Measuring Standard Electrode Potential (Edexcel International A Level (IAL) Chemistry): Revision Note
Exam code: YCH11
Measuring Standard Electrode Potential
There are three different types of half-cells that can be connected to a standard hydrogen electrode to measure standard electrode potential
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 electrode
Oxidation occurs at the negative electrode
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
Br2 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 (aq) + 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

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) 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 = (+ 1.52) - (0.00) = + 1.52 V

Ions in solution half cell
The Salt Bridge
The salt bridge completes the circuit in a voltaic (electrochemical) cell by allowing ions to move between the two half-cells
It helps to balance the charges as electrons flow through the external wire
A metal wire must not be used as a salt bridge
Metals allow electrons to flow, but a salt bridge must carry ions, not electrons
Potassium chloride and potassium nitrate are commonly used to make the salt bridge because chloride and nitrate ions are:
Highly soluble in water
Unreactive with most ions in the half-cells
Unlikely to form precipitates
This ensures smooth ion flow and prevents disruption of the redox reactions
Electromotive Force
Standard cell potential
Once the Eꝋ of a half-cell is known, the potential difference or voltage or emf of an electrochemical cell made up of any two half-cells can be calculated
These could be any half-cells and neither have to be a standard hydrogen electrode
The standard cell potential (Ecellꝋ) can be calculated by subtracting the less positive Eꝋ from the more positive Eꝋ value
The half-cell with the more positive Eꝋ value will be the positive pole
By convention this is shown on the right hand side in a conventional cell diagram, so is termed Erightꝋ
The half-cell with the less positive Eꝋ value will be the negative pole
By convention this is shown on the left hand side in a conventional cell diagram, so is termed Eleftꝋ
Ecellꝋ = Erightꝋ - Eleftꝋ
Since oxidation is always on the left and reduction on the right, you can also use this version
Ecellꝋ = Ereductionꝋ - Eoxidation
Worked Example
Calculating the standard cell potential
Calculate the standard cell potential for the electrochemical cell below and explain why the Cu2+ / Cu half-cell is the positive pole. The half-equations are as follows:
Cu2+(aq) + 2e- ⇌ Cu(s) Eꝋ = +0.34 V
Zn2+(aq) + 2e- ⇌ Zn(s) Eꝋ = −0.76 V

Answer
Step 1: Calculate the standard cell potential. The copper is more positive so must be the right hand side.
Ecellꝋ = Erightꝋ - Eleftꝋ
Ecellꝋ = (+0.34) - (-0.76)
= +1.10 V
The voltmeter will therefore give a value of +1.10 V
Step 2: Determine the positive and negative poles
The Cu2+ / Cu half-cell is the positive pole as its Eꝋ is more positive than the Eꝋ value of the Zn2+ / Zn half-cell
Examiner Tips and Tricks
A helpful mnemonic for remembering redox in cells

Lio the lion goes Roor!
Lio stands for 'Left Is Oxidation' and he is saying ROOR because that is the order of species in the cell:
Reduced/Oxidised (salt bridge) Oxidised/Reduced
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