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4hv.org :: Forums :: Chemistry
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Al + Fe in NH3

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Heiders
Sun May 28 2006, 06:33PM
Heiders Registered Member #268 Joined: Tue Feb 28 2006, 02:44AM
Location: Ontario, Canada
Posts: 48
So from what I can see, the general consensus is that the iron is just conducting, and the half reactions are

3 [2 NH4+(aq) + 2 e- → 2 NH3(g) + H2(g)]
2 [Al(s) → Al3+(aq) + 3 e-]
6 NH4+(aq) + 2 Al (s) → 6 NH3(g) + H2(g) + 2 Al3+(aq)

This looks logical, and it makes sense. Corrections?
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Bored Chemist
Mon May 29 2006, 08:08AM
Bored Chemist Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
The 6 NH 3 in the last equation will be (aq) rather than (g). Otherwise I think it's OK.

(In reality, it will be more complicated- CO2 and O2 from the air will probably be involved and some of the Al will form various hydrated oxides and these may, in turn, incorporate NH3.)

The story with the different counter-electrode (Zn, Cu) is also more complicated due to overvoltage effects, which should be small, and the real effects of the different metals.
Granted that, in this case, the other electrode isn't doing anything but conducting. On the other hand, it doesn't know that- it will still produce a potential when placed in the solution.
With the ammonia there, the concentration of H+ is very small, I guess about 10^-8M. Nevertheless the zinc will still react with it to give Zn++ and H2 (very slowly). In the cell you have set up this reaction is prevented from taking place, so that affects the cell voltage overall. The effect of Fe would be similar, but smaller because Fe is less reactive.

The trouble with electrochemistry is that it looks nice and simple but really it's quite complicated.
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Heiders
Tue May 30 2006, 02:29AM
Heiders Registered Member #268 Joined: Tue Feb 28 2006, 02:44AM
Location: Ontario, Canada
Posts: 48
Ah, BC, you are awesome! To the whole lot of you, thanks from my group and me!
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