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23.2: Cell Potentials

Created by: CK-12

Lesson Objectives

  • Describe how an electrical potential is generated in an electrochemical cell.
  • Describe the standard hydrogen electrode and how it is used to determine the standard cell potentials of other half-cells.
  • Calculate the standard cell potentials from a table of standard reduction potentials.
  • Predict the behavior of oxidizing and reducing agents based on their position in the table of standard reduction potentials.

Lesson Vocabulary

  • cell potential
  • electrical potential
  • reduction potential
  • standard cell potential
  • standard hydrogen electrode

Check Your Understanding

Recalling Prior Knowledge

  • What are the parts of an electrochemical cell and how does it work?
  • What is the difference between a spontaneous and a nonspontaneous reaction?

Voltaic cells harness energy from spontaneous redox reactions to produce electrical energy. However, not all redox reactions have the same ability to generate an electric current. In this lesson, you will learn about electrical potential and how to determine the potential of various electrochemical cells.

Reduction Potential

Electrical potential is a measurement of the ability of a voltaic cell to produce an electric current. Electrical potential is typically measured in volts (V). Like energy, electrical potential is a relative term; it can only be measured by comparison with something else. The voltage that is produced by a given voltaic cell is the difference in electrical potential between the two half-cells, but it is not possible to measure the electrical potential of an isolated half-cell. For example, if only a zinc half-cell were constructed, no complete redox reaction can occur, so no electrical potential can be measured. It is only when another half-cell is combined with the zinc half-cell that an electrical potential difference, or voltage, can be measured.

The electrical potential of a cell results from a competition for electrons. In the zinc-copper voltaic cell described in the previous lesson, the copper(II) ions were reduced to copper metal. That is because the Cu2+ ions have a greater attraction for electrons than the Zn2+ ions in the other half-cell. Instead, the zinc metal is oxidized. A reduction potential measures the tendency of a given half-reaction to occur as a reduction in an electrochemical cell. In a given voltaic cell, the half-cell that has the greater reduction potential is the one in which reduction will occur. In the half-cell with the lower reduction potential, the reverse process (oxidation) will occur. The cell potential (Ecell) is the difference in reduction potential between the two half-cells in an electrochemical cell.

Standard Cell Potentials

The standard cell potential (E0cell) is the potential of an electrochemical cell when the temperature is 25°C, all aqueous components are present at a concentration of 1 M, and all gases are at the standard pressure of 1 atm. The standard cell potential can be calculated by finding the difference between the standard reduction potentials of the two half-cells.

\text{E}^0_{\text{cell}}=\text{E}^0_{\text{red}}-\text{E}^0_{\text{oxid}}

Since the reduction potentials for half-cells cannot be measured independently, it is necessary to establish a standard to serve as a reference. This reference is given a reduction potential of 0 volts by definition. Every other half-cell can then be compared to this standard electrode in order to determine the reduction potential for any half-cell. The standard hydrogen electrode is a reference electrode that is used with another electrode (half-cell) to determine its standard reduction potential. The standard hydrogen electrode (SHE) is shown in Figure below.

The standard hydrogen electrode is an arbitrary reference cell that is assigned a standard reduction potential of 0.00 V.

The electrode itself is made of platinum, which serves as an inert surface upon which the oxidation or reduction reaction takes place. The electrode is then placed in contact with both hydrogen gas (at a pressure of 1 atm) and an acidic solution in which the concentration of H+ is 1.0 M. Written as a reduction, the following half-reaction takes place in a SHE:

2\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{H}_2(g) \;\;\;\;\;\;\;\;\; \text{E}^0 = 0.00 \ \text{V}

Depending on the relative electrical potential of other half-cell that the SHE is combined with, the hydrogen ions may be reduced or the hydrogen gas may be oxidized. In general, reversing a reaction will also reverse the sign of the corresponding electrical potential. However, reversing the above reaction has no effect on the standard potential, because the opposite of zero is still zero.

\text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2\text{e}^- \;\;\;\;\;\;\;\;\; \text{E}^0 = 0.00 \ \text{V}

Determining Standard Reduction Potentials

When a standard hydrogen half-cell is connected to a standard copper half-cell and connected to a voltmeter (Figure below (A)), the reading is 0.34 V.

