M18Q3: Voltaic Cells

Learning Objectives

  • Identify the half-reactions occurring at the anode and the cathode in a voltaic cell, as well as the polarity of the electrodes, the direction of electron flow in the external circuit, and the direction of ion flow in the salt bridge.
  • Use and interpret line notation for a voltaic cell.
    | Line Notation |

| Key Concepts and Summary | Glossary | End of Section Exercises |

Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure 1). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. Electrons spontaneously move from the copper metal to the silver ions.

This figure includes three photographs. In a, a test tube containing a clear, colorless liquid is shown with a loosely coiled copper wire outside the test tube to its right. In b, the wire has been submerged into the clear colorless liquid in the test tube and the surface of the wire is darkened. In c, the liquid in the test tube is a bright blue-green color, the wire in the solution appears dark near the top, and a grey “fuzzy” material is present at the bottom of the test tube on the lower portion of the copper coil, giving a murky appearance to the liquid near the bottom of the test tube.
Figure 1. When a clean piece of copper metal is placed into a clear solution of silver nitrate (a), an oxidation-reduction reaction occurs that results in the exchange of Cu2+ for Ag+ ions in solution. As the reaction proceeds (b), the solution turns blue (c) because of the copper ions present, and silver metal is deposited on the copper strip as the silver ions are removed from solution. (credit: modification of work by Mark Ott)

Galvanic or voltaic cells involve spontaneous oxidation reduction reactions in which the half-reactions are separated so that electrons, or current, can flow through an external wire (Figure 2).

This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a blue solution and is labeled below as “1 M solution of copper (II) nitrate ( C u ( N O subscript 3 ) subscript 2 ).” The beaker on the right contains a colorless solution and is labeled below as “1 M solution of silver nitrate ( A g N O subscript 3 ).” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. The plug in the left beaker is labeled “Porous plug.” At the center of the diagram, the tube is labeled “Salt bridge ( N a N O subscript 3 ). Each beaker shows a metal strip partially submerged in the liquid. The beaker on the left has an orange brown strip that is labeled “C u anode negative” at the top. The beaker on the right has a silver strip that is labeled “A g cathode positive” at the top. A wire extends from the top of each of these strips to a rectangular digital readout indicating a reading of positive 0.46 V that is labeled “Voltmeter.” An arrow points toward the voltmeter from the left which is labeled “Flow of electrons.” Similarly, an arrow points away from the voltmeter to the right which is also labeled “Flow of electrons.” A curved arrow extends from the C u strip into the surrounding solution. The tip of this arrow is labeled “C u superscript 2 plus.” A curved arrow extends from the salt bridge into the beaker on the left into the blue solution. The tip of this arrow is labeled “N O subscript 3 superscript negative.” A curved arrow extends from the solution in the beaker on the right to the A g strip. The base of this arrow is labeled “A g superscript plus.” A curved arrow extends from the colorless solution to salt bridge in the beaker on the right. The base of this arrow is labeled “N O subscript 3 superscript negative.” Just right of the center of the salt bridge on the tube an arrow is placed on the salt bridge that points down and to the right. The base of this arrow is labeled “N a superscript plus.” Just above this region of the tube appears the label “Flow of cations.” Just left of the center of the salt bridge on the tube an arrow is placed on the salt bridge that points down and to the left. The base of this arrow is labeled “N O subscript 3 superscript negative.” Just above this region of the tube appears the label “Flow of anions.”
Figure 2. An example of a voltaic cell where the half-reactions are occurring in separate beakers.

The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode, a solid electrical conductor that carries electric current into non-metallic substances, such as aqueous solutions, gases, or non-metallic liquids and solids. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire.

The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO3). The silver is undergoing reduction; therefore, the silver electrode is the cathode.

At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions. The salt bridge consists of a concentrated, nonreactive, electrolyte solution, such as the sodium nitrate (NaNO3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral. Without the salt bridge, the compartments would not remain electrically neutral and no significant current would flow.

The instant the circuit is completed, the voltmeter reads +0.46 V, this is called the cell potential. The cell potential is created when the two dissimilar metals are connected, and is a measure of the energy per unit charge available from the oxidation-reduction reaction.

When the electrochemical cell is constructed in this fashion, a positive cell potential indicates a spontaneous reaction and that the electrons are flowing from the anode to the cathode.

To summarize, in Figure 2:

  • Electrons flow from the anode to the cathode: left to right in Figure 2.
  • The electrode in the left half-cell is the anode because oxidation occurs here. The anions from the salt bridge flow towards the anode.
  • The electrode in the right half-cell is the cathode because reduction occurs here. The cations from the salt bridge flow towards the cathode.
  • In Figure 2, the cell potential of +0.46 V results from the inherent differences in the nature of the materials used to make the two half-cells.
  • The salt bridge must be present to complete the circuit and both an oxidation and reduction must occur for current to flow.

