Redox reaction: Fe(s) + 2HCl(aq) ---> FeCl2 (aq) + H2 (g)?

Question

Sorry to keep bothering, I just seem to have a problem on each exercise *hides*

I got the more simple reactions, but when I'm thrown HCL and FCl2 I'm lost...
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Fe(s) + HCl(aq) ---> FeCl2 (aq) + H2 (g)

Balanced: Fe(s) + 2HCl(aq) ---> FeCl2 (aq) + H2 (g)

Oxidation numbers for each:
Fe = 0
2HCl === H=2(+1)=+2, Cl=2(-1)=-2
FeCl2 === Cl2 = 2(-1)=-2 therefore Fe= +2?
H2 = 2(1) === 2

If I'm right... only Fe is oxidized. Which is reduced? H stays 2 and Cl too.

Fe ---> Fe2+ - 2e-

OOOH ok ok, H2 is by itself so it's oxidation number is zero.

Now, does H go from +1 to 0. Or 2 to 0. Because it is 2HCl. The 2 from the balancing modifies the oxidation, doesn't it?

Answer 1

Fe goes from 0 to +2 (oxidation)
In HCl H has oxidation number +1 and Cl has o.n. -1
So H goes from +1 to 0 ( reduction)

Answer 2

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RE:
Redox reaction: Fe(s) + 2HCl(aq) ---> FeCl2 (aq) + H2 (g)?
Sorry to keep bothering, I just seem to have a problem on each exercise *hides*

I got the more simple reactions, but when I'm thrown HCL and FCl2 I'm lost...
-------------------------------

Fe(s) + HCl(aq) ---> FeCl2 (aq) + H2 (g)

Balanced: Fe(s) + 2HCl(aq) ---> FeCl2 (aq) +...

Answer 3

Agreed re IRON (Fe),
It is hydrogen which is reduced - it gains an electron.

'H' in HCl may be thought of as 'H+'. By becoming a diatomic molecule H2, each 'H' atom has gained an electron, hence it is reduced.
'Cl' in HCl may be thought of as 'Cl-' and remains as 'Cl-' in FeCl2, hence the oxid'n state of chlorine has not altered.

Answer 4

the protons are reduced to hydrogen gas

2H+ + 2e- -> H2