Electrochemical Reactions
Electrical Work From Spontaneous Oxidation-Reduction Reactions Voltaic Cells Standard-State Cell Potentials for Voltaic Cells
Predicting Spontaneous Redox Reactions from the Sign of E Standard-State Reduction Half-Cell Potentials Predicting Standard-State Cell Potentials
Line Notation for Voltaic Cells The Nernst Equation Using the Nernst Equation to Measure Equilibrium Constants
Table: Standard State Reduction Potentials

Electrical Work From Spontaneous Oxidation-Reduction Reactions

The following rule can be used to predict whether an oxidation-reduction reaction should occur. Oxidation-reduction reactions should occur when they convert the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent.
Practice Problem 1:

Predict whether zinc metal should dissolve in acid.

Click here to check your answer to Practice Problem 1

We can test this prediction by adding a few chunks of mossy zinc to a beaker of concentrated hydrochloric acid. Within a few minutes, the zinc metal dissolves, and significant amounts of hydrogen gas are liberated.

The reaction in Practice Problem 1 has some of the characteristic features of oxidation-reduction reactions.

It is exothermic, in this case giving off 153.89 kilojoules per mole of zinc consumed.
The equilibrium constant for the reaction is very large (Kc = 6 x 1025), and chemists often write the equation for this reaction as if essentially all of the reactants were converted to products.

Zn(s) + 2 H+(aq) —–> Zn2+(aq) + H2(g)

It can be formally divided into separate oxidation and reduction half-reactions.

Oxidation: Zn —–> Zn2+ + 2 e-
Reduction: 2 H+ + 2 e- —–> H2

By separating the two half-reactions, the energy given off by this reaction can be used to do work.

According to the first law of thermodynamics, the energy given off in a chemical reaction can be converted into heat, work, or a mixture of heat and work. By running the half-reactions in separate containers, we can force the electrons to flow from the oxidation to the reduction half-reaction through an external wire, which allows us to capture as much as possible of the energy given off in the reaction as electrical work.

We can start by immersing a strip of zinc metal into a 1 M Zn2+ ion solution, as shown in the figure below. We then immerse a piece of platinum wire in a second beaker filled with 1 M HCl and bubble H2 gas over the Pt wire. Finally, we connect the zinc metal and platinum wire to form an electric circuit.

diagram

We’ve now made a system in which electrons can flow from one half-reaction, or half-cell, to another. The same driving force that makes zinc metal react with acid when the two are in contact should operate in this system. Zinc atoms on the metal surface lose electrons to form Zn2+ ions, which go into solution.
Oxidation: Zn —–> Zn2+ + 2 e-

The electrons given off in this half-reaction flow through the circuit and eventually accumulate on the platinum wire to give this wire a net negative charge. The H+ ions from the hydrochloric acid are attracted to this negative charge and migrate toward the platinum wire. When the H+ ions touch the platinum wire, they pick up electrons to form hydrogen atoms, which immediately combine to form H2 molecules.
Reduction: 2 H+ + 2 e- —–> H2

The oxidation of zinc metal releases Zn2+ ions into the Zn/Zn2+ half-cell. This half-cell therefore picks up a positive charge that interferes with the transfer of more electrons. The reduction of H+ ions in the H2/H+ half-cell leads to a net negative charge as these H+ ions are removed from the solution. This negative charge also interferes with the transfer of more electrons.

To overcome this problem, we complete the circuit by adding a U-tube filled with a saturated solution of a soluble salt such as KCl. Negatively charged Cl- ions flow out of one end of the U-tube to balance the positive charge on the Zn2+ ions created in one half-cell. Positively charged K+ ions flow out of the other end of the tube to replace the H+ ions consumed in the other half cell. The U-tube is called a salt bridge, because it contains a solution of a salt that literally serves as a bridge to complete the electric circuit.

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Voltaic Cells

Electrochemical cells that use an oxidation-reduction reaction to generate an electric current are known as galvanic or voltaic cells. Because the potential of these cells to do work by driving an electric current through a wire is measured in units of volts, we will refer to the cells that generate this potential from now on as voltaic cells.

Let’s take another look at the voltaic cell in the figure below.

diagram

Within each half-cell, reaction occurs on the surface of the metal electrode. At the zinc electrode, zinc atoms are oxidized to form Zn2+ ions, which go into solution. The electrons liberated in this reaction flow through the zinc metal until they reach the wire that connects the zinc electrode to the platinum wire. They then flow through the platinum wire, where they eventually reduce an H+ ion in the neighboring solution to a hydrogen atom, which combines with another hydrogen atom to form an H2 molecule.

