Electrochemistry
Oxidation Numbers
In order to keep
track of what loses electrons and what gains them, we
assign oxidation
Oxidation and Reduction
• A species is oxidized when it loses electrons.
Electrochemistry
Oxidation and Reduction
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron, and they
Oxidation and Reduction
• What is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
Electrochemistry
Assigning Oxidation Numbers
1. Elements in their elemental form have an oxidation number of 0.
Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are positive in certain compounds or ions.
– Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1.
Electrochemistry
Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are positive in certain compounds or ions.
– Fluorine always has an oxidation number of −1.
– The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation
Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a neutral compound is 0.
5. The sum of the oxidation numbers in a polyatomic ion is the charge on the
Electrochemistry
Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction
Balancing Oxidation-Reduction
Equations
This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the
Electrochemistry
The Half-Reaction Method
1. Assign oxidation numbers to
determine what is oxidized and what is reduced.
The Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O. b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
Electrochemistry
The Half-Reaction Method
5. Add the half-reactions, subtracting things that appear on both sides.
6. Make sure the equation is balanced according to mass.
The Half-Reaction Method
Consider the reaction between MnO4− and C
2O42− :
MnO4−
Electrochemistry
The Half-Reaction Method
First, we assign oxidation numbers.
MnO
4−+ C
2
O
42- Mn
2++ CO
2+7 +3 +2 +4
Oxidation Half-Reaction
C2O42− CO 2
To balance the carbon, we add a coefficient of 2:
Electrochemistry
Oxidation Half-Reaction
C2O42− 2 CO 2
The oxygen is now balanced as well. To balance the charge, we must add 2
electrons to the right side.
C2O42− 2 CO
Reduction Half-Reaction
MnO4− Mn2+
The manganese is balanced; to balance the oxygen, we must add 4 waters to
the right side.
MnO4− Mn2+ + 4 H
Electrochemistry
Reduction Half-Reaction
MnO4− Mn2+ + 4 H
2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO
Reduction Half-Reaction
8 H+ + MnO
4− Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO
Electrochemistry
Combining the Half-Reactions
Now we evaluate the two half-reactions together:
C2O42− 2 CO
2 + 2 e−
5 e− + 8 H+ + MnO
4− Mn2+ + 4 H2O
Combining the Half-Reactions
5 C2O42− 10 CO
2 + 10 e−
10 e− + 16 H+ + 2 MnO
4− 2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO
4− + 5 C2O42−
2 Mn2+ + 8 H
Electrochemistry
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO
4− + 5 C2O42−
2 Mn2+ + 8 H
2O + 10 CO2 +10 e−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H+ + 2 MnO
4− + 5 C2O42−
2 Mn2+ + 8 H
Balancing in Basic Solution
• If a reaction occurs in basic solution, one can balance it as if it occurred in acid.
• Once the equation is balanced, add OH−
to each side to “neutralize” the H+ in the
equation and create water in its place.
Electrochemistry
Example
• MnO4- + Br- MnO
2 + BrO3
Voltaic Cells
In spontaneous
oxidation-reduction (redox) reactions, electrons are
transferred and
Electrochemistry
Voltaic Cells
• We can use that
energy to do work if we make the
electrons flow
through an external device.
• We call such a
Voltaic Cells
• A typical cell looks like this.
• The oxidation occurs
at the anode.
• The reduction occurs
at the cathode.
Electrochemistry
Voltaic Cells
Once even one
electron flows from the anode to the cathode, the
charges in each
beaker would not be balanced and the
Voltaic Cells
• Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt
solution, to keep the charges balanced.
– Cations move toward the cathode.
Electrochemistry
Voltaic Cells
• In the cell, then,
electrons leave the anode and flow
through the wire to the cathode.
• As the electrons leave the anode, the cations formed dissolve into the solution in the anode
Voltaic Cells
• As the electrons reach the cathode, cations in the
cathode are
attracted to the now negative cathode. • The electrons are
taken by the cation, and the neutral
Electrochemistry
Electromotive Force (emf)
• Water only spontaneously flows one way in a waterfall. • Likewise, electrons
Electromotive Force (emf)
• The potential difference between the
anode and cathode in a cell is called the
electromotive force (emf).
Electrochemistry
Cell Potential
Cell potential is measured in volts (V).
