redox reaction
In redox questions do I first balance normally and then find n factor of each compound/element like this for example?
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@Dexter
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and i clicked it again and welp
its ok
N factor is the no of electron moles required to be transferred for oxidation/reduction for a species, you can find it without balancing
Notice the oxidation state of Mg changed from 0 to +2 therefore two mole of electrons were transferred per mole of Mg hence the n factor is 2 in case of Mg
ik but shd i first balance it normally
Good evening sir, thank you sir but my q is different, im asking shd I first balance normally that is for eg Mg + O2----> MgO
where i balance it by doing 2Mg + O2 ----> 2MgO
like that shd i normally balance it first and then find n factor? or am I doing something wrong
Hi good evening!
in case of this reaction you do whatever you think is easier to do
but if you have a redox reaction it is better to split into ox red halves
here, doing so overcomplicates a simple setup
redox here would be
Mg -> Mg 2+
O2 -> O2-
then balance out number of atoms, followed by charge
you could just balance the equation in a fraction of all this :)
in this case as well, there is no chagne in oxidation number that is heavy. sure, Mg is going to Mg 2+ and H+ is going to H
but just putting a 2 in a couple of places balances this :)
its a redox reaction right
yep
the 'balancing normally' if you think it is easy to do, go for it. if it seems complicated, go the balancing of half reaction way!
;-;
Im getting more confused now.
What i meant was first shd I balance it normally and then proceed with the other
now I get what the problem was, the q has a h2O in it.. ;-; sorry my bad
it shd have been NO2 + H2O not just H2
H2 na.
no i mean the q shs have Mg + HNO3 ----> NO2+ H20 + Mg(NO3)2
We good?
kinda
+solved @iTeachChem @¹¹⁷sos @ᴘɪᴄᴄʜɪ | 🎧 🎶
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