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AQA Chemistry- Electrochemistry Summary Notes
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ElectrochemistryAt the anode the metal ion is oxidised so there are more electrons in the anode than in
the cathode so a potential difference is created between the 2 electrodes
Voltmeter needs to be of high resistance to stop its current from flowing in the circuit
preventing a reaction
If the voltmeter is replaced by a bulb, reactions will occur separately at the electrodes.
Over time the voltage would fall to 0 as the reactants get used up.
A salt bridge is a filter paper soaked in KNO3 solution (as it is unreactive). The salt
bridge completes the circuit and allows the flow of ions.
The positive half-cell favours reduction and the negative favours oxidation
If a cell uses gases or 2 solutions then a platinum electrode is used to provide an inert
and conducting surface for a reaction to take place
To measure the potential difference of a half-cell it has to be connected to another
half-cell with known potential
Conventional Cell Diagram
Reduced(s.s) | Oxidised(s.s) || Oxidised(s.s) | Reduced(s.s)
The more positive half-cell is written on the right.
Left-Hand Side: flip the reduced equation to oxidised i.e. X(s) → X+(aq) +e-
Right-Hand Side: keep the reduced equation i.e Y+(aq) +e- → Y(s)
The electrodes go at the end of each side
∴ cell diagram is X(s) | X+(aq) || Y+(aq) | Y(s)
Standard Hydrogen Electrode (SHE)
Conditions: H2 gas 100kPa, Hydrogen ion solution 1 moldm-3, 298K, Platinum Electrode
In a cell diagram : Pt(s) | H2(g) | H+(aq)
Emf = Erhs - Elhs
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