Electrode Potentials Flashcards
If it is more reactive where with it go
To the anode
The more negative the E value is the more reactive, so more easily it goes under oxidation, so it gives up its electrons
AO let’s go
Anode is oxidisation
More positive E value will go to anode/cathode?
Cathode
More positive E value the more it gains electrons, so the more reductive.
CR
Christiano Renaldo
Cathode is reduction
Calculate emf value
Right - left
More positive - more negative
Units of emf value
Volts
What is the salt bridge for
Allows ions to move across
Standard condition of Electrochemical cell
Conc - 1 mol dm-3 but be careful if H2SO4 used as this needs 0.5 mol em-3 as there are 2H+ this equals to 1 Mol dm-3
Temp - 298K
Pressure - 100kPa
Why is Hydrogen 0V
By definition
Everything is relative to hydrogen
Why is Pt used sometimes
Inert metal that can conduct electricity for electrons to move
I think usually used when reactions only contain aqueous substances - double check
Electrode potential equation
R O || O R
Only compounds going through redox is put into this
Solid, aqueous || aqueous, solid
Left - anode
Right - cathode
This is because you have to lose electrons before you gain electrons so oxidation comes before reduction
How does changing the concentration of Zn2+ affect E value
Zn2+ + 2e- <—-> Zn
-0.76V
Increasing Zn2+ will shift equilibrium to the right towards Zn
This will increase the Zn E value making it less negative so the overall E value will decrease (between Zn and another ion)
- _________+
-0.76 —— -0.50 - this makes Zn E value less negative
Looking at Zn and another ion eg Cu which has E value +0.34V
Original E value = 0.34 + 0.76 = 1.1V
New E value = 0.34 + 0.50 = 0.84V so overall E value decreases.
Vice versa
Increase Zn equilibrium shifts to the left towards Zn2+
Makes Zn E value more negative to overall E value will Increase.
Which is at the anode/cathode
What is the emf value
What is the sentence used at the end of this
Reduction equations:
Zn2+ + 2e- <—-> Zn / E value = -0.76V
Cu2+ + 2e- <—-> Cu / E value = +0.34V
- Anode
- Zn <—> Zn2+ + 2e- the equation flips as oxidation occurs here AO
Cathode
- Cu2+ +2e- <—> Cu the equation remains as reduction occurs here CR
- emf = 1.1V
- E value (Cu2+/Cu) > E value (Zn2+/Zn) so Cu2+ gains electrons from Zn
Different type of cells
- Non-rechargeable
- Rechargeable
- Fuel cells
Lithium cell equation
What type of cell is a lithium cell
Rechargeable
Fuel cells: Hydrogen-Oxygen fuel cell
The equations involved