(A) The standard hydrogen half-cell is paired with a Cu/Cu2+ half-cell. H2 is oxidized, while Cu2+ is reduced. (B) The standard hydrogen half-cell is paired with a Zn/Zn2+ half-cell. Zn is oxidized, while H+ is reduced.

Observation of the cell shows that the copper(II) ion is reduced to copper metal, while the hydrogen gas is oxidized to hydrogen ions. This is shown below along with the overall reaction taking place in the cell.

&\text{Oxidation:} && \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2\text{e}^- \\&\text{Reduction:} && \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}(s) \\\hline&\text{Overall:} && \text{H}_2(g) + \text{Cu}^{2+}(aq) \rightarrow 2\text{H}^+(aq) + \text{Cu}(s)

In this particular voltaic cell, the SHE is the anode (where oxidation takes place), while the copper half-cell is the cathode. Electrons flow from the SHE to the copper electrode. The standard cell potential (\text{E}^0_{\text{cell}}) is the measured value of 0.34 V, while the potential of the SHE is defined to be zero. This allows us to determine the reduction potential of the copper half-cell.

& \text{E}^0_{\text{cell}} = \text{E}^0_{\text{red}} - \text{E}^0_{\text{oxid}} \\& 0.34 \ \text{V} = \text{E}^0_{\text{Cu}} - 0.00 \ \text{V} \\& \text{E}^0_{\text{Cu}} = 0.34 \ \text{V} - 0.00 \ \text{V} = +0.34 \ \text{V}

The standard reduction potential for the Cu2+|Cu half-cell is thus equal to +0.34 V. In a similar way, the reduction potential for any half-cell can be determined by connecting it to a SHE and measuring the voltage.

When a standard hydrogen half-cell is connected to a standard zinc half-cell (Figure above (B)), the measured voltage is 0.76 V. However, it is observed that the zinc electrode is oxidized to zinc ions while the hydrogen ion is reduced to hydrogen gas.

& \text{Oxidation:} && \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2\text{e}^- \\& \text{Reduction:} && 2\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{H}_2(g) \\\hline& \text{Overall:} && \text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g)

The SHE is now the cathode, while the zinc electrode is the anode. Now it is the E0 for the oxidation half-cell that is the unknown in the standard cell potential equation.

&\text{E}^0_{\text{cell}} = \text{E}^0_{\text{red}} - \text{E}^0_{\text{oxid}} \\&0.76 \ \text{V} = 0.00 \ \text{V} - \text{E}^0_{\text{Zn}} \\&\text{E}^0_{\text{Zn}} = 0.00 \ \text{V} - 0.76 \ \text{V} = -0.76 \ \text{V}

The standard reduction potential for the Zn2+/Zn half-cell is equal to −0.76 V. A negative standard reduction potential for a particular species means that is easier to reduce H+ than to reduce that species. A positive standard reduction potential for a species means that it reduces more easily than H+. The Table below lists many standard reduction potentials under standard conditions. From top to bottom, they are listed in decreasing order of their tendency to occur in the forward direction, as a reduction. Fluorine gas is the most easily reduced, while lithium ions are the most difficult to reduce. Note that this table is the exact opposite of the activity series. Lithium ions are very difficult to reduce, which means that lithium metal is very easy to oxidize.

