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Summary AQA Chemistry A-Level - Electrode Potentials $4.81   Add to cart

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Summary AQA Chemistry A-Level - Electrode Potentials

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I completed the A-Level in one year teaching most of the course to myself, earning an A overall. This is a summary of the textbook chapter with all details you will need for the exam, I found this most useful as it streamlines the information and keeps everything I needed in one place. Comparing ma...

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  • November 8, 2023
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Electrode Potentials, Electrochemical cells
20.1 Electrode potentials and the electrochemical series: 20.2 Predicting the direction of redox reactions:

If you out 2 different metals in a salt solution and When connecting 2 electrodes the electrons will only flow
connect them it allows an electrical current to flow, from the more negative to the more positive electrode, not to
electrons pass from the more reactive metal to the least the other direction. The signs of the electrodes show the
reactive metal. direction of the redox reaction.
A reaction is only feasible when showing the redox
Half cell: when a metal rod is dipped into a solution of reaction agreeing with the flow of electrons, the reverse
it’s own ions to create an equilibrium. would not be feasible.
M(s) ———> M2+(aq) + 2e-
if measuring the potential, it would show how readily
electrons are released by the metal / how good of a 20.3 Electrochemical cells:
reducing agent it is. Yet yet you can’t measure this
directly and you have to measure potential difference Zinc/Carbon cells, the zinc electrode acts as usual but the
instead, using a voltmeter. carbon electrode with ammonium chloride and a mixture
Circuit completed with a salt bridge (paper soaked in salt of manganese oxide with powered carbon.
solution), this allows the transfer of electrons and ions,
while a wire would only allow electrons. Gives the half equations,
A negative potential difference shows the metal loses Zn(s) ——-> Zn2+ + 2e-
electrons more readily than the other metal, is a better 2NH4+(aq) + 2e- ——> 2NH3(g) + H2(g)
reducing agent.
Overall,
Hydrogen electrode is used to compare the different 2NH4+(aq)+Zn(s -> 2NH3(g) + H2(g) + Zn2+(aq)
metals ability to release electrons, means each half cell
equation can be compared. 1. The hydrogen gas is oxidised to water by manganese
oxide, ammonia dissolves in water
2. As the zinc is used the walls thin and leak, the
ammonium chloride is acidic and so can damage the
equipment

1. Hydrogen gas is bubbled into a solution of H+ ions Rechargeable batteries are recharged by reversing the cell
2. Hydrogen doesn’t conduct so the electrode has a piece reactions, you add a voltmeter with a voltage greater than
of platinum the emf voltage of the cell, forcing electrons to flow in the
3. Used under standard conditions (H+ concentration opposite direction
= 1.00moldm-3, 100kPa, 298K) Lead-acid batteries, each cell has 2 plates dipped in
The potential of the hydrogen electrode is defined as 0 so sulphuric acid, positive plate is made of lead coated with
when connected to the other metal the reading on the lead oxide, the negative plate is made of lead.
voltmeter is the emf. Value for that metal. Lead plate,
Negative emf. Values means the metal is a better Pb(s) + SO42-(aq) ——-> PbSO4(s) + 2e-
reducing agent At the lead oxide plate,
Positive emf. Values show the metal is a better oxidising PbO2(s)+4H+(aq)+—--> 2PbSO4(s)+2H2O(l)
agent SO42-(aq)+Pb (s)
Changing concentration of ions, temperature or an
electrode will change the emf value. Nickel/Cadium batteries contain an alkaline electrolyte,
Cd(OH)2(s) + 2e- ——> Cd(s) + 2OH-(aq)
Electrochemical series: a list of the emf. values of the NiO(OH)(s) + H2O(l) ——> Ni(OH)2(s) + OH-(aq)
metals compared to the hydrogen electrode, written as Overall,
reduction equations. 2NiO(OH) + Cd + 2H2O ---> 2Ni(OH)2 + Cd(OH)2
Potential difference is found by connecting two metal
electrodes in the electrochemical series and calculating Lithium ion batteries the electrolyte is a solid polymer so
the difference between the emf values. cannot leak, positive electrode made of lithium colbalt
oxide, negative plate is carbon.
Representing cells- Li+ + e- ——> Li(s)
• a vertical solid line is used as a phase boundary Li+(CoO2)- ——> Li+ + CoO2 + e-
• a double vertical line is the salt bridge
• The specialties with the highest oxidation state is

next to the salt bridge

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