1. Chapter 15 Oxidation and Reduction
15.4 Electrical Energy from Oxidation−Reduction Reactions
Learning Goal Use the activity series to determine if oxidation−reduction reaction is spontaneous. Write the half-reactions that occur in a voltaic cell and the cell notation.

2. Spontaneous Oxidation−Reduction Reactions
When we place a zinc metal strip in a solution of Cu2+, a reddish-brown Cu metal accumulates on the Zn strip.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Spontaneous
The activity series predicts which reactions will be spontaneous by ranking metals and H2 in terms of how easily they lose electrons.

3. Activity Series

4. Activity Series, Spontaneous Reactions
We use the activity series to predict the direction of the spontaneous reaction. Mg(s)  Mg2+(aq) + 2 e− Ni(s)  Ni2+ (aq) + 2 e− Since Mg is more reactive than Ni, the reaction will be spontaneous for the oxidation of Mg and reduction of Ni2+. Ni2+ (aq) + 2 e−  Ni(s) reduction of Ni2+

5. Activity Series, Spontaneous Reactions
Combining the half-reactions gives the overall spontaneous reaction.

6. Learning Check
Use the activity series to determine if the following reactions are spontaneous. A. Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s) B. Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)

7. Solution
Use the activity series to determine if the following reactions are spontaneous. A. Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s) Spontaneous, Zn is more reactive than Fe. B. Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g) Not spontaneous, Cu is less reactive than H.

8. Voltaic Cells
A voltaic cell
uses a spontaneous oxidation−reduction reaction
generates electric current
separates the half-reactions into half cells
has electron flow through an external circuit
is also called an electrochemical cell

9. Voltaic Cell
In the following voltaic cell, the oxidation of zinc provides electrons for the reduction of copper
Zn(s)  Zn2+(aq) + 2 e− Oxidation
Cu2+(aq) + 2 e−  Cu (s) Reduction
The overall reaction is
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu (s)
As long as the zinc and copper are in the same container, the electrons are transferred directly from Zn to Cu2+.

10. Half-Cells
Voltaic cells contain
two half-cells, in separate containers, one for oxidation, one for reduction
an electrode in each container in contact with the ionic solution
an anode, the electrode where oxidation takes place
a cathode, the electrode where reduction takes place

11. Half-Cells, Anode
In the voltaic cell,
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
oxidation takes place at the anode, producing electrons for Cu2+
a zinc metal strip is placed in a Zn2+ (ZnSO4) solution
Zn(s)  Zn2+(aq) + 2 e−

12. Half-Cells, Cathode
In the voltaic cell,
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
reduction takes place at the cathode, using the electrons produced at the anode
a copper metal strip is placed in a Cu2+ (CuSO4) solution
Cu2+(aq) + 2 e−  Cu(s) + 2 e−

13. Half-Cells, Anode and Cathode
In the voltaic cell,
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
electrons are transported from the anode to the cathode by a wire connecting the two half-cells
the two half-cells are also connected by a salt bridge containing positive and negative ions also found in the half-cell solutions
electrons move from the cathode through the salt bridge to the anode, completing the circuit

14. Voltaic Cell Diagram
Figure 15.2 In this voltaic cell, the Zn anode is in a Zn2+ solution, and the Cu cathode is in a Cu2+ solution. Electrons produced by the oxidation of Zn flow from the anode through the wire to the cathode where they reduce Cu2+ to Cu. The circuit is completed by the flow of SO42– through the salt bridge.

15. Voltaic Cell Notation
Voltaic cells can be described using a shorthand. For the cell, Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

16. Learning Check
Write the half-cell reactions at the anode and cathode and the shorthand cell notation for Cr(s) + 3Ag+(aq)  Cr3+(aq) + 3Ag(s)

17. Solution
Write the half-cell reactions at the anode and cathode and the shorthand cell notation for Cr(s) + 3Ag+(aq)  Cr3+(aq)+ 3Ag(s) Anode, oxidation Cr(s)  Cr3+(aq) + 3 e− Cathode, reduction Ag+(aq) + 1 e−  Ag(s)

18. Learning Check
Provide the reaction for the following cell: Mn(s)│Mn2+(aq) || Cd2+(aq)│Cd(s)

19. Solution
Provide the reaction for the following cell: Mn(s)│Mn2+(aq) || Cd2+(aq)│Cd(s) Anode, oxidation Mn(s)  Mn2+(aq) + 2 e− Cathode, reduction Cd2+(aq) + 2 e−  Cd(s) Voltaic cell reaction: Mn(s) + Cd2+(aq)  Mn2+(aq) + Cd (s)

20. Lead Storage Batteries
A lead storage battery for a car consists of
H2SO4 and lead (Pb) plates as anodes
Pb(s) + SO42−(aq) 
PbSO4(s) + 2 e−
PbO2 plates as cathodes
PbO2(s) + 2H2SO4(aq) + 2 e−  PbSO4(s)

21. Dry Cell Battery
An acidic dry cell battery has
a zinc metal case (anode)
Zn(s)  Zn2+(aq) + 2 e−
an electrolyte paste with MnO2 (cathode)
2MnO2(s) + 2 e−  Mn2O3 + H2O

22. Learning Check
The following half-reaction takes place in a cadmium battery: Cd(s) + 2OH− (aq)  Cd(OH)2(s) + 2 e− A. Is this an oxidation or a reduction? B. Does this half-reaction take place at the anode or cathode? C. What substance is reduced or oxidized?

23. Solution
The following half-reaction takes place in a cadmium battery:
Cd(s) + 2OH− (aq)  Cd(OH)2(s) + 2 e−
Is this an oxidation or a reduction?
Oxidation, with a loss of electrons.
B. Does this half-reaction take place at the anode or cathode? Anode
C. What substance is reduced or oxidized?
Cd(s) is oxidized.