Standard Reduction Potentials at 25°C
Half Reaction Eo (V)
F2 + 2e- → 2F +2.87
PbO2 + 4H++ SO42− + 2e- → PbSO4 + 2H2O +1.70
MnO4 + 8H++ 5e- → Mn2+ + 4H2O +1.51
Au3+ + 3e- → Au +1.50
Cl2 + 2e- → 2Cl +1.36
Cr2O72− + 14H++ 6e- → 2Cr3+ + 7H2O +1.33
O2 + 4H++ 4e- → 2H2O +1.23
Br2 + 2e- → 2Br +1.07
NO3 + 4H++ 3e- → NO + 2H2O +0.96
2Hg2+ + 2e- → Hg2 2+ +0.92
Hg2+ + 2e- → Hg +0.85
Ag+ + e- → Ag +0.80
Fe3+ + e- → Fe2+ +0.77
I2 + 2e- → 2I +0.53
Cu+ + e- → Cu +0.52
O2 + 2H2O + 4e- → 4OH +0.40
Cu2+ + 2e- → Cu +0.34
Sn4+ + 2e- → Sn2+ +0.13
2H++ 2e- → H2 0.00
Pb2+ + 2e- → Pb −0.13
Sn2+ + 2e- → Sn −0.14
Ni2+ + 2e- → Ni −0.25
Co2+ + 2e- → Co −0.28
PbSO4 + 2e- → Pb + SO42− −0.31
Cd2+ + 2e- → Cd −0.40
Fe2+ + 2e- → Fe −0.44
Cr3+ + 3e- → Cr −0.74
Zn2+ + 2e- → Zn −0.76
2H2O + 2e- → H2 + 2OH −0.83
Mn2+ + 2e- → Mn −1.18
Al3+ + 3e- → Al −1.66
Be2+ + 2e- → Be −1.70
Mg2+ + 2e- → Mg −2.37
Na+ + e- → Na −2.71
Ca2+ + 2e- → Ca −2.87
Sr2+ + 2e- → Sr −2.89
Ba2+ + 2e- → Ba −2.90
Rb+ + e- → Rb −2.92
K+ + e- → K −2.92
Cs+ + e- → Cs −2.92
Li+ + e- → Li −3.05

Calculating Standard Cell Potentials

In order to function, any electrochemical cell must consist of two half-cells. Table above can be used to determine the reactions that will occur and the standard cell potential for any combination of two half-cells without actually constructing the cell. The half-cell with the higher reduction potential according to the table will undergo reduction, while the half-cell with the lower reduction potential will undergo oxidation. If those specifications are followed, the overall cell potential will be a positive value. The cell potential must be positive in order for the redox reaction in the cell to be spontaneous. If a negative cell potential were calculated, the reaction would not be spontaneous. However, that reaction would be spontaneous in the reverse direction.

Sample Problem 23.1: Calculating Standard Cell Potentials

Calculate the standard cell potential of a voltaic cell that uses the Ag|Ag+ and Sn|Sn2+ half-cell reactions. Write the balanced equation for the overall cell reaction that occurs. Identify the anode and the cathode.

Step 1: List the known values and plan the problem.

Known

  • \text{E}^0_{\text{Ag}} = +0.80 \ \text{V}
  • \text{E}^0_{\text{Sn}} = -0.14 \ \text{V}

Unknown

  • \text{E}^0_{\text{cell}} = ? \ \text{V}

The silver half-cell will undergo reduction because its standard reduction potential is higher. The tin half-cell will undergo oxidation. The overall cell potential can be calculated by using the equation \text{E}^0_{\text{cell}} = \text{E}^0_{\text{red}} - \text{E}^0_{\text{oxid}}.

Step 2: Solve.

& \text{Oxidation (anode):} && \text{Sn}(s)\rightarrow \text{Sn}^{2+}(aq) + 2\text{e}^- \\& \text{Reduction (cathode):} && \text{Ag}^+(aq) + \text{e}^- \rightarrow \text{Ag}(s)

Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. The silver half-cell reaction must be multiplied by two. After doing that and adding to the tin half-cell reaction, the overall equation is obtained.

Overall Equation: Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)

The cell potential is calculated.

\text{E}^0_{\text{cell}} = \text{E}^0_{\text{red}} - \text{E}^0_{\text{oxid}} = +0.80 \ \text{V} - (-0.14 \ \text{V}) = +0.94 \ \text{V}

Step 3: Think about your result.

The standard cell potential is positive, so the reaction is spontaneous as written. Tin is oxidized at the anode, while silver ion is reduced at the cathode. Note that the voltage for the silver ion reduction is not doubled even though the reduction half-reaction had to be doubled to balance the overall redox equation.

