Redox help

[Mattamz]

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hey guys this is from pg88 of the success one hsc chemistry book. (Q1 e part iv)

Using the table of reduction potentials demonstrate with equations how the metal you have chosen prevents or slows the corrosion of iron in seawater.

i chose zinc.

according to the answers:
Cathode: 2H2O + 2e- --> 2OH- + H2 E = -0.83V
Anode: Zn --->Zn2+ + 2e- E = +0.76V
eventaully the zinc will oxidies away and the iron will begin to corrode

my problem is that the overall E nought value is negative.
ie E = -0.83 + 0.76 = -0.07V which means its not spontaneous.

does anyone agree that the cathode reaction should instead be
0.5O2 + H2O +2e- --> 2OH- E = 0.40V
giving the overall e nough value: E = 0.4 + 0.76 = 1.16V

thank in advanced

[Forbidden.]

Is this an older edition ?
I've got the 2007 edition.

[wrxsti]

youve got your values mixed up

your 2H20 + 2e- -> H2 + 2OH- with an e* of -0.83
whilst Zn -> Zn2+ + 2e- e* -0.76

this Zinc has a HIGHER reduction potential and will get reduced.........

what i do..... is that if they are asking how Iron gets protected... i dont bother with the H20 equations .. if you get what i mean... what i would rite is...

Zn -> Zn2+ + 2e- e* -0.76V
Fe -> Fe 2+ + 2e- e* -o.44V

this Iron has a higher reduction potential and will be protectedt by excess electrong

Zn -> Zn2+ + 2e-
Fe + 2e- -> Fe

thats how your supposed to solve it anyway..

[Mattamz]

Originally Posted by Forbidden.

Is this an older edition ?
I've got the 2007 edition.

sry i meant q1 (c) part (iv), its the last question pg 88

ive got no prob getting that zinc will be oxided over iron, but the reduction reaction is really annoying me.

[wrxsti]

i have the same probelm with reduction of water and stuf....
you can juse say that Iron will be protected by excess electrons...

so Iron will ge reduced.... works the same way