2 syllabus points i need help with (1 Viewer)

[Pace_T]

Hey can someone help me with these 2 syllabus points. im not sure what to do for them...


Use available evidence to predict the rate of corrosion of a metal wreck at great depths in the oceans and give reasons for the prediction made

Discuss the range of chemical procedures which can be used to clean, preserve and stabilise artefacts from wrecks, and where possible, provide an example of the use of each procedure.

Oh and also, how do I know which is the reduction reaction, when it is out of these two:

2H2O +2e- -> 2OH- +H2

O2 +2H2O +4e- <--> 4OH-

Thanks!

 

[Templar]

Use available evidence to predict the rate of corrosion of a metal wreck at great depths in the oceans and give reasons for the prediction made

Well it's cold, has low oxygen concentration, so it should slow down corrosion.

Discuss the range of chemical procedures which can be used to clean, preserve and stabilise artefacts from wrecks, and where possible, provide an example of the use of each procedure.

Iron is placed in sodium hydroxide to remove chloride, the solution is regularly replaced until chlorine concentration drops below 50ppm. PEG is used on organics, but it is somewhat damaging to the artefacts. Can't remember other treatments.

Oh and also, how do I know which is the reduction reaction, when it is out of these two:

2H2O +2e- -> 2OH- +H2

O2 +2H2O +4e- <--> 4OH-

I think it's the latter, check which has lowest energy.

 

[Jumbo Cactuar]

Depends, in electrolysis;


2H₂O ---> O₂ + 4H⁺ + 4e^- E^o = -1.23V oxidation
2H₂O + 2e^----> H₂ + 2OH^- E^o = -0.83V reduction
2H₂O ---> O₂ + 2H₂ E^o = -2.89V


as a cell;


O₂ + 4H⁺ + 4e^- ---> 2H₂O E^o = 1.23V reduction
H₂ + 2OH^----> 2H₂O + 2e^- E^o = 0.83V oxidation
O₂ + 2H₂ ---> 2H₂O E^o = 2.89V

 

[Pace_T]

Hey, thanks for your replies.

Depends, in electrolysis;


2H₂O ---> O₂ + 4H⁺ + 4e^- E^o = -1.23V oxidation
2H₂O + 2e^----> H₂ + 2OH^- E^o = -0.83V reduction
2H₂O ---> O₂ + 2H₂ E^o = -2.89V


as a cell;


O₂ + 4H⁺ + 4e^- ---> 2H₂O E^o = 1.23V reduction
H₂ + 2OH^----> 2H₂O + 2e^- E^o = 0.83V oxidation
O₂ + 2H₂ ---> 2H₂O E^o = 2.89V

Hey could you please explain this for me? I'm so confused, lol. How do you know which equation goes to reaction, etc.. Thanks!

 

[Jumbo Cactuar]

Whoa! Sorry dude.

You know a electochemical reaction is spontaneous if the net potential (E⁰) is positive. However, under electrolysis you are applying a voltage, so that if the applied potential is strong enough (such that E⁰ becomes positive) the reaction becomes spontaneous. Remember the E⁰ are for 298.15K & 0.1MPa. Furthermore that they are all equilibria.