Electrochemical Cells (Edexcel International A Level (IAL) Chemistry): Revision Note
Exam code: YCH11
Core Practical 12: Investigating Electrochemical Cells
Measuring the EMF of a cell
To measure a cell EMF you will need
Two small beakers, around 75 cm3 capacity
Strips of suitable metals such as copper, zinc, iron and silver
1.0 mol dm-3 solutions of the metal ions (nitrates, chlorides or sulfates depending on their solubility)
A high resistance voltmeter (usually a digital multimeter has this)
Two sets of wires with crocodile clips
A salt bridge consisting of a strip of filter paper soaked in saturated potassium nitrate

The experimental set up for measuring the EMF of a cell made of two metal / metal ion half cells
The procedure
The strips of metals need to be freshly cleaned to remove any oxide coatings
This can be done with a piece of sandpaper
To support the metals, it is easiest to have long strips that can be folded over the side of the beaker and held in place with the crocodile clips
Fill up the beakers to about two thirds of the way with the metal ion solutions
Using tongs, dip a strip of filter paper into a beaker of saturated potassium nitrate solution and then place it between the two beakers making sure the ends of the strip are well immersed in the solutions
Connect the crocodile clips to the voltmeter, wait for a steady reading and record the measurement
Practical tips
If you don't get a positive reading on the voltmeter swap the terminals around
Voltmeters will have marked positive and negative terminals (usually in red and black, respectively), so when you get a positive reading this tells you the relative polarity of the metals in the cell
Change the salt bridge each time, to prevent cross contamination of ions between half cells
Calculating the Theoretical EMF of a Cell
The voltage you measure in your experiment can be compared to a theoretical value calculated using standard electrode potentials (E°)
These values are found in the electrochemical series in your data book.
The EMF of a cell is the difference between the E° values of the two half-cells
The formula is:
E°(cell) = E°(positive electrode) - E°(negative electrode)
A simple way to remember this is "most positive E° value minus most negative E° value
Worked Example
Calculate the theoretical EMF for a cell made of zinc and iron.
Find the E° values from a data book:
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) has an E° = -0.44 V
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) has an E° = -0.76 V
Identify the positive and negative electrodes:
The half-cell with the more positive (less negative) E° value will be the positive electrode
In this case, that is the Fe²⁺/Fe half-cell.
The half-cell with the more negative E° value will be the negative electrode
This is the Zn²⁺/Zn half-cell.
Apply the formula:
E°(cell) = E°(positive) - E°(negative)
E°(cell) = (-0.44) - (-0.76)
E°(cell) = +0.32 V
This calculation is how the theoretical values in the specimen results table are determined.
Specimen Results
Here is a set of typical results for this experiment:
Negative electrode | Positive electrode | EMF / V |
|---|---|---|
Zn (s) / Zn2+ (aq) | Cu2+ (aq) / Cu (s) | 1.10 |
Zn (s) / Zn2+ (aq) | Fe2+ (aq) / Fe (s) | 0.32 |
Fe (s) / Fe2+ (aq) | Cu2+ (aq) / Cu (s) | 0.78 |
Zn (s) / Zn2+ (aq) | Ag+ (aq) / Ag (s) | 1.56 |
Cu (s) / Cu2+ (aq) | Ag+ (aq) / Ag (s) | 0.46 |
Analysis
It is unlikely you will get very close to the theoretical results as these would be obtained under standard conditions (1.0 mol dm-3 concentration, 298 K temperature)
This is because these conditions are hard to achieve in a school laboratory
However, the relative EMF of cells you construct should match the theoretical values
The higher the EMF, the larger the difference in reactivity ('electron pushing power') between the metals
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