6.1.3 Measuring Standard Electrode Potential

• There are three different types of half-cells that can be connected to a standard hydrogen electrode to measure standard electrode potential 
  • A metal / metal ion half-cell
  • A non-metal / non-metal ion half-cell
  • An ion / ion half-cell (the ions are in different oxidation states)

Metal / metal-ion half-cell

Example of a metal / metal ion half-cell connected to a standard hydrogen electrode

• An example of a metal/metal ion half-cell is the Ag⁺/ Ag half-cell
  • Ag is the metal
  • Ag⁺ is the metal ion

• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Ag⁺ (aq) + e^- ⇌ Ag (s)        E^ꝋ = + 0.80 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)        E^ꝋ = 0.00 V 

• Since the Ag⁺/ Ag half-cell has a more positive E^ꝋ value, this is the positive pole and the H⁺/H₂ half-cell is the negative pole
• The standard cell potential (E_cell^ꝋ) is E_cell^ꝋ = (+ 0.80) - (0.00) = + 0.80 V
• The Ag⁺ ions are more likely to get reduced than the H⁺ ions as it has a greater E^ꝋ value
  • Reduction occurs at the positive electrode
  • Oxidation occurs at the negative electrode

Non-metal / non-metal ion half-cell

• In a non-metal / non-metal ion half-cell, platinum wire or foil is used as an electrode to make electrical contact with the solution
  • Like graphite, platinum is inert and does not take part in the reaction
  • The redox equilibrium is established on the platinum surface

• An example of a non-metal / non-metal ion is the Br₂ / Br^- half-cell
  • Br₂ is the non-metal
  • Br^- is the non-metal ion

• The half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Br₂ (aq) + 2e^- ⇌ 2Br^- (aq)        E^ꝋ = +1.09 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)        E^ꝋ = 0.00 V   

• The Br₂ / Br^- half-cell is the positive pole and the H⁺ / H₂ is the negative pole
• The E_cell^ꝋ is: E_cell^ꝋ = (+ 1.09) - (0.00) = + 1.09 V
• The Br₂ molecules are more likely to get reduced than H⁺ as they have a greater E^ꝋ value

Example of a non-metal / non-metal ion half-cell connected to a standard hydrogen electrode

Ion / Ion half-cell

• A platinum electrode is again used to form a half-cell of ions that are in different oxidation states
• An example of such a half-cell is the MnO₄^- / Mn²⁺ half-cell
  • MnO₄^- is an ion containing Mn with oxidation state +7
  • The Mn²⁺ ion contains Mn with oxidation state +2

• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

MnO₄^- (aq) + 8H⁺ (aq) + 5e^- ⇌ Mn²⁺ (aq) + 4H₂O (l)       E^ꝋ = +1.52 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g)       E^ꝋ = 0.00 V   

• The H⁺ ions are also present in the half-cell as they are required to convert MnO₄^- into Mn²⁺ ions
• The MnO₄^- / Mn²⁺ half-cell is the positive pole and the H⁺ / H₂ is the negative pole
• The E_cell^ꝋ is E_cell^ꝋ = (+ 1.09) - (0.00) = + 1.09 V

Ions in solution half cell

Standard cell potential

• Once the E^ꝋ of a half-cell is known, the potential difference or voltage or emf of an electrochemical cell made up of any two half-cells can be calculated
  • These could be any half-cells and neither have to be a standard hydrogen electrode

• The standard cell potential (E_cell^ꝋ) can be calculated by subtracting the less positive E^ꝋ from the more positive E^ꝋ value
  • The half-cell with the more positive E^ꝋ value will be the positive pole
    • By convention this is shown on the right hand side in a conventional cell diagram, so is termed  E_right^ꝋ

  • The half-cell with the less positive E^ꝋ value will be the negative pole
    • By convention this is shown on the left hand side in a conventional cell diagram, so is termed  E_left^ꝋ

E_cell^ꝋ = E_right^ꝋ - E_left^ꝋ   

• • Since oxidation is always on the left and reduction on the right, you can also use this version

E_cell^ꝋ = E_reduction^ꝋ - E_oxidation

Answer

Step 1: Calculate the standard cell potential. The copper is more positive so must be the right hand side.

E_cell^ꝋ = E_right^ꝋ - E_left^ꝋ   

E_cell^ꝋ = (+0.34) - (-0.76)

= +1.10 V

The voltmeter will therefore give a value of +1.10 V

Step 2: Determine the positive and negative poles

The Cu²⁺ / Cu  half-cell is the positive pole as its E^ꝋ is more positive than the E^ꝋ value of the Zn²⁺ / Zn half-cell