Standard Electrode Potentials (AQA A Level Chemistry): Revision Note

Standard Electrode Potentials

Standard electrode potential

• The position of equilibrium and therefore the electrode potential depends on factors such as:

  • Temperature

  • Pressure of gases

  • Concentration of reagents

• So, to be able to compare the electrode potentials of different species, they all have to be measured against a common reference or standard

• Standard conditions also have to be used when comparing electrode potentials

• These standard conditions are:

  • Ion concentration of 1.00 mol dm^-3

  • A temperature of 298 K

  • A pressure of 100 kPa

• Standard measurements are made using a high resistance voltmeter so that no current flows and the maximum potential difference is achieved

• The electrode potentials are measured relative to a standard hydrogen electrode

• The standard hydrogen electrode is given a value of 0.00 V, and all other electrode potentials are compared to this standard

• This means that the electrode potentials are always referred to as a standard electrode potential (E^ꝋ)

• The standard electrode potential (E^ꝋ) is the potential difference ( sometimes called voltage) produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions

• For example, the standard electrode potential of bromine suggests that relative to the hydrogen half-cell it is more likely to get reduced, as it has a more positive E^ꝋ value

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

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

• The standard electrode potential of sodium, on the other hand, suggests that relative to the hydrogen half-cell it is less likely to get reduced as it has a more negative E^ꝋ value

Na⁺ (aq) + e^– ⇌ Na(s)        E^ꝋ = -2.71 V

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

Standard Hydrogen Electrode

• The standard hydrogen electrode is a half-cell used as a reference electrode and consists of:

  • Hydrogen gas in equilibrium with H⁺ ions of concentration 1.00 mol dm^-3 (at 100 kPa)

2H⁺ (aq) + 2e^- ⇌ H₂ (g)

• An inert platinum electrode that is in contact with the hydrogen gas and H⁺ ions

• When the standard hydrogen electrode is connected to another half-cell, the standard electrode potential of that half-cell can be read off a high resistance voltmeter

The standard electrode potential of a half-cell can be determined by connecting it to a standard hydrogen electrode

• There are three different types of half-cells that can be connected to a standard hydrogen electrode

  • 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 = (+ 1.52) - (0.00) = + 1.52 V

Ions in solution half cell

Calculating EMF

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