Electrode Potentials & Half Cells | A-level Chemistry | OCR, AQA, Edexcel

Electrode Potentials & Half Cells in a Snap!
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The key points covered in this video include:
1. Electricity and redox reactions
2. Splitting up redox reactions
3. Half cells with metals in solution
4. Half cells with gases
5. Half cells with only aqueous ions
6. Which half-cell is which?
7. The standard hydrogen electrode
Electricity and Redox Reactions
In our redox videos we see that redox reactions involve the transfer of electrons. The flow of electrons is associated with electrical energy. If we can understand and utilise the flow of electrons in redox reactions then we have a source of electrical energy.
Splitting up Redox Reactions
In redox reactions we can split the full reaction into two half-equations. An oxidation equation where electrons are generated. Along with a reduction equation where those electrons are used up. To utilise the flow of electrons to make a cell, we separate these processes into half-cells so electrons have to flow between.
Metal-Solution Half Cells
The key feature of all half-cells is that they contain an element in two different oxidation states. The most simple one to look at is a metal placed in a solutions of its ions. The ions are being reduced and the metal is being oxidised so that an equilibrium is set up. It’s very important to realise all of the half-cells are equilibriums not one way reactions. The equation shown is the half-cell equation and the forward reaction is always gain of electrons (reduction).
Half Cells with Gases
Sometimes one of the oxidation states present is a gas. We construct half-cells by bubbling the gas through the liquid and giving the reaction a surface to take place on. The chlorine gas is being reduced and the ions are being oxidised so that an equilibrium is set up. Remember both changes are happening - it’s a dynamic equilibrium. Points about the platinum electrode: a. It’s inert, so does not react and affect the redox reaction. b. It conducts electricity, so provides a way to add or remove electrons. c. The electrode is coated in platinum black - this is a porous coating which gives a large surface area for the redox reaction to happen on.
Half-Cells with Only Ions
Sometimes both oxidation states are present as aqueous ions. We construct the half-cells with an equimolar solution and a platinum electrode to provide or remove electrons from the half-cell. An equilibrium is set up between the different oxidation states. Remember it’s a dynamic equilibrium.
Which Half-Cell is Which?
We’ve made it clear that within each half-cell both reduction and oxidation take place to set up an equilibrium. So if we connect two, which cell provides the electrons and which uses them up? The equilibriums of the half-cell equations all have different positions. If the equilibrium lies further to the left then the half-cell is better at releasing electrons. If the equilibrium lies further to the right then the half-cell is better at accepting electrons. The standard electrode potential, Eθ, is what we use to measure this.
The Hydrogen Half-Cell
The standard electrode potential tells us about a half-cell’s tendency to accept or release electrons. But half-cells can only release or accept electrons if they are attached to something to release them to or accept them from. To measure the standard electrode potentials we use a special half-cell as a reference - the hydrogen half cell. We connect the two half-cells into a circuit. The salt bridge contains free ions to complete the circuit so charge can flow - it is usually made from paper soaked in KNO3(aq) or NH4NO3(aq). The high resistance volt meter tells us the difference in standard potentials of the half-cells measured in volts. The hydrogen half-cell is assigned a standard electrode potential value of 0V as it is a reference. This means the above set up gives us the value of the standard electrode potential of the half-cell on the right. The standard electrode potential is the emf generated by a half-cell when it is connected to the standard hydrogen electrode at 298K, 100KPa and all solutions having a concentration of 1 mol dm^-3.

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Equilibrium on the right or on the left???

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