Cell Potentials

Redox Equations and Cell Potentials

Redox reactions are divided into individual equations for the oxidation and reduction processes separately. These are called "half equations":

In the oxidation process, a substance loses one or more electrons: "Loss of Electrons is Oxidation" (OIL)
In the reduction process, a substance gains one or more electrons: "Reduction Is Gain" (RIG)

In Galvanic cells (batteries), the two processes are physically separate and are connected only a wire and a salt bridge. The two reactions happen in separate containers (such as beakers) and the electrons travel from one to the other via the wire, providing the possibility of producing electrical energy for us to use. The two processes together make the Galvanic cell and each is called a "half cell". In fact, the two half cells can be two different parts of an enzyme and the wire then becomes the atoms that connect them in the biomolecule.

Each half cell has a characteristic ability to give or take electrons. This is called its cell potential. The voltage (or potential) of the whole cell depends on the half cell potentials.

Redox reactions always involve an oxidation half cell and a reduction half cell: the electrons produced by the oxidation have to be used by a simultaneous reduction and the electrons required by the reduction have to come from an oxidation.

Standard Reduction Potentials

Standard reduction potentials:

The cell potential, E, of many half cells have been measured and tabulated. For convenience, these are tabulated using "standard" conditions: 1 atm pressure, 1 M solution and 25 °C (= 298 K) in which case they are labelled as E°. They are always written as reductions (Reduction Is Gain). The potential for the opposite, oxidation reaction is simply (-1 × the reduction potential).

Standard Reduction Potentials for Common Metal Ions

Standard reduction potentials for common metal ions:

Reaction E° / V

Au³⁺(aq) + 3e^- → Au(s) +1.50

Au³⁺(aq) + 3e^- → Au(s) +1.50

Pd²⁺(aq) + 2e^- → Pd(s) +0.92

Ag⁺ + e^- → Ag(s) +0.80

Cu⁺ + e^- → Cu(s) +0.53

Cu²⁺ + 2e^- → Cu(s) +0.34

Fe³⁺ + 3e^- → Fe(s) –0.04

Pb²⁺ + 2e^- → Pb(s) –0.13

Sn²⁺ + 2e^- → Sn(s) –0.14

Ni²⁺ + 2e^- → Ni(s) –0.24

Co²⁺ + 2e^- → Co(s) –0.28

Fe²⁺ + 2e^- → Fe(s) –0.44

Cr³⁺ + 3e^- → Cr(s) –0.74

Zn²⁺ + 2e^- → Zn(s) –0.76

Cr²⁺ + 2e^- → Cr(s) –0.89

Al³⁺ + 3e^- → Al(s) –1.68

Mg²⁺ + 2e^- → Mg(s) –2.36

Na⁺ + e^- → Na(s) –2.71

Ca²⁺ + 2e^- → Ca(s) –2.87

Li⁺ + e^- → Li(s) –3.04

Reaction	E° / V
Au³⁺(aq) + 3e^- → Au(s)	+1.50
Au³⁺(aq) + 3e^- → Au(s)	+1.50
Pd²⁺(aq) + 2e^- → Pd(s)	+0.92
Ag⁺ + e^- → Ag(s)	+0.80
Cu⁺ + e^- → Cu(s)	+0.53
Cu²⁺ + 2e^- → Cu(s)	+0.34
Fe³⁺ + 3e^- → Fe(s)	–0.04
Pb²⁺ + 2e^- → Pb(s)	–0.13
Sn²⁺ + 2e^- → Sn(s)	–0.14
Ni²⁺ + 2e^- → Ni(s)	–0.24
Co²⁺ + 2e^- → Co(s)	–0.28
Fe²⁺ + 2e^- → Fe(s)	–0.44
Cr³⁺ + 3e^- → Cr(s)	–0.74
Zn²⁺ + 2e^- → Zn(s)	–0.76
Cr²⁺ + 2e^- → Cr(s)	–0.89
Al³⁺ + 3e^- → Al(s)	–1.68
Mg²⁺ + 2e^- → Mg(s)	–2.36
Na⁺ + e^- → Na(s)	–2.71
Ca²⁺ + 2e^- → Ca(s)	–2.87
Li⁺ + e^- → Li(s)	–3.04

Notice that all the reactive metals (Na, Li, Ca etc) are at the bottom of the table and have low or even negative E° values: they are easy to oxidize but hard to reduce. The unreactive metals (Ag, Au, Pd etc) are at the top of the table and have high, positive E° values: they are easy to reduce but hard to oxidize.