There are many possible reactions used in galvanic cells, so a shorthand notation can be used to describe them. The line notation (sometimes called a cell notation) provides information about the various species involved in the reaction. This notation also works for other types of cells. A vertical line, │, denotes a phase boundary and a double line, ‖, the salt bridge. Information about the anode is written to the left, followed by the anode solution, then the salt bridge (when present), then the cathode solution, and, finally, information about the cathode to the right. The line notation for the galvanic cell in Figure 2 is then:

Cu(s) | Cu2+(aq, 1 M) || Ag+(aq, 1 M) | Ag(s)

Note that spectator ions are not included and that the simplest form of each half-reaction was used. When known, the initial concentrations of the various ions are usually included.

Some oxidation-reduction reactions involve species that are poor conductors of electricity, and so an inert electrode is used that does not participate in the reactions. Frequently, the electrode is platinum, gold, or graphite, all of which are inert to many chemical reactions. One such system is shown in Figure 3. Magnesium undergoes oxidation at the anode on the left in the figure and hydrogen ions undergo reduction at the cathode on the right. The reaction may be summarized as

Oxidation, in anode cell: Mg(s) Mg2+(aq)  +  2 e
Reduction, in cathode cell: 2 H+(aq)  +  2 e H2(g)
Overall Reaction: Mg(s)  +  2 H+(aq) Mg2+(aq)  +  H2(g)

The cell used an inert platinum wire for the cathode. The platinum does not participate in the chemistry and is not included in the chemical equation, however, it is included in the line notation as follows:

Mg(s) | Mg2+(aq) || H+(aq) | H2(g) | Pt(s)

This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a colorless solution and is labeled “Solution of M g C l subscript 2.” The beaker on the right contains a colorless solution and is labeled “Solution of H C l.” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. At the center of the diagram, the tube is labeled “Salt bridge. Each beaker shows a metal coils submerged in the liquid. The beaker on the left has a thin grey coiled strip that is labeled “M g coil.” The beaker on the right has a black wire that is oriented horizontally and coiled up in a spring-like appearance that is labeled “Non reactive platinum wire.” A wire extends across the top of the diagram that connects the ends of the M g strip and platinum wire just above the opening of each beaker. This wire is labeled “Conducting wire.” At the center of this wire above the two beakers near the center of the diagram is a right pointing arrow with the label “e superscript negative” at its base. Bubbles appear to be rising from the coiled platinum wire in the beaker. These bubbles are labeled “Bubbles of H subscript 2.” An arrow points down and to the right from a plus sign at the upper right region of the salt bride. An arrow points down and to the left from a negative sign at the upper left region of the salt bride. A curved arrow extends from the grey plug at the left end of the salt bridge into the surrounding solution in the left beaker. The label “Oxidation M g right pointing arrow M g superscript 2 plus plus 2 e superscript negative” appears beneath the left beaker. The label “Reduction 2 H superscript plus plus 2 e superscript negative right pointing arrow H subscript 2” appears beneath the right beaker.
Figure 3. The oxidation of magnesium to magnesium ion occurs in the beaker on the left side in this apparatus; the reduction of hydrogen ions to hydrogen occurs in the beaker on the right. A nonreactive (inert) platinum wire conducts electrons into the right beaker.

Example 1: Using Line Notation

1.  Consider a galvanic cell consisting of

2 Cr(s)  +  3 Cu2+(aq)   →   2 Cr3+(aq)  +  3 Cu(s)

Write the oxidation and reduction half-reactions and write the reaction using line notation. Which reaction occurs at the anode? The cathode?

Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr3+, and Cu2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives:

Oxidation, in anode cell: 2 Cr(s)   →   2 Cr3+(aq)  +  6 e
Reduction, in cathode cell: 3 Cu2+(aq)  +  6 e   →   3 Cu(s)
Overall Reaction: 2 Cr(s)  +  3 Cu2+(aq)   →   2 Cr3+(aq)  +  3 Cu(s)

Line notation uses the simplest form of each of the equations (without coefficients), and starts with the reaction at the anode. No concentrations were specified so:

Cr(s) | Cr3+(aq) || Cu2+(aq) | Cu(s)

2.  Consider a galvanic cell consisting of:

5 Fe2+(aq)  +  MnO4(aq)  +  8 H+(aq)   →   5 Fe3+(aq)  +  Mn2+(aq)  +  4 H2O(l)

Write the oxidation and reduction half-reactions and write the reaction using line notation. Which reaction occurs at the anode? The cathode?

Solution
By inspection, Fe2+ undergoes oxidation when one electron is lost to form Fe3+, and MnO4 is reduced as it gains five electrons to form Mn2+. Balancing the charge gives:

Oxidation, in anode cell: 5 Fe2+(aq)   →   5 Fe3+(aq)  +  5 e
Reduction, in cathode cell: MnO4(aq)  +  8 H+(aq)  +  5 e   →   Mn2+(aq)  +  4 H2O(l)
Overall Reaction: 5 Fe2+(aq)  +  MnO4(aq)  +  8 H+(aq)   →   5 Fe3+(aq)  +  Mn2+(aq)  +  4 H2O(l)

Line notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode.

