5.2.3: Redox and electrode potentials Flashcards
Standard electrode potential
the e.m.f. (electron motive force) of a half cell connected to a standard hydrogen half cell under the standard conditions of 298K, 100kPa and 1 moldm⁻³.
What is the electrode made out of for a hydrogen half cell?
Platinum
How to set up a half cell with two ions of the same metal?
Two ions in solution at 1 molmoldm⁻³ EACH
Platinum electrode
Reducing agent
Is itself oxidised
Oxidising agent
Is itself reduced
Why might a reaction that is feasible by its Eᶿ values not actually happens?
- Conditions are non-standard (state how…)
* Activation energy is too high
A positive Eᶿ value means
the metal is readily reduced (REDuction Is More Positive)
A negative Eᶿ value means
the metal is readily oxidised
What happens at the electrode with the more reactive metal?
The more reactive metal loses electrons and is oxidised –> chucking out electrons, XS of e- –> is the negative electrode
What happens at the electrode with the less reactive metal?
The less reactive metal gains electrons and is reduced –> deficit of e- –> is the positive electrode
Charge carriers in the salt bridge
Ions
Charge carriers in the wire
electrons
Predicting feasibility from Eᶿ value
Positive Eᶿ value means the redox reaction is feasible.
Features of a primary cell
- Non-rechargeable
* Electrical energy produced by reduction and oxidation at the electrodes –> however reactions cannot be reversed
Features of a secondary cell
- Rechargeable
* Cell reactions can be reversed during charging –> regenerates chemicals so that cell can be used again
Types of cell
- Primary
- Secondary
- Fuel cells
What are redox titrations? Why do them?
Redox titration: a titration that uses transition metal elements
- Transition metals have variable oxidation states so useful as oxidising/reducing agents because they readily give accept/give out electrons.
- Often change colour with oxidation state; easy to see when reaction has finished
What are redox titrations used to calculate?
The concentration of the oxidising/reducing agent
How to calculate the concentration of reducing agent (redox titration)?
Use manganate ions as oxidising agent
1) Pipette fixed volume unknown reducing agent into conical
2) Add excess H₂SO₄ to the flask to provide any H⁺ needed
3) Gradually add MnO₄⁻ (oxidising agent) from burette, swirling conical
4) Stop when purple MnO₄⁻ colour becomes permanent; = endpoint because no more reducing agent to react with MnO₄⁻
5) Repeat to calculate mean titre
How to calculate the concentration of an oxidising agent (redox titration)?
Where IO₃⁻ is the oxidising agent in Stage 1
Stage 1
1) Measure out 25cm³ of potassium iodate solution
2) Add excess KI. This oxidises the iodide to iodine
Stage 2
1) Add sodium thiosulfate dropwise to flask from stage 1. End point is pale yellow
2) Add 2cm³ starch. Dark blue if iodine present.
3) When dark blue colour disappears, all the iodine has been reduced to iodide
4) Use titre of sodium thiosulfate to calculate moles of iodine. Then ➗3 for the moles (hence conc.) of iodate