1 V = 1 J C
A Coulomb is a measure of charge: symbol=Q
Standard Reduction Potentials
The reduction potentials in your book are listed alphabetically, the ones listed here are by decreasing
potential and the ones listed on the AP Exam are listed by increasing
Electrochemistry
Standard Hydrogen Electrode
• Their values are referenced to a standard hydrogen electrode (SHE).
• By definition, the reduction potential for hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e− H
Standard Cell Potentials
The cell potential at standard conditions can be found through this equation:
Ecell = Ered (cathode) − Ered (anode)
Because cell potential is based on the potential energy per unit of charge, it is an intensive property: doesn’t matter how you get there.
Electrochemistry
Cell Potentials
• For the oxidation in this cell, note that since Zn is being oxidized, we reversed the sign on the E.
• For the reduction,
• The Cell E is 1.10V
Eo = 0.76 V
Cell Potentials
Ecell = Ered (cathode) + E (anode)
Electrochemistry
Sample Exercise 20.6 Calculating E°cell from E°red
Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction
1. Write the two half reactions with their correctly oriented E cell from table:
2. Note that you do NOT multiply E (voltage).
Cr2O72- + 14H+ + 6e- 2 Cr3+ + 7H
2O + 6e- 1.33 V
6 I- 3 I2 + 6e- Note sign! -.53 V
Sample Exercise 20.6 Calculating E°cell from E°red
Using data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:
Practice Exercise
2 Al 2 Al 3+ + 6e- Don’t multiply!!! And note sign:
3 I2 + 6e- 6I- Don’t multiply!!:
Electrochemistry
Oxidizing and Reducing Agents
• The strongest
oxidizers have the most positive
reduction potentials. • The strongest
reducers have the most negative
Oxidizing and Reducing Agents
The greater the
Electrochemistry
Sample Exercise 20.8 Determining the Relative Strengths of Oxidizing Agents
Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3–(aq), Ag+
(aq), Cr2O72–(aq).
Solution
Analyze: We are given several ions and asked to rank their abilities to act as oxidizing agents.
Plan: The more readily an ion is reduced (the more positive its E°red value), the stronger it is as an oxidizing
agent.
Solve: From Table 20.1, we have
Because the standard reduction potential of Cr2O72– is the most positive, Cr
2O72– is the strongest oxidizing
agent of the three. The rank order is Ag+ < NO
Helpful Conversions
• The charge on 1 electron is: 1.602 X 10-19 C
• A Coulomb is a fundamental charge. If there are 1 mole of electrons: then the charge on 1 mole of e- is (1.602 X 10-19 C)(6.022 x 1023)
• This is known as a Faraday: 96,500 C/mol e- (rounded) or 9.65 x 104 C.
• A volt (electromotive force) is a J/C.
Electrochemistry
Free Energy
G for a redox reaction can be found by using the equation
G = −nFE
where n is the number of moles of electrons transferred, and F is a constant, the Faraday.
Free Energy
Under standard conditions,
G = −nFE
Electrochemistry
Sample Exercise 20.10 Determining ΔG° and K
(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, , and the equilibrium constant, K, at 298 K for the reaction
(b) Suppose the reaction in part (a) was written
Nernst Equation
• Remember that
G = G + RT ln Q
• This means
Electrochemistry
Nernst Equation
Dividing both sides by −nF, we get the
Nernst equation:
E = E − RT
nF ln Q
or, using base-10 logarithms,
E = E − 2.303 RT
Nernst Equation
At room temperature (298 K),
Thus the equation becomes
E = E − 0.0592
n log Q
2.303 RT
Electrochemistry
Sample Exercise 20.11 Voltaic Cell EMF under Nonstandard Conditions
Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 When
Solution
Analyze: We are given a chemical equation for a voltaic cell and the concentrations of reactants and products under which it operates. We are asked to calculate the emf of the cell under these nonstandard conditions.
Plan: To calculate the emf of a cell under nonstandard conditions, we use the Nernst equation in the form of Equation 20.16.
Solve: We first calculate E° for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E° = 0.79 V. As you will see if you refer back to that exercise, the balanced equation shows six electrons transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is
Concentration Cells
• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
• For such a cell, Ecell would be 0, but Q would not.
• Therefore, as long as the concentrations are different, E will not be 0.
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrolytic Cell
• In an electrolytic cell, an current does work on the system. Electoplating and other processes cause nonspontaneous redox reactions to occur.
• Two electrodes in a molten salt or
Electrolytic Cell
Electrochemistry
Electroplating
Nickel dissolves at the anode, but is plated out on the steel cathode to produce a
nickel-plated steel.
Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.
