21.2    Measuring Standard Electrode Potential

The big picture

• The standard electrode potential of the following system will be considered:
  1. A metal/metal ion half-cell with SHE
  2. A non-metal/non-metal ion half-cell with SHE
  3. An ion/ion(with different oxidation states) half-cell with SHE
  4. The combination of the above three half-cells.

 

Metal/metal ion half-cell

• Take magnesium as an example of a metal. The equilibrium set up on both electrodes are:

 

• From the voltmeter reading, the E°cell = 2.34 Thus, the standard electrode potential of Mg/Mg²⁺ half-cell is -2.34 V.

• • Cathode, which is the positive terminal is platinum of the H2/H⁺ half-cell. This is because more electrons are being accepted here due to the position of equilibrium of the H2/H⁺ being further to the right.
  • Anode, which is the negative terminal is magnesium of the Mg/Mg²⁺ half-cell. This is because more electrons are being released here due to the position of equilibrium of the Mg/Mg²⁺ half-cell being further to the left.

• The full cell can be represented using a cell diagram. A simple cell diagram is as shown beside
  Note:
  i) The positive electrode is always written on the left side.ii) The species with lowest oxidation state is written next to the inert electrode

• A negative E° value implies that magnesium loses electrons more readily than hydrogen does.

 

• If magnesium is replaced by copper, the value of E° will be +0.34 Copper of Cu/Cu²⁺ half-cell will be the positive terminal while platinum of H2/H⁺ half-cell will be the negative terminal.

 

• A positive E° value implies that copper ions gain electrons more readily than hydrogen ions does.

 

Non-metal/non-metal ion half-cell

• Take chlorine as an example of a non-metal. The equilibrium set up on both electrodes are:
• Since chlorine is not a metal, the electrical contact with the solution is made by using platinum as the electrode(same as SHE).
• From the voltmeter reading, the E°cell = 1.36 Thus, the standard electrode potential of Cl2/Cl⁻ half-cell is +1.36 V.
• • Cathode, which is the positive terminal is platinum of the Cl2/Cl⁻ half-cell. This is because more electrons are being accepted here due to the position of equilibrium of  the Cl2/Cl⁻ being further to the right.
  • Anode, which is the negative terminal is platinum of the H2/H⁺ half-cell. This is because more electrons are being released here due to the position of equilibrium of the H2/H⁺ half-cell being further to the left.

• A positive E° value implies that chloride ions gain electrons more readily than hydrogen ions does.

 

• The cell diagram of this cell is:

 

Ion/ion half-cell

• Two different ions of the same element with two different oxidation states can also be used as a half-cell. Take iron(II) and iron(III) ions as example. The equilibrium set up on both electrodes are:

• Since there is no solid metal as an electrode, electrical contact the solution is made by using platinum as the electrode

 

• From the voltmeter reading, the E°cell = 0.77 Thus, the standard electrode potential of Fe²⁺/Fe³⁺ half-cell is +0.77 V.
• • Cathode, which is the positive terminal is platinum of the Fe²⁺/Fe³⁺ half-cell. This is because more electrons are being accepted here due to the position of equilibrium of the Fe²⁺/Fe³⁺ being further to the right.
  • Anode, which is the negative terminal is platinum of the H2/H⁺ half-cell. This is because more electrons are being released here due to the position of equilibrium of the H2/H⁺ half-cell being further to the left.

• A positive E° value implies that iron(III) ions gain electrons more readily than hydrogen  ions does

 

• The cell diagram of this cell is:

 

Combination of half-cells

• In fact, the SHE shown in previous examples can be replaced by other half-cells.

 

• When this happens, the standard cell potential, E°cell will also change, the new E°cell is simply the difference between the E° values of the two half-cells.
  E°cell = E°(bigger value) – E°(smaller value)

• For example, when Zn/Zn²⁺ half-cell is connected to Cu/Cu²⁺ half-cell, the equilibrium set up on both electrodes are:
  Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)    ; E° = -0.76 VCu²⁺(aq) + 2e⁻ ⇌ Cu(s)   ; E° = +0.34 V

• From the voltmeter reading, the E°cell = 1.10 This can also be calculated, using:
  E°cell = 0.34 – (-0.76) = 1.10 V
• 1. Cathode, which is the positive terminal is copper of the Cu/Cu²⁺ half-cell. This is because more electrons are being accepted here due to the position of equilibrium of the Cu/Cu²⁺ being further to the right.
  2. Anode, which is the negative terminal is platinum of the Zn/Zn²⁺ half-cell. This is because more electrons are being released here due to the position of equilibrium of the Zn/Zn²⁺ half-cell being further to the left.