• 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^??= + 0.80 V

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

• Since the Ag⁺/ Ag half-cell has a more positive?E^??value, this is the?positive pole?and the H⁺/H₂?half-cell is the?negative?pole
• The?standard cell potential (E_cell^?) is?E_cell^??= (+ 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^??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^??= +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
• The?E_cell^??is:?E_cell^??= (+ 1.09) - (0.00)?= + 1.09 V
• The Br₂?molecules are more likely to get?reduced?than H⁺?as they have a greater?E^??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 Mn^2+?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^??= +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 Mn^2+?ions
• The MnO₄^-/Mn²⁺?^-?half-cell is the?positive pole?and the H⁺/H₂?is the?negative?pole
• The?E_cell^??= (+ 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^??values 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
  • Cu^2+?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^??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^??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?E^??value of -1.02 ((+0.34) - (+1.36))

A reaction is feasible when the standard cell potential E^??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^??= +1.36 V

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

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

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