The Electrochemical Series (SQA National 5 Chemistry): Revision Note
Exam code: X813 75
The electrochemical series
The electrochemical series is a list of elements and ions arranged in order of their ability to gain or lose electrons
Metals higher up the series are more reactive and lose electrons more easily
The electrochemical series is found on page 10 of the SQA Data Booklet
It allows us to predict two key things about any simple cell:
The direction of electron flow
The relative size of the voltage
Direction of electron flow
The rule:
In a cell, electrons always flow from the substance that is higher in the electrochemical series to the substance that is lower
The substance higher up loses electrons (oxidation).
The substance lower down gains electrons (reduction)
For example, in the Mg and Cu cell:
Magnesium is higher than copper
Therefore, electrons will flow from the magnesium electrode to the copper electrode
Size of the voltage
The rule:
The further apart the two substances are in the electrochemical series, the larger the voltage the cell will produce
For example:
A cell made from magnesium and copper will produce a higher voltage than a cell made from zinc and copper
This is because the distance between Mg and Cu in the series is much greater than the distance between Zn and Cu
Summary of processes in a cell
Position in electrochemical series | Process | What happens to the electrode | Direction of electron flow |
|---|---|---|---|
Higher substance | Oxidation (loses e⁻) | The electrode gets smaller / dissolves / erodes | Away from this electrode |
Lower substance | Reduction (gains e⁻) | The electrode gets bigger / plates out | Towards this electrode |
Electrochemical cell ion-electron equations
Every electrochemical cell involves both an oxidation and a reduction reaction happening simultaneously in the two half-cells
We can write three equations to describe what is happening:
The oxidation ion-electron equation
This is for the substance losing electrons
The reduction ion-electron equation
This is for the substance gaining electrons
The overall redox equation
This combines the two ion-electron equations and shows the full reaction
How to write the equations
We can combine the two ion-electron equations to find the overall redox equation using the same 5-step method as for any redox reaction
Identify the oxidised and reduced substances using the Electrochemical Series (page 10)
The substance higher up is oxidised
The substance lower down is reduced
Write the two ion-electron equations
Remember to reverse the equation for the substance that is oxidised
Balance the electrons
The number of electrons lost must equal the number of electrons gained
Multiply one or both equations if necessary
Combine the equations
Add all the reactants together on one side and all the products together on the other side.
Cancel out the electrons
Cancel electrons that appear on both sides to get the final, overall equation
Worked Example
Write the three key equations for a cell made from zinc and copper.
[3]
Answer:
Identify the oxidised and reduced substances
Zinc is higher than copper, so it is oxidised (reversed equation)
Copper is lower, so it is reduced (equation as written)
Write the two ion-electron equations
Zn (s) → Zn2+ (aq) + 2e⁻ [1 mark]
Cu2+ (aq) + 2e⁻ → Cu (s) [1 mark]
Balance the electrons
Zn loses 2e- and Cu2+ gains 2e-
Combine the equations
Zn (s) + Cu2+ (aq) + 2e⁻ → Zn2+ (aq) + Cu (s) + 2e⁻
Cancel out the electrons
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) [1 mark]
Worked Example
Write the three key equations for a cell made from aluminium and iron(II).
[3]
Answer:
Identify the oxidised and reduced substances
Aluminium is higher than iron(II), so it is oxidised (reversed equation)
Iron(II) is lower, so it is reduced (equation as written)
Write the two ion-electron equations
Al → Al3+ + 3e- [1 mark]
Fe2+ + 2e- → Fe [1 mark]
Balance the electrons
Al loses 3e- and Fe2+ gains 2e-
The numbers are not balanced
The lowest common multiple of 3 and 2 is 6
Multiply the aluminium equation by 2:
2Al → 2Al3+ + 6e-
Multiply the iron(II) equation by 3:
3Fe2+ + 6e- → 3Fe
Now, 6 electrons are lost and 6 electrons are gained
Combine the equations
2Al (s) + 3Fe2+ (aq) + 6e⁻ → 2Al3+ (aq) + 3Fe (s) + 6e⁻
Cancel out the electrons
2Al (s) + 3Fe2+ (aq) → 2Al3+ (aq) + 3Fe (s) [1 mark]
Worked Example
Write the three key equations for a cell made from an iron half-cell and an iodine half-cell.
[3]
Answer:
Identify the oxidised and reduced substances
Iron is higher than iodine, so it is oxidised (reversed equation)
Iodine is lower, so it is reduced (equation as written)
Write the two ion-electron equations
Fe (s) → Fe2+ (aq) + 2e⁻ [1 mark]
I2 (aq) + 2e⁻ → 2I⁻ (aq) [1 mark]
Balance the electrons
Fe loses 2e- and I2 gains 2e-
Combine the equations
Fe (s) + I2 (aq) + 2e⁻ → Fe2+ (aq) + 2I⁻ (aq) + 2e⁻
Cancel out the electrons
Fe (s) + I2 (aq) → Fe2+ (aq) + 2I⁻ (aq) [1 mark]
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