Connecting Half Cells

Connecting half cells:

Redox reactions always involve an oxidation half cell and a reduction half cell. Using the standard cell potentials, it is possible to work out which will be which when two cells are connected:

If two half cells are connected:

the one with the larger E° value (more positive or less negative) is the one where reduction occurs. Its reaction is the same as in the standard reduction table.

the one with the smaller E° value (less positive or more negative) is the one where oxidation occurs. As the standard reduction table has the reaction as a reduction, it must be reversed.

Once we know which half cell involves the reduction and which involves the oxidation, we can combine them to make the overall redox reaction. We can also work out the maximum voltage that our cell will deliver. The cell potential is given by:

E°_cell = E°_{reduction half cell} - E°_{oxidation half cell}

Example 1

Example 1:

A zinc half cell has a strip of zinc metal in a beaker containing Zn²⁺(aq) ions. A copper half cell has a strip of copper metal in a beaker containing Cu²⁺ ions. The two half cells are connected together to make a Galvanic cell. What reaction will occur and what will the cell potential be?

There are two possible reduction processes:

²⁺

There are two possible oxidation processes:

²⁺

Redox reactions always involve an oxidation half cell and a reduction half cell.

The standard reduction potentials from the table are E° = -0.76 V for Zn²⁺ / Zn(s) and +0.34 V for Cu²⁺ / Cu(s).

From above, the half cell with the larger E° value (more positive or less negative) is the one where reduction occurs. This is the Cu²⁺ / Cu(s) half cell. Hence, the reduction reaction is:

Cu²⁺(aq) + 2e^- → Cu(s)

From above, the half cell with the smaller E° value (less positive or more negative) is the one where oxidation occurs. This is the Zn²⁺ / Zn(s) half cell. Hence, the oxidation reaction is:

Zn(s) → Zn²⁺(aq) + 2e^-

The overall, balanced equation for the redox reaction is then:

Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq)

Finally, the cell potential is:

E°_cell = E°_{reduction half cell} - E°_{oxidation half cell} = (+0.34 V) - (-0.76 V) = +1.10 V

Example 2

Example 2:

A silver half cell has a strip of silver metal in a beaker containing Ag⁺(aq) ions. A copper half cell has a strip of copper metal in a beaker containing Cu²⁺ ions. The two half cells are connected together to make a Galvanic cell. What reaction will occur and what will the cell potential be?

There are two possible reduction processes:

⁺

²⁺

There are two possible oxidation processes:

⁺

²⁺

Redox reactions always involve an oxidation half cell and a reduction half cell.

The standard reduction potentials from the table are E° = +0.80 V for Ag⁺ / Ag(s) and +0.34 V for Cu²⁺ / Cu(s).

From above, the half cell with the larger E° value (more positive or less negative) is the one where reduction occurs. This is the Ag⁺ / Ag(s) half cell. Hence, the reduction reaction is:

Ag⁺(aq) + e^- → Ag(s)

From above, the half cell with the smaller E° value (less positive or more negative) is the one where oxidation occurs. This is the Cu²⁺ / Cu(s) half cell. Hence, the oxidation reaction is:

Cu(s) → Cu²⁺(aq) + 2e^-

The overall, balanced equation for the redox reaction is then:

2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq)

Notice, that in order for the redox reaction to be charged balanced, the Ag⁺(aq) / Ag(s) reaction has been multipled by two: revise balancing redox equations if you are unsure about this!