The electrode at which oxidation takes place in a electrochemical cell is called the anode. The electrode at which reduction occurs is called the cathode. The identity of the cathode and anode can be remembered by recognizing that positive ions, or cations, flow toward the cathode, while negative ions, or anions, flow toward the anode. In the voltaic cell shown above, H+ ions flow toward the cathode, where they are reduced to H2 gas. On the other side of the cell, Cl- ions are released from the salt bridge and flow toward the anode, where the zinc metal is oxidized.

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Standard-State Cell Potentials for Voltaic Cells

The cell potential for a voltaic cell is literally the potential of the cell to do work on its surroundings by driving an electric current through a wire. By definition, one joule of energy is produced when one coulomb of electrical charge is transported across a potential of one volt.

1V = 1 J/1 C

The potential of a voltaic cell depends on the concentrations of any species present in solution, the partial pressures of any gases involved in the reaction, and the temperature at which the reaction is run. To provide a basis for comparing the results of one experiment with another, the following set of standard-state conditions for electrochemical measurements has been defined.

All solutions are 1 M.
All gases have a partial pressure of 0.1 MPa (0.9869 atm).

Although standard-state measurements can be made at any temperature, they are often taken at 25oC.

Cell potentials measured under standard-state conditions are represented by the symbol Eo. The standard-state cell potential, Eo, measures the strength of the driving force behind the chemical reaction. The larger the difference between the oxidizing and reducing strengths of the reactants and products, the larger the cell potential. To obtain a relatively large cell potential, we have to react a strong reducing agent with a strong oxidizing agent.

Example: The experimental value for the standard-state cell potential for the reaction between zinc metal and acid is 0.76 volts.
Zn(s) + 2 H+(aq) —-> Zn2+(aq) + H2(g) Eo = 0.76 V

The cell potential for this reaction measures the relative reducing power of zinc metal compared with hydrogen gas. But it doesn’t tell us anything about the absolute value of the reducing power for either zinc metal or H2.

We therefore arbitrarily define the standard-state potential for the reduction of H+ ions to H2 gas as exactly zero volts.
2 H+ + 2 e- —-> H2 Eo = 0.000… V

We will then use this reference point to calibrate the potential of any other half-reaction.

The key to using this reference point is recognizing that the overall cell potential for a reaction must be the sum of the potentials for the oxidation and reduction half-reactions.

Eooverall = Eoox + Eored

If the overall potential for the reaction between zinc and acid is 0.76 volts, and the half-cell potential for the reduction of H+ ions is 0 volts, then the half-cell potential for the oxidation of zinc metal must be 0.76 volts.
Zn —-> Zn2+ + 2 e- Eoox = 0.76 V
+ 2 H+ + 2 e- —-> H2 Eored = 0.00 V

Zn + 2 H+ —-> Zn2+ + H2 Eo = Eoox + Eored = 0.76 V

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Predicting Spontaneous Redox Reactions From the Sign of Eo

The magnitude of the cell potential is a measure of the driving force behind a reaction. The larger the value of the cell potential, the further the reaction is from equilibrium. The sign of the cell potential tells us the direction in which the reaction must shift to reach equilibrium.

Consider the reaction between zinc and acid, for example.
Zn(s) + 2 H+(aq) —-> Zn2+(aq) + H2(g) Eo = 0.76 V

The fact that Eo is positive tells us that when this system is present at standard-state conditions, it has to shift to the right to reach equilibrium. Reactions for which Eo is positive therefore have equilibrium constants that favor the products of the reaction. It is tempting to describe these reactions as “spontaneous.”

What happens to the cell potential when we reverse the direction in which a reaction is written? Turning the reaction around doesn’t change the relative strengths of the oxidizing or reducing agents. The magnitude of the potential must remain the same. But turning the equation around changes the sign of the cell potential, and can therefore turn an unfavorable reaction into one that is spontaneous, or vice versa.
Practice Problem 2:

Use the overall cell potentials to predict which of the following reactions are spontaneous.
(a) Cu(s) + 2 Ag+(aq)arrow.gif (874 bytes)Cu2+(aq) + 2 Ag(s) Eo = 0.46 V
(b) 2 Fe3+(aq) + 2 Cl-(aq)arrow.gif (874 bytes)2 Fe2+(aq) + Cl2(g) Eo = -0.59 V
(c) 2 Fe3+(aq) + 2 I-(aq)arrow.gif (874 bytes)2 Fe2+(aq) + I2(aq) Eo = 0.24 V
(d) 2 H2O2(aq)arrow.gif (874 bytes)2 H2O(l) + O2(aq) Eo = 1.09 V
(e) Cu(s) + 2 H+(aq)arrow.gif (874 bytes)Cu2+(aq) + H2(g) Eo = -0.34 V

Click here to check your answer to Practice Problem 2

Practice Problem 3:

Use the standard-state cell potential for the following reaction
Cu(s) + 2 H+(aq) —–> Cu2+(aq) + H2(g) Eo = -0.34 V

to predict the standard-state cell potential for the opposite reaction.
Cu2+(aq) + H2(g) —–> Cu(s) + 2 H+(aq) Eo = ?

