Standard Electrode Potentials (AQA A Level Chemistry): Revision Note
Exam code: 7405
Standard Electrode Potentials
Standard electrode potential
The position of equilibrium and therefore the electrode potential depends on factors such as:
Temperature
Pressure of gases
Concentration of reagents
So, to be able to compare the electrode potentials of different species, they all have to be measured against a common reference or standard
Standard conditions also have to be used when comparing electrode potentials
These standard conditions are:
Ion concentration of 1.00 mol dm-3
A temperature of 298 K
A pressure of 100 kPa
Standard measurements are made using a high-resistance voltmeter so that no current flows and the maximum potential difference is achieved
The electrode potentials are measured relative to a standard hydrogen electrode
The standard hydrogen electrode is given a value of 0.00 V, and all other electrode potentials are compared to this standard
This means that the electrode potentials are always referred to as a standard electrode potential (Eꝋ)
The standard electrode potential (Eꝋ) is the potential difference (sometimes called voltage) produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
For example, the standard electrode potential of bromine suggests that relative to the hydrogen half-cell, it is more likely to get reduced, as it has a more positive Eꝋ value
Br2(aq) + 2e– ⇌ 2Br–(aq) Eꝋ = +1.09 V
2H+(aq) + 2e– ⇌ H2(g) Eꝋ = 0.00 V
The standard electrode potential of sodium, on the other hand, suggests that relative to the hydrogen half-cell, it is less likely to get reduced as it has a more negative Eꝋ value
Na+ (aq) + e– ⇌ Na(s) Eꝋ = -2.71 V
2H+ (aq) + 2e– ⇌ H2(g) Eꝋ = 0.00 V
Electrochemical Series
Standard electrode potentials can be listed as an electrochemical series:
Reduction Half Equation | E ꝋ/V |
Li+(aq) + e- Li(s) | -3.03 |
Ca2+(aq) + 2e- Ca(s) | -2.87 |
Al3+(aq) + 3e- Al(s) | -1.66 |
Zn2+(aq) + 2e- Zn(s) | -0.76 |
Pb2+(aq) + 2e- Pb(s) | -0.13 |
2H+(aq) + 2e- H2(g) | 0.00 |
Cu2+(aq) + 2e- Cu(s) | +0.34 |
I2(aq) + 2e- 2I-(aq) | +0.54 |
Fe3+(aq) + e- Fe2+(aq) | +0.77 |
CI2(aq) + 2e- 2CI-(aq) | +1.36 |
MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) | +1.51 |
A more positive Eꝋ means the species is more easily reduced
Standard Hydrogen Electrode
The standard hydrogen electrode is a half-cell used as a reference electrode and consists of:
Hydrogen gas in equilibrium with H+ ions:
2H+ (aq) + 2e- ⇌ H2 (g)
The four critical conditions are
hydrogen gas (bubbled through)
1.0 mol dm⁻³ H⁺ (HCl is accepted)
298 K
100 kPa
An inert platinum electrode that is in contact with the hydrogen gas and H+ ions allows the transfer of electrons / provides a reaction surface
When the standard hydrogen electrode is connected to another half-cell, the standard electrode potential of that half-cell can be read off a high-resistance voltmeter

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

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

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 an oxidation state of +7
The Mn2+ ion contains Mn with an oxidation state of +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

Calculating EMF
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 has 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, this version can also be used:
Ecellꝋ = Ereductionꝋ - Eoxidation
Related topics
Examiner Tips and Tricks
Never multiply Eꝋ by the number of electrons. AQA mark schemes subtract the two E° values directly even when the half-equations have different electron counts ( Eꝋ is an intensive property) — a very common place where students lose marks.
The emf of a spontaneous or commercial cell is positive — examiners do not allow a negative emf. Always write the + or − sign on the final value.
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
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