Finally, the cell potential is:

E°_cell = E°_{reduction half cell} - E°_{oxidation half cell} = (+0.80 V) - (+0.34 V) = +0.46 V

Notice, that although the Ag⁺(aq) / Ag(s) reaction was multiplied by two to balance the equation, we do NOT multiply its cell potential by two!

Example 3

Example 3:

A nickel half cell has a strip of nickel metal in a beaker containing Ni²⁺(aq) ions. An iron half cell has a strip of iron metal in a beaker containing Fe²⁺ ions. The two half cells are connected together to make a Galvanic cell. What reaction will occur and what will the cell potential be?

There are two possible reduction processes:

²⁺

There are two possible oxidation processes:

²⁺

Redox reactions always involve an oxidation half cell and a reduction half cell.

The standard reduction potentials from the table are E° = -0.24 V for Ni²⁺ / Ni(s) and -0.44 V for Fe²⁺ / Fe(s).

From above, the half cell with the larger E° value (more positive or less negative) is the one where reduction occurs. This is the Ni²⁺ / Ni(s) half cell. Hence, the reduction reaction is:

Ni²⁺(aq) + 2e^- → Ni(s)

From above, the half cell with the smaller E° value (less positive or more negative) is the one where oxidation occurs. This is the Fe²⁺ / Fe(s) half cell. Hence, the oxidation reaction is:

Fe(s) → Fe²⁺(aq) + 2e^-

The overall, balanced equation for the redox reaction is then:

Ni⁺(aq) + Fe(s) → Ni(s) + Fe²⁺(aq)

Finally, the cell potential is:

E°_cell = E°_{reduction half cell} - E°_{oxidation half cell} = (-0.24 V) - (-0.44 V) = +0.20 V

Example 4

Example 4:

A half cell has a strip of nickel metal in a beaker containing Ni²⁺(aq) ions. A different iron half cell to that in example 3 has a strip of iron metal in a beaker containing Fe³⁺ ions. The two half cells are connected together to make a Galvanic cell. What reaction will occur and what will the cell potential be?

There are two possible reduction processes:

²⁺

³⁺

There are two possible oxidation processes:

²⁺

³⁺

Redox reactions always involve an oxidation half cell and a reduction half cell.

The standard reduction potentials from the table are E° = -0.24 V for Ni²⁺ / Ni(s) and -0.04 V for Fe³⁺ / Fe(s).

From above, the half cell with the larger E° value (more positive or less negative) is the one where reduction occurs. This is the Fe³⁺ / Fe(s) half cell. Hence, the reduction reaction is:

Fe³⁺(aq) + 3e^- → Fe(s)

From above, the half cell with the smaller E° value (less positive or more negative) is the one where oxidation occurs. This is the Ni²⁺ / Ni(s) half cell. Hence, the oxidation reaction is:

Ni(s) → Ni²⁺(aq) + 2e^-

The overall, balanced equation for the redox reaction is then:

2Fe³⁺(aq) + 3Ni(s) → 2Fe(s) + 3Ni²⁺(aq)

As in example 2, notice, that in order for the redox reaction to be charged balanced, the Fe³⁺(aq) / Fe(s) reaction has been multipled by two and the Ni²⁺ / Ni(s) reaction has been multiplied by three: revise balancing redox equations if you are unsure about this!

Finally, the cell potential is:

E°_cell = E°_{reduction half cell} - E°_{oxidation half cell} = (-0.04 V) - (-0.24 V) = +0.20 V

As in example 2, notice, that although the half reactions were multiplied in order to balance the equation, we do NOT multiply their cell potentials when they are combined!

Self Test Quizzes

Test your understanding using the quizzes:

Each quiz is made up of 5 questions randomly drawn from a large set so you can repeat the quiz many times over. The results are not stored!
Practice multiple choice questions (not an assessment)
Note: you will need to use the standard cell potentials to complete the quiz.

Cell Potentials

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Redox Equations and Cell Potentials