the shipwrecks q *again.. ^^ (1 Viewer)

+:: $i[Q]u3 ::+

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soz frigid - it got too messy wit all the edits.

if i have beaker with copper and iron

then iron will corrode preferentialy to copper.
anodic reaction: fe --> fe2+ & 2e-

then the cathodic reaction will be
o2 + 2h2o + 4e- --> 4oh- ?

does it change if acidified?
 

inasero

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yes oit does....macmillan has a good discussion of this,.....
 

Frigid

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the only change if the water was acidified would be the reduction eqn, which will become:

1/2 O2 + 2H+ + 2e- ---> H2O E = +1.23V

which means the reaction will occur faster.

edit: typo before.
 
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+:: $i[Q]u3 ::+

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can u put it up plz? i only have conq and the big orange one.. thickett... and oten.
 

inasero

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your cathode equation will become O2 + 4H+ + 4e- ----->2H2O
 

Frigid

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gimme a page reference inas.

errrr... wilson, that's what i said...
 

inasero

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opps looks like you could be rite....btw i got this from any ol standard reduction potentials table
 

Constip8edSkunk

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dun think it changes when acidified, H+ removes OH form the equilibrium... which causes the change in the written eqn, but is essentially the same thing. although the direct oxidation of iron by hydrogen ions, which also occurs
ie Fe -> Fe2+ + 2e-
2H+ + 2e- ->H2
----------------------
Fe+2H+ -> Fe2+ + H2


EDIT: Thickett=Pathways=Macmillan :p
 
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inasero

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my impression is that it doesnt matter which one is the one which actualy occurs because they both have a more positive E0 than the original electrolysis of water...therefore acidified electrolyte (addition of H+) wil serve to enhace the rate of corrosion
 

Frigid

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Originally posted by Frigid
1/2 O2 + 2H+ + 2e- ---> 2OH- E = +1.23V
Originally posted by inasero
your cathode equation will become O2 + 4H+ + 4e- ----->2H2O
tell me what's the difference.

edit: yeap, i see it :(
 
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Skywalker

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The products of the reaction?

I think inasero's got the right one btw.
 

Frigid

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Originally posted by Skywalker
The products of the reaction?

I think inasero's got the right one btw.
damn sorry, typo...
yeah, its what inas said... but the E naught value is +1.23V...
 

inasero

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hey wenhui is it true you're bringing nancy to d formal?
 

+:: $i[Q]u3 ::+

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thaaannkk you.... =P so much controversy.
btw.. u know how with electrolytic cells, the anode/cathode reactions will be the ones closest together on the potentials sheet.. for galvanic cells, will they be the ones furthest away? like.. if u have a galvanic cell capable of producing 2V and the same cell is capable of producing 3V through a different reaction, then will it go ahead with the 3V reaction?
 

+:: $i[Q]u3 ::+

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u know how wit the acidified one, the reduction of oxygen and hydrogen ions occur (E=1.23V).. which i guess is implying that this reaction is preferred over the reduction of O2 and H2o to oh- ions (e=0.40V)

Does that apply for all equations.. would the galvanic cell produce max voltage...

hang on.. yes it will. dwdw.. ^^ i'm okay.
 

mercury

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For electrolytical cells, the reaction that requires lowest Ecell potential would go ahead first, then as the voltage applied increases, the other potential reactions would go ahead as well.

For galvanic cells, any reaction with postive Ecell potential will occur spontaneously... the one with greatest Ecell would dominate but doesn't mean the other ones don't occur.
 
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Originally posted by +:: $i[Q]u3 ::+
soz frigid - it got too messy wit all the edits.

if i have beaker with copper and iron

then iron will corrode preferentialy to copper.
anodic reaction: fe --> fe2+ & 2e-

then the cathodic reaction will be
o2 + 2h2o + 4e- --> 4oh- ?

does it change if acidified?
If the two metals aren't touching each other and are far apart, both metals will corrode. But if they are touching, or are very close together (i think), both the water and the Cu (2+) will be reduced. The reason the copper won't corrode is because the electrons from the iron go into reducing it back to its elemental form.
 

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