CIE A Level Chemistry复习笔记5.3.4 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)

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 pole
  • Oxidation occurs at the negative pole

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
• 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^-/Mn^2+ - half-cell is the positive pole and the H⁺/H₂ is the negative pole
• The E_cell^ꝋ = (+ 1.52) - (0.00) = + 1.52 V

Ions in solution half cell

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

2Cl₂ (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

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))

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

• The redox equations of an electrochemical cell can be constructed using the relevant half-equations of the two half-cells

Constructing redox equations

• Step 1: Determine in which half-cell the oxidation and in which half-cell the reduction  reaction takes place

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

Zn²⁺ (aq) + 2e^- ⇌ Zn (s)        E^ꝋ = -0.76 V

• • Reduction occurs in the Cl₂/Cl^- half-cell as it has the more positive E^ꝋ value
  • Oxidation occurs in the Zn⁺/Zn half-cell as it has the least positive E^ꝋ value

• Step 2: Write down the half equations for each half-cell
  • Half-equation of the Cl₂/Cl^- half-cell

Cl₂ (g) + 2e^-  → 2Cl^- (aq)         

• • Half-equation of the Zn⁺/Zn half-cell

Zn (s)  → Zn²⁺ (aq) + 2e^-

• Step 3: Balance the number of electrons in both half-equations
  • The number of electrons is already balanced in both half-equations as they both contain two electrons

• Step 4 - Add up the two half-equations

Cl₂ (g) + 2e^-  → 2Cl^- (aq)         

Zn (s)  → Zn²⁺ (aq) + 2e^-

______________________________________ +

Cl₂ (g) + Zn (s) + 2e → 2Cl^- (aq) + Zn²⁺ (aq) + 2e^-

• Step 5 - Cancel out the electrons (and H⁺ ions and H₂O molecules if any present) to find the overall redox reaction

Cl₂ (g) + Zn (s) → 2Cl^- (aq) + Zn²⁺ (aq)

OR

Cl₂ (g) + Zn (s) → ZnCl₂ (s)

Worked example: Constructing redox reactions

Answer

• Step 1:Determine in which cell oxidation and in which cell reduction takes placeReduction occurs in the Ag⁺/Ag half-cell as it has the most positive E^ꝋ value

  Oxidation occurs in the MnO₄^-/ Mn²⁺ half-cell as it has the least positive E^ꝋ value

• Step 2:Write down the half-equations for each cellThe half-equation for the Ag⁺/Ag half-cell is:

Ag⁺ (aq) + e^- → Ag (s)             

The half-equation for the MnO₄^-/ Mn²⁺  half-cell is:

Mn²⁺ (aq) + 4H₂O (l) → MnO₄^- (aq) + 8H⁺ (aq) + 5e^-

• Step 3:Balance the number of electrons in both half-equations

  Multiply the half-equation for the Ag⁺/Ag half-cell by 5 so that both half-equations contain 5 electronsThis gives: 5Ag⁺ (aq) + 5e^- → 5Ag (s)

• Step 4:Add the two half-equations

5Ag⁺ (aq) + 5e^-  + Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO₄^- (aq) + 8H⁺ (aq) + 5e^-

• Step 5:Cancel out the electrons to find the overall redox equation

5Ag⁺ (aq) + 5e^-+ Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO₄^- (aq) + 8H⁺ (aq) + 5e^-The fully balanced redox equation is:

5Ag⁺ (aq) + Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO₄^- (aq) + 8H⁺ (aq)