Measuring the Standard Electrode Potential (Cambridge International A-Level Chemistry)

Spec point — 24.2.3

Measuring the Standard Electrode Potential

• 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)

• When a half-cell is connected to a standard hydrogen electrode, or when two half-cells are connected, a salt bridge is required

  • A salt bridge has mobile ions that complete the circuit
  • A salt bridge is typically a strip of filter paper soaked in a saturated solution of potassium nitrate or potassium chloride as nitrates and chlorides are usually soluble
  • This should ensure that no precipitates form which can affect the equilibrium position of the half-cells

Metal / metal ion half-cell

• 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 pole
  • Oxidation occurs at the negative pole

Example of an electrochemical cell containing a metal / metal ion half-cell

Under standard conditions, a metal / metal ion half-cell is connected to the standard hydrogen electrode to measure the cell potential

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

  • Br 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(l) + 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
• 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 an electrochemical cell containing a non-metal / non-metal ion half-cell

Under standard conditions, a non-metal / non-metal ion half-cell is connected to the standard hydrogen electrode to measure the cell potential

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 an oxidation number of +7
  • The Mn²⁺ ion contains Mn with an oxidation number of +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^- / Mn^{2+ -} half-cell is the positive pole and the H⁺ / H₂ is the negative pole
• E_cell^ꝋ= (+ 1.52) - (0.00) = + 1.52 V

Example of an electrochemical cell containing an ion / ion half-cell

Under standard conditions, an ion / ion half-cell is connected to the standard hydrogen electrode to measure the cell potential

Standard Cell Potential: Direction of Electron Flow & Feasibility

Spec point — 24.2.5

Direction of electron flow

• The direction of electron flow can be determined by comparing the E^ꝋ values of two half-cells in an electrochemical cell

Cl₂(g) + 2e^- ⇌ 2Cl^-(aq) E^ꝋ = +1.36 V

Cu²⁺(aq) + 2e^- ⇌ Cu (s) E^ꝋ = +0.34 V

• The Cl₂ more readily accept electrons from the Cu²⁺/Cu half-cell

  • This is the positive pole
  • Cl₂ gets more readily reduced

• The Cu²⁺ more readily loses electrons to the Cl₂/Cl^- half-cell

  • This is the negative pole
  • Cu²⁺ gets more readily oxidised

• The electrons flow from the Cu²⁺/Cu half-cell to the Cl₂/Cl^- half-cell

• The flow of electrons is from the negative pole to the positive pole

Flow of electrons through an electrochemical cell

The electrons flow through the wires from the negative pole to the positive pole

Feasibility

• The E^ꝋ values of a species indicate how easily they can get oxidised or reduced

• The more positive the value, the easier it is to reduce the species on the left of the half-equation

  • The reaction will tend to proceed in the forward direction

• The less positive the value, the easier it is to oxidise the species on the right of the half-equation

  • The reaction will tend to proceed in the backward direction
  • A reaction is feasible(likely to occur) when the E_cell^ꝋ is positive

• For example, two half-cells in the following electrochemical cell are:

Cl₂(g) + 2e^- ⇌ 2Cl^-(aq) E^ꝋ = +1.36 V

Cu²⁺(aq) + 2e^- ⇌ Cu (s) E^ꝋ = +0.34 V

• Cl₂ molecules are reduced as they have a more positive E^ꝋ value
• The chemical reaction that occurs in this half-cell is:

Cl₂(g) + 2e^- → 2Cl^-(aq)

• Cu²⁺ ions are oxidised as they have a less positive E^ꝋ value
• The chemical reaction that occurs in this half-cell is:

Cu (s) → Cu²⁺(aq) + 2e^-

• The overall equation of the electrochemical cell is (after cancelling out the electrons):

Cu (s) + Cl₂(g) → 2Cl^- (aq) + Cu²⁺(aq)

OR

Cu (s) + Cl₂(g) → CuCl₂(s)

• The forward reaction is feasible (spontaneous) as it has a positive E^ꝋ value of +1.02 V ((+1.36) - (+0.34))
• The backward reaction is not feasible (not spontaneous) as it has a negative E^ꝋ value of -1.02 ((+0.34) - (+1.36))

Reaction feasibility and the standard cell potential

A reaction is feasible when the standard cell potential Eꝋ is positive

Exam Hint Remember that the electrons only move through the wires in the external circuit and not through the electrolyte solution.

Redox Equations

Spec point — 24.2.7

• The main ways to construct redox equations using the relevant half-equations:

  • Using changes in oxidation numbers to help balance chemical equations

  • Using the number of electrons for each half-cell

    • Both of these are discussed in the Redox Reactions topic

  • Interpreting the information given to you and predicting any other chemicals involved in the reaction

Worked Example Write the balanced chemical equation for hydrogen iodide reacting with sulfuric acid to form hydrogen sulfide, iodine and one other product

Answer:

• There are 2 possible methods:

  1. Using ionic half-equations
  2. Interpreting the information and predicting any other chemicals involved

Using ionic half-equations method:

• Step 1: Identify possible half-equations from the information

  • Iodide → iodine
  • Sulfuric acid → hydrogen sulfide

• Step 2: Construct the half equations

  • Iodide → iodine

    • I^– → I₂
    • This requires 2 iodide ions and 2 electrons as products to balance the equation
    • 2I^– → I₂ + 2e^–

  • Sulfuric acid → hydrogen sulfide

    • H₂SO₄ → H₂S
    • The 4 oxygen atoms will form 4 water molecules as products
    • H₂SO₄ → H₂S + 4H₂O
    • The 8 hydrogen atoms in the water molecules will require 8 protons as reactants
    • H₂SO₄ + 8H⁺ → H₂S + 4H₂O
    • The 8 protons will require 8 electrons to balance the charge
    • H₂SO₄ + 8H⁺ + 8e^– → H₂S + 4H₂O

• Step 3: Combine the half-equations

  • Iodide → iodine

    • 2I^– → I₂ + 2e^–
    • The iodide half-equation needs to be multiplied by 4 to have the same number of electrons as the sulfuric acid half-equation
    • 8I^– → 4I₂ + 8e^–

  • Sulfuric acid → hydrogen sulfide

    • H₂SO₄ + 8H⁺ + 8e^– → H₂S + 4H₂O

  • Combining the half-equations

    • 8I^– + H₂SO₄ + 8H⁺ + 8e^– → 4I₂ + 8e^– H₂S + 4H₂O

  • Cancelling, where appropriate

    • The electrons cancel on both sides
    • The 8I^– and 8H⁺ can be re-written as 8HI
    • 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O

Interpreting and predicting method:

• Step 1: Start with what you know:

  • HI + H₂SO₄ → H₂S + I₂

• Step 2: Consider any elements that are not accounted for

  • The only element that is not currently considered is oxygen

• Step 3: Make a common and appropriate suggestion for the missing product

  • Most of these questions are in solution so there is always H₂O, H⁺ and OH^- available
  • Missing product suggestion = 4H₂O
  • HI + H₂SO₄ → H₂S + I₂ + 4H₂O

• Step 4: Balance the remaining chemicals

  • 8HI + H₂SO₄ → H₂S + 4I₂ + 4H₂O

Exam Hint

• Similar approaches can be used to balance more complicated ionic half-equations
• In these situations, you will have H₂O, H⁺, OH^- and electrons available