Pt(s) | Fe2+(aq), Fe3+(aq) || MnO4(aq), H+(aq), Mn2+(aq) | Pt(s)

Oxidation occurs at the anode and reduction at the cathode.

Check Your Learning
Use line notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.

Answer:

From the information given in the problem:

Oxidation, in anode cell: Zn(s)   →   Cr2+(aq)  +  2 e
Reduction, in cathode cell: Cu2+(aq)  +  2 e   →   Cu(s)
Overall Reaction: Zn(s)  +  Cu2+(aq)   →   Zn2+(aq)  +  Cu(s)

Using line notation:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Key Concepts and Summary

A common type of electrochemical cell is a galvanic or voltaic cell, which involves spontaneous redox reaction with electrons flowing through an external wire. Each half-cell has metal electrodes that carry the electric current from the anode, where oxidation occurs, to the cathode, where reduction occurs. In order to complete the circuit, a salt bridge is used to connect the aqueous phases of the anode and cathode and a flow of ions, anions toward the anode and cations toward the cathode, maintain charge neutrality. Line notation is often used as a shorthand way to describe a voltaic cell, rather than having to draw a full diagram.

Glossary

anode
the electrode where oxidation, or loss of electrons, occurs

cathode
the electrode where reduction, or gain of electrons, occurs

cell potential
a measure of the energy per unit charge available from the redox reaction

electrode
solid electrical conductor that carries electric current into non-metallic substances, such as aqueous solutions, gases, or non-metallic liquids and solids

galvanic or voltaic cells
electrochemical cell that involves a spontaneous redox reaction in which the half-reactions are separated so that electrons (or current) can flow through an external wire

line notation
a shorthand notation that provides information about the half-cells and which species is oxidized or reduced in an electrochemical cell

salt bridge
a concentrated, non-reactive electrolyte solution used to close the electrochemical circuit in a voltaic cell by transmitting current with the flow of ions

Chemistry End of Section Exercises

  1. Why is a salt bridge necessary in a voltaic cell?
  2. An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.
  3. A voltaic cell has been assembled to use the following overall chemical reaction:

    Fe(s) + Cu(NO3)2(aq) → Fe(NO3)2(aq) + Cu(s)

    1. Write the half reactions for each cell and indicate which reaction is occurring at the anode and which is occurring at the cathode.
    2. Fe(NO3)2 solutions are pale green and Cu(NO3)2 solutions are blue. What visual observations do you predict you will see as the voltaic cell reaction progresses?
    3. The two half-cells are connected by a salt bridge containing concentrated KNO3. Into which cell will each ion flow? Explain.
    4. Write the cell notation for this voltaic cell.
  4. A voltaic cell has been assembled to use the following overall chemical reaction:

    Ni(s) + 2 Ag+(aq) → Ni2+(aq) + 2 Ag(s)

    1. Write the half reactions for each cell and indicate which reaction is occurring at the anode and which is occurring at the cathode.
    2. Determine the electron flow over an external wire
    3. The two half-cells are connected by a salt bridge containing concentrated KNO3. Into which cell will each ion flow? Explain.
    4. Write the cell notation for this voltaic cell.

Answers to Chemistry End of Section Exercises

  1. Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.
  2. Oxidation reactions convert metal to metal ions, which pass into solution, so mass loss will be seen at the anode in a voltaic cell.
    1. Anode: Fe(s) → Fe2+(aq) + 2 e ; Cathode: Cu2+(aq) + 2 e → Cu(s)
      Note that the NO3 ions are left out of the half-reactions because they are spectator ions in this process.
    2. The anode half-cell will contain a pale green solution that will become more green as the reaction progresses and more Fe2+ ions are produced. The cathode reaction will contain a blue solution that will become more colorless as the reaction progresses and Cu2+ ions are consumed.
    3. As the reaction progresses, positive Fe ions are being produced so anions, NO3, will flow from the salt bridge into the anode cell to keep the solution neutrally charged. At the same time, positive Cu ions are being consumed, so cations will flow from the salt bridge into the cathode cell to keep the solution neutrally charged.
    4. Fe(s) | Fe2+(aq) || Cu2+(aq) | Cu(s)
  3. Oxidation reactions convert metal to metal ions, which pass into solution, so mass loss will be seen at te anode in a voltaic cell.
    1. Anode: Ni(s) → Ni2+(aq) + 2 e ; Cathode: Ag+(aq) + e → Ag(s)
    2. Electrons will flow from the oxidation half-cell, containing the Ni/Ni2+, to the reduction cell, containing the Ag/Ag+.
    3. As the reaction progresses, Ni cations are being produced so anions, NO3, will flow from the salt bridge into the anode cell to keep the solution neutrally charged. At the same time, Ag cations are being consumed, so cations will flow from the salt bridge into the cathode cell to keep the solution neutrally charged.
    4. Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
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