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) Eꝋ = + 0.80 V
2H+ (aq) + 2e– ⇌ H2 (g) Eꝋ = 0.00 V
- Since the Ag+/ Ag half-cell has a more positive Eꝋ value, this is the positive pole and the H+/H2 half-cell is the negative pole
- The standard cell potential (Ecellꝋ) is Ecellꝋ = (+ 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 Eꝋ 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) Eꝋ = +1.09 V
2H+ (aq) + 2e– ⇌ H2 (g) Eꝋ = 0.00 V
- The Br2/Br– half-cell is the positive pole and the H+/H2 is the negative pole
- The Ecellꝋ is: Ecellꝋ = (+ 1.09) – (0.00) = + 1.09 V
- The Br2 molecules are more likely to get reduced than H+ as they have a greater Eꝋ value
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) Eꝋ = +1.52 V
2H+ (aq) + 2e– ⇌ H2 (g) Eꝋ = 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 Ecellꝋ is Ecellꝋ = (+ 1.09) – (0.00) = + 1.09 V
Standard Cell Potential: Direction of Electron Flow & Feasibility
Direction of electron flow
- The direction of electron flow can be determined by comparing the Eꝋ values of two half-cells in an electrochemical cell
2Cl2 (g) + 2e– ⇌ 2Cl– (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e– ⇌ Cu (s) Eꝋ = +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
Feasibility
- The Eꝋ 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 Ecellꝋ is positive
- For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e– ⇌ 2Cl– (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e– ⇌ Cu (s) Eꝋ = +0.34 V
- Cl2 molecules are reduced as they have a more positive Eꝋ 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 Eꝋ 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 Eꝋ value of +1.02 V ((+1.36) – (+0.34))
- The backward reaction is not feasible (not spontaneous) as it has a negative Eꝋ value of -1.02 ((+0.34) – (+1.36))
Exam Tip
Remember that the electrons only move through the wires in the external circuit and not through the electrolyte solution.
Redox Equations
- 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) Eꝋ = +1.36 V
Zn2+ (aq) + 2e– ⇌ Zn (s) Eꝋ = -0.76 V
-
- Reduction occurs in the Cl2/Cl– half-cell as it has the more positive Eꝋ value
- Oxidation occurs in the Zn+/Zn half-cell as it has the least positive Eꝋ 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)
Answer
- Step 1:
Determine in which cell oxidation and in which cell reduction takes place
Reduction occurs in the Ag+/Ag half-cell as it has the most positive Eꝋ value
Oxidation occurs in the MnO4–/ Mn2+ half-cell as it has the least positive Eꝋ value - Step 2:
Write down the half-equations for each cell
The 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 electrons
This 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)