8.42 - Reactions at the electrodes

IB Syllabus ref: 9.4.1, 2

9.4.1 Explain how a redox reaction is used to produce electricity in a voltaic cell. This should include a diagram to show how two half cells can be connected by a salt bridge. Examples of half cells are Mg. Zn, Fe, and Cu in solutions of their ions

9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode)

As we have seen, the electrical cell is a controlled form of chemical reaction. Here we examine the chemical reactions occurring at the electrodes in electrochemical cells.

 

The negative electrode - anode

This is the electrode that actually produces the electrons that would flow around the external circuit. It is always the electrode of the most reactive metal (SL), the half cell with the most negative electrode potential (HL only).

The metal ions in the electrode dissolve as ions, leaving their electrons behind on the electrode. These electrons are then able to flow around the external circuit. For example in the zinc-copper voltaic cell the zinc half cell is the negative electrode: In voltaic cells this electrode is called the anode.

Zn(s) Zn2+(aq) + 2e

The reactions at the negative electrode always involve electrons being 'dropped off' by metals in the electrodes dissolving as ions. This is a process of oxidation; the metal atoms are getting oxidised to ions (and releasing electrons).

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The positive electrode

This is electrode towards which the electrons in the external circuit flow. It is the electrode that has positive ions (from the solution) removing the electrons as they arrive. In the case of the zinc-copper voltaic cell, the copper half cell is the positive electrode:

Cu2+(aq) + 2e Cu(s)

In voltaic cells this electrode is called the cathode. It is the electrode at which the electrons are 'picked up' by the ions from the solution. The reactions occurring at this electrode are always reductions. The ions from the solution collect electrons to become atoms.

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The cell reaction

The overall cell reaction can be obtained by adding together the reactions occurring at the positive and negative electrodes. If the number of electrons involved at both electrodes is different then the equation must be manipulated by mutiplication to make the number of electrons involved in each the same.

Example 1: The copper-zinc voltaic cell

  • Reaction at the negative electrode: Zn(s) Zn2+(aq) + 2e
  • Reaction at the positive electrode: Cu2+(aq) + 2e Cu(s)

In this case the number of electrons involved at both electrodes is the same, so the equations can be added together:

Zn(s) Zn2+(aq) + 2e
Cu2+(aq) + 2e Cu(s)

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

 

Example 2: The copper-silver voltaic cell

  • Reaction at the negative electrode: Cu(s) Cu2+(aq) + 2e
  • Reaction at the positive electrode: Ag+(aq) + 1e Ag(s)

In this case the number of electrons involved at both electrodes is not the same, so before the the equations can be added together the silver equation must be doubled (multiplied by 2):

Cu(s) Cu2+(aq) + 2e
2Ag+(aq) + 2e 2Ag(s)

Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

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Cell representation conventions

Descriptions of voltaic cells can be drawn using a standard convention. The vertical lines represent barriers between different phases and the salt bridge is represented by a double vertical line. The chemical components of the cell are written as drawn in a cell diagram. The external circuit is not drawn. By convention, the most negative half cell is written first.

The copper-zinc voltaic cell can be represented as:

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

This literally means reading from left to right: There is a solid zinc electrode dipped in a solution containing zinc 2+ ions. This zinc 2+ ion solution is connected by means of a salt bridge to a solution containing copper 2+ ions. This copper 2+ ion solution has a solid copper electrode immersed in it. The external circuit joins the solid copper electrode to the solid zinc electrode.

If two different phases are present in the same half cell, then these are usually shown with a semi-colon between them. This is the case in the standard hydrogen electrode (see section )

Pt|H2(g);H+(aq)||Cu2+(aq)|Cu(s)

Here, the salt bridge joins the copper 2+ ion solution to a solution containing hydrogen ions. These hydrogen ions are also in contact with hydrogen gas at the surface of a platinum electrode.

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Worked examples

Q842-01 A voltaic cell is made from magnesium and iron half cells. Magnesium is a more reactive metal than iron. Which statement is correct when the cell produces electricity?

  1. Electrons are lost from magnesium atoms
  2. The concentration of the Fe2+ ions increases
  3. Electrons flow from the iron half-cell to the magnesium half-cell
  4. Negative ions flow through the salt bridge from the magnesium half-cell to the iron half-cell
Answer

As magnesium is the most reactive metal it preferentially becomes ions according to the equation:

Mg(s) Mg2+(aq) + 2e

The electrons produced flow around the external circuit to the iron half cell. Statement A is correct.
Q842-02 A voltaic cell made from copper and zinc half cells has the equation shown below:

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

Which statement is correct when the cell produces electricity?

  1. Electrons are lost from zinc atoms
  2. The mass of the copper electrode decreases
  3. Electrons flow from the copper half-cell to the zinc half-cell
  4. Negative ions flow through the salt bridge from the zinc half-cell to the copper half-cell
Answer

The equation shows us that the zinc provides electrons for the copper (II) ions, therefore statement A, electrons are lost from zinc atoms, is correct.
Q842-03 What occurs during the operation of a voltaic cell based on the following reaction?

