# 5.3.4 Measuring the Standard Electrode Potential

### 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= + 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

#### 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 Ecellis: 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

#### 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 = +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 MnO4into Mn2+ ions
• The MnO4/Mn2+ half-cell is the positive pole and the H+/H2 is the negative pole
• The Ecell is Ecell = (+ 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 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

#### 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))

#### 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)        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 Evalue
• Oxidation occurs in the Zn+/Zn half-cell as it has the least positive Evalue
• 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 Evalue
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 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) ### Author: Francesca

Fran has taught A level Chemistry in the UK for over 10 years. As head of science, she used her passion for education to drive improvement for staff and students, supporting them to achieve their full potential. Fran has also co-written science textbooks and worked as an examiner for UK exam boards.
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