Measuring the Standard Electrode Potential (CIE A Level Chemistry)

Revision Note

Philippa Platt

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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= + 0.80 V

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

  • Since the Ag+ / Ag half-cell has a more positive Evalue, this is the positive pole and the H+ / H2 half-cell is the negative pole
  • The standard cell potential (Ecell) is Ecell = (+ 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 Evalue
    • Reduction occurs at the positive pole
    • Oxidation occurs at the negative pole

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

Principles of Electrochemistry - Example of a Metal_Metal Ion Half-Cell, downloadable AS & A Level Chemistry revision notes

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 = +1.09 V

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

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

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

Principles of Electrochemistry - Example of a Non-Metal_Non-Metal Ion Half-Cell, downloadable AS & A Level Chemistry revision notes

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 = +1.52 V

2H+ (aq) + 2e- ⇌ H2 (g)       E= 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
  • Ecell = (+ 1.52) - (0.00) = + 1.52 V

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

Principles of Electrochemistry - Example of an Ion_ Ion Half-Cell, downloadable AS & A Level Chemistry revision notes

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

Direction of electron flow

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

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

Cu2+ (aq) + 2e- ⇌ Cu (s)        E = +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

Principles of Electrochemistry - Direction of Flow of Electrons, downloadable AS & A Level Chemistry revision notes

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

Feasibility

  • The Evalues 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 Ecell is positive
  • For example, two half-cells in the following electrochemical cell are:

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

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

  • Cl2 molecules are reduced as they have a more positive E 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 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 value of +1.02 V ((+1.36) - (+0.34))
  • The backward reaction is not feasible (not spontaneous) as it has a negative Evalue of -1.02 ((+0.34) - (+1.36))

Reaction feasibility and the standard cell potential

Principles of Electrochemistry - Reaction Feasibility (1), downloadable AS & A Level Chemistry revision notesPrinciples of Electrochemistry - Reaction Feasibility (2), downloadable AS & A Level Chemistry revision notes

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

Examiner Tip

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

Redox Equations

  • 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
    • 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

Examiner Tip

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

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.