Click here to check your answer to Practice Problem 3

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Standard-State Reduction Half-Cell Potentials

The standard-state cell potentials for some common half-reactions are given in the table below.

Standard-State Reduction Potentials, Eored
Half-Reaction Eored
K+ + e- K -2.924 Best
Ba2+ + 2 e- Ba -2.90 reducing
Ca2+ + 2 e- Ca -2.76 agents
Na+ + e- Na -2.7109
Mg2+ + 2 e- Mg -2.375
H2 + 2 e- 2 H- -2.23
Al3+ + 3 e- Al -1.706
Mn2+ + 2 e- Mn -1.04
Zn2+ + 2 e- Zn -0.7628
Cr3+ + 3 e- Cr -0.74
S + 2 e- S2- -0.508
2 CO2 + 2 H+ + 2 e- H2C2O4 -0.49
Cr3+ + e- Cr2+ -0.41
Fe2+ + 2 e- Fe -0.409
Co2+ + 2 e- Co -0.28
Ni2+ + 2 e- Ni -0.23
Sn2+ + 2 e- Sn -0.1364
Pb2+ + 2 e- Pb -0.1263
Fe3+ + 3 e- Fe -0.036
2 H+ + 2 e- H2 0.0000…
S4O62- + 2 e- 2 S2O32- 0.0895
Oxidizing Sn4+ + 2 e- Sn2+ 0.15 up
power Cu2+ + e- Cu+ 0.158 Reducing
increases Cu2+ + 2 e- Cu 0.3402 power
down O2 + 2 H2O + 4 e- 4 OH- 0.401 increases
Cu+ + e- Cu 0.522
I3- + 2 e- 3 I- 0.5338
MnO4- + 2 H2O + 3 e- MnO2 + 4 OH- 0.588
O2 + 2 H+ + 2 e- H2O2 0.682
Fe3+ + e- Fe2+ 0.770
Hg22+ + 2 e- Hg 0.7961
Ag+ + e- Ag 0.7996
Hg2+ + 2 e- Hg 0.851
H2O2 + 2 e- 2 OH- 0.88
HNO3 + 3 H+ + 3 e- NO + 2 H2O 0.96
Br2(aq) + 2 e- 2 Br- 1.087
2 IO3- + 12 H+ + 10 e- I2 + 6 H2O 1.19
CrO42- + 8 H+ + 3 e- Cr3+ + 4 H2O 1.195
Pt2+ + 2 e- Pt 1.2
MnO2 + 4 H+ + 2 e- Mn2+ + 2 H2O 1.208
O2 + 4 H+ + 4 e- 2 H2O 1.229
Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O 1.33
Cl2(g) + 2 e- 2 Cl- 1.3583
PbO2 + 4 H+ + 2 e- Pb2+ + 2 H2O 1.467
MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O 1.491
Au+ + e- Au 1.68
H2O2 + 2 H+ + 2 e- 2 H2O 1.776
Co3+ + e- Co2+ 1.842
Best S2O82- + 2 e- 2 SO42- 2.05
oxidizing O3(g) + 2 H+ + 2 e- O2(g) + H2O 2.07
agents F2(g) + 2 H+ + 2 e- 2 HF(aq) 3.03

There is no need to remember that reducing agents become stronger toward the upper right corner of this table, or that the strength of the oxidizing agents increases toward the bottom left corner. All you have to do is remember some of the chemistry of the elements at the top and bottom of this table.

Take a look at the half-reaction at the top of the table.
K+ + e- K Eored = -2.924 V

What do we know about potassium metal? Potassium is one of the most reactive metals–it bursts into flame when added to water, for example. Furthermore, we know that metals are reducing agents in all of their chemical reactions. When we find potassium in this table, we can therefore conclude that it is listed among the strongest reducing agents.

Conversely, look at the last reaction in the table.
F2 + 2 e- 2 F- Eored = 3.03 V

Fluorine is the most electronegative element in the periodic table. It shouldn’t be surprising to find that F2 is the strongest oxidizing agent in the above table.

Referring to either end of this table can also help you remember the sign convention for cell potentials. The previous section introduced the following rule: Oxidation-reduction reactions that have a positive overall cell potential are spontaneous. This is consistent with the data in the above table. We know that fluorine wants to gain electrons to form fluoride ions, and the half-cell potential for this reaction is positive.
F2 + 2 e- 2 F- Eored = 3.03 V

We also know that potassium is an excellent reducing agent. Thus, the potential for the reduction of K+ ions to potassium metal is negative
K+ + e- K Eored = -2.924 V

but the potential for the oxidation of potassium metal to K+ ions is positive.
K K+ + e- Eoox = -(-2.924 V) = 2.924 V