 2e– Eooxidation = V
Ni2+(aq) + 2e– → Ni(s) Eoreduction = –0.23 V

38. The half-cells and their half-cell potentials:
Determine whether the reaction Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) is spontaneous under standard conditions.
Asked: The spontaneity of the reaction under standard conditions
Given: The reaction and the table of standard reduction potentials
Relationships: Eocell = Eoreduction + Eooxidation
The half-cells and their half-cell potentials:
Zn(s) → Zn2+(aq) + 2e– Eooxidation = V
Ni2+(aq) + 2e– → Ni(s) Eoreduction = –0.23 V
Solve: Eocell = –0.23 V V = V > 0
Answer: Since the total cell voltage is a positive number
the reaction is spontaneous and proceeds as indicated.

39. Three main components of an electrochemical cell:
1) Two electrodes
2) The electrolyte
3) A conducting path that connects the electrodes externally (like a salt bridge)

40. Faraday constant = 96,500 C/mole
cell emf
(the driving force of the cell reaction)
electrical energy output
moles of electrons
Faraday constant = 96,500 C/mole
Eocell = Eoreduction + Eooxidation
Standard conditions:
Temperature = 25oC
Pressure = 1 atm (for gases)
Concentration = 1 M (for solutions)

41. standard cell potential
Nernst equation
Batteries (a voltaic cell) run out of energy
Nernst equation:
standard cell potential
moles of electrons
Under standard conditions:
Q = 1 so that Ecell = Eocell

42. Nernst equation Batteries (a voltaic cell) run out of energy

43. The nonspontaneous reaction that is induced is called electrolysis
Electrolytic cells
There are two types of electrochemical cells:
Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell)
Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte
(or galvanic cells)
The nonspontaneous reaction that is induced is called electrolysis

44. Electrolytic cells There are two types of electrochemical cells:
Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell)
(or galvanic cells)
When we use a battery it acts as a voltaic cell

45. Electrolytic cells There are two types of electrochemical cells:
Voltaic cells Spontaneous chemical reactions at the electrodes generate electrical current (A battery is a voltaic cell)
Electrolytic cells An external electric current drives nonspontaneous reactions between the electrodes and the electrolyte
(or galvanic cells)
When we recharge a battery it acts as an electrolytic cell

46. Electrolytic cells
Decomposition of water:
2H2O(l) → 2H2(g) + O2(g)

47. Electrolytic cells Decomposition of water: 2H2O(l) → 2H2(g) + O2(g)
+1 –
oxidation numbers
reduction
oxidation

48. nonspontaneous reaction
Electrolytic cells
Decomposition of water:
2H2O(l) → 2H2(g) + O2(g)
+1 –
oxidation numbers
reduction
oxidation
Reduction:
2H2O(l) + 2e– → H2(g) + 2OH–(aq) Eored = –0.83 V
Oxidation:
2H2O → O2(g) + 4H+(aq) + 4e– Eoox = –1.23 V
Eocell = –2.06 V
nonspontaneous reaction

49. nonspontaneous reaction
Electrolytic cells
Decomposition of water:
2H2O(l) → 2H2(g) + O2(g)
+1 –
oxidation numbers
reduction
oxidation
Hydrogen (H2) is being considered as a source of energy for the future.
The production of hydrogen from water is the cleanest but also the most expensive way of producing hydrogen.
Eocell = –2.06 V
nonspontaneous reaction