Practice Problems
  1. For the following cell combinations, write the overall cell reaction and calculate the standard cell potential.
    1. Cd|Cd2+ and Cu|Cu2+
    2. Al|Al3+ and Mg|Mg2+

Oxidizing and Reducing Agents

A substance that is capable of being reduced very easily is a strong oxidizing agent. Conversely, a substance that is capable of being oxidized very easily is a strong reducing agent. Of the substances found in Table above, fluorine (F2) is the strongest oxidizing agent. It will spontaneously oxidize any of the products from the reduction reactions below it on the table. For example, fluorine will oxidize gold metal according to the following reaction:

3\text{F}_2(g) + 2\text{Au}(s) \rightarrow 6\text{F}^-(aq) + 2\text{Au}^{3+}(aq)

Lithium metal (Li) is the strongest reducing agent. It is capable of reducing any of the reactants above it on the table. For example, lithium will reduce water according to the following reaction:

2\text{Li}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{Li}^+(aq) + 2\text{OH}^-(aq) + \text{H}_2(g)

Using Table above will allow you to predict whether reactions will occur or not. For example, nickel metal is capable of reducing copper(II) ions, but it is not capable of reducing zinc ions. Nickel (Ni) is below Cu2+ but above Zn2+ on the table. This means Ni2+ can outcompete Zn2+ for electrons, but not Cu2+. As a result, Cu2+ can pull electrons away from neutral Ni, but Zn2+ cannot. In order for two species to react spontaneously, they must be in an upper-left to lower-right diagonal orientation on a table of standard reduction potentials, as shown below.

Lesson Summary

  • The ability of a particular electrochemical cell to generate an electric current is called its electrical potential. Reduction potentials measure the tendency of a substance to be reduced in a redox reaction.
  • The standard hydrogen electrode is arbitrarily assigned a standard reduction potential of 0.00 V, and it serves as a reference by which all other half-cell potentials are measured.
  • The standard cell potential for any electrochemical cell can be determined by finding the difference in reduction potentials between the two half-cells. The cell potential must be positive for the overall reaction to be spontaneous.
  • Reduction potentials can be used to make predictions about whether reactions will occur and whether a particular oxidizing or reducing agent is strong enough for a given purpose.

Lesson Review Questions

Reviewing Concepts

  1. How can the standard reduction potential of a half-cell be determined?
  2. The reduction potential of A+
  3. Use Table above to rank the following reducing agents from strongest to weakest: Pb, Cl, Ca, Fe2+, Au, and Cs.
  4. Rank the following oxidizing agents from strongest to weakest: NO3, Al3+, Na+, Br2, MnO4, and H+.

Problems

  1. Determine whether the following redox reactions will occur spontaneously or not. Calculate the standard cell potential in each case.
    1. Co(s) + Ni2+(aq) → Co2+(aq) + Ni(s)
    2. 3Sn2+(aq) + 2Cr3+(aq) → 3Sn4+(aq) + 2Cr(s)
    3. 6Ag(s) + Cr2O72-(aq) + 14H+(aq) → 6Ag+(aq) + 2Cr3+(aq) + 7H2O(l)
  2. For the following cell combinations, write the overall cell reaction and calculate the standard cell potential.
    1. Zn|Zn2+ and Mn|Mn2+
    2. Hg|Hg2+ and Cr|Cr3+
  3. Which of the following is capable of oxidizing bromide ions (Br) to bromine (Br2) under standard state conditions? List all that apply. Pb2+(aq), Cl2(g), Ag+(aq), Cl(aq), H+(aq), MnO4(aq) in acid
  4. Which of the following metals would react with hydrochloric acid? Ni, Co, Cu, Au, Ba
  5. Nitric acid (HNO3) is considered to be an oxidizing acid because the nitrate ion (NO3) is a relatively strong oxidizing agent. Use Table above to write a balanced equation for the reaction that occurs when nitric acid reacts with silver metal. Calculate the standard reduction potential.
  6. A Mg|Mg2+ half-cell is constructed under standard state conditions and connected to another half-cell. The voltage registered on a voltmeter is 1.97 V. What is a possible identity for the other half-cell?

Further Reading / Supplemental Links

Points to Consider

Nonspontaneous redox reactions can be driven to completion by the application of an electric current in a process called electrolysis.

  • How does an electrolytic cell work?
  • What is electroplating?

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Mar 29, 2013

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CK.SCI.ENG.SE.1.Chemistry-Intermediate.23.2

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