Ni(s) + Pb2+(aq) --> Ni2+(aq) + Pb(s)

 External circuit Ion movement in solution
A.
 Electrons move from Ni to Pb Pb2+(aq) moves away from Pb(s)
B.
 Electrons move from Ni to Pb Pb2+(aq) moves towards Pb(s)
C.
 Electrons move from Pb to Ni Ni2+(aq) moves away from Ni(s)
D.
 Electrons move from Pb to Ni Ni2+(aq) moves towards Ni(s)

Answer

The cell equation shows us that nickel atoms lose electrons and pass them around the external circuit to the lead ions. In the solutions the lead ions in the lead half cell pick up the electrons to become lead atoms. For this to happen, the Pb2+(aq) ions must move towards the Pb(s) electrode. The correct combination is selection B.
Q842-04 In a voltaic cell, oxidation occurs at which of the following?
  1. anode
  2. cathode
  3. salt bridge
  4. electrode at which electrons enter from the outside
Answer

Oxidation is the process of electron removal (OILRIG = oxidation is loss, reduction is gain).

The process of oxidation happens at the most reactive metal half-cell electrode: M(s) M2+(aq) + 2e

These electrons then flow arond the external circuit to the other electrode where they become available for reaction. This electrode, where the electrons are available is called the cathode. It is the positive side of the electro chemical cell. Hence oxidation occurs at the other electrode - the anode.
Q842-05 Mercury batteries, like those used in electric watches, provide a voltage of 1.35 V. If the overall oxidation-reduction equation taking place is:

Zn(s) + HgO(s) + 2H2O(l) Zn(OH)2(s) + Hg(l)

the anode reaction must be which of the following?

  1. HgO(s) + 2H2O(l) + 2e Hg(l) + 2OH-(aq)
  2. Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2e
  3. Hg(l) + 2OH-(aq) HgO(s) + 2H20(l) + 2e
  4. Zn(OH)2(s) + 2e Zn(s) + 2OH-(aq)
Answer

The equation shows us that the zinc is turning from zinc atoms to zinc ions, i.e. its oxidation state is increasing from 0 to II. It is being oxidised. The Zn atoms are becoming Zn(OH)2(s) so the half equation at the anode is:

Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2e

The anode is the electrode where the electrons are produced by the metal and the metal is oxidised, it is the negative electrode, which in this case is the zinc.
Q842-06 A battery consists of which type of cells?
  1. electrolytic
  2. electrochemical
  3. electroplating
  4. electromagnetic
Answer

 
Q842-07 Given the lead-acid battery reaction:
Pb + PbO2 + H2SO4
Discharge
Charge
2PbSO4 + 2H2O

Which species is oxidized during battery discharge?

  1. Pb
  2. PbO2
  3. SO42-
  4. H2O
Answer

When the battery discharges it is producing electricity. Therefore it is an electrochemical cell. The forward direction is when the cell discharges, thus the Pb is becoming PbSO4. However, lead is present in two forms, Pb and PbO2. This means that there are two equations involving lead, one is oxidation and the other is reduction:

Pb Pb2+ + 2e

Pb(IV) + 2e Pb2+

In the first equation the lead is oxidised.
Q842-08 Given the reaction for the nickel-cadmium battery:

2NiO(OH) + Cd +2H2O 2Ni(OH)2 + Cd(OH)2

Which of the following species is oxidised during the discharge of the battery?

  1. Ni3+
  2. Ni2+
  3. Cd
  4. Cd2+
Answer

From the equation nickel changes from NiO(OH) in which it has an oxidation state of (III) to Ni(OH)2 in which it has an oxidation state of (II). The nickel has been reduced. Cadmium changes from the element to cadmium (II) - the cadmium has been oxidised.
Q842-09 Which statement best describes how a salt bridge maintains electrical neutrality in the half cells of an electrochemical cell?
  1. It prevents the migration of electrons.
  2. It permits the migration of ions.
  3. It permits the two solutions to mix completely.
  4. It prevents the reaction from occurring spontaneously.
Answer

The salt bridge allows ions to flow in either direction to prevent any build up of charge in the half cells. The correct response is that it permits the migration of ions.
Q842-10 The apparatus shown may be used to carry out a redox reaction.

State the function of the salt bridge. [1]
Write the half equation for the oxidation reaction. [1]
The above reaction are carried out under standard conditions. State what the standard conditions are for the cell. [2]
Using the data given calculate the cell potential. [2]

Eº of Zn2+(aq)|Zn(s) = -0.76 V
Eº of Cu2+(aq)|Cu(s) = +0.34 V
State and explain what happens to the concentration of copper (II) ions when the cell is producing an electrical current. [2]
State two observations that could be made if the zinc rod were to be placed in the solution of copper (II) ions. [2]
Answer

The salt bridge completes the circuit by allowing ions to flow in both directions to equalise charge in the two half cells.

The oxidation reaction is:

Zn Zn2+ + 2e

Standard conditions are a temperature of 25ºC and 1.0 mol dm-3 solution concentrations.

As the cell produces current, the copper ions in solution pick up electrons from the cathode and become copper atoms. The concentration of copper (II) ions decreases.

If the zinc rod were placed in the copper (II) sulphate solution it would start to dissolve, the copper would be directly deposited on the surface of the zinc and the solution's blue colour would fade.
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