Question

What does it mean when chemists say that a reactant is reduced?

Answer 1

See Below.

Explanation:

There are several meanings.

To say a reactant is reduced means, it has oxidised other reactant.

More precisely, The Reactant has undergone such a reaction that it has gained a hydrogen or some electropostive atom or lost an oxygen or some electronegative atom.

Even more precisely, The Reactant has gained electrons.

In the Other Way, The Reactant has gained Oxidation Number.

By Example,

`2Na(s) + dil. H_2SO_4(aq) = Na_2SO_4(aq) + H_2uparrow`

At First, `H_2SO_4` had the `SO_4^(2-)` radical (which is negatively charged), but after the reaction, it got displaced, and `H_2` was liberated. So, Loss of Electronagtive Radical, That Means the reactant is Reduced.

But see, what it has done.

`Na` was oxidised, as it donated an electron to become a `Na^+` cation. And Then Two `Na^+` cations joined with the escaped `SO_4^(2-)` radical to form `Na_2SO_4`.

And In the Concept Of Oxidation Number,

In `H_2SO_4`, `H` had an oxidation number of `+1`. But after the reaction, `H_2` is liberated, so it is a free molecule, and has the oxidation number `0`. So, The Oxidation Number Decreased, and it is a reduction.

Hope this explains my point.

Answer 2

A reactant is formally `"reduced"` when it accepts electrons...and its formal oxidation number is numerically DECREASED or `"REDUCED"`...

Explanation:

Electrons and oxidation numbers are introduced into redox equations as a conceptual aid in order to help us balance and mass and charge. As in any chemical reaction, mass and charge are ABSOLUTELY conserved, and garbage out must necessarily equal garbage in....

And so for every oxidation, every loss of electrons, there is a corresponding reduction, a corresponding electron gain. We invoke the electron as a virtual particle of convenience to help us write redox reactions.

And so when we say that calcium metal is oxidized by dioxygen gas....we split up the oxidation and reduction reactions, and introduce the electrons....

`Ca(s)→Ca^(2+)+2e^−` `(i)`

Now this is balanced with respect to mass and charge, and the electron is presumed to GO SOMEWHERE, i.e. where it causes a reduction; and here it reduces dioxygen according to the following reaction....

`1/2O_2(g)+2e^(-)→O^(2-)` `(ii)`

And we add the individual equations together in such a way as to ELIMINATE the electrons, i.e.

`2xx(i)+(ii)`

to give....

`Ca(s)+1/2O_2(g) rarrCaO(s)`
as the final balanced equation.....

I will provide other examples....but I still do not know at which level you study....

Ammonia gas CAN be oxidized to nitrate ion, and this is a formal 8 electron oxidation....

`NH_3(g) +3H_2O(l) rarr NO_3^(-) + 9H^+ + 8e^(-)`

Nitrate ion, `N(V)` COULD be reduced to dinitrogen, a five electron reduction...

`NO_3^(-) + 6H^+ + 5e^(-) rarr 1/2N_2(g) + 3H_2O(l)`

And you would need a fairly fierce reductant to accomplish this reduction....may be five equiv of alkali metal..

And for another example....

Carbon undergoes a 4 electron oxidation........

`H_3stackrel(-III)C-stackrel(-I)CH_2OH +H_2Orarr H_3stackrel(-III)C-stackrel(+III)C(=O)OH +4H^+ + 4e^(-)" "(i)`

But permanganate ion undergoes a THREE electron reduction.....`Mn(+VII)rarrMn(+IV)`

`MnO_4^(-) +4H^+ + 3e^(-) rarr MnO_2(s) + 2H_2O" "(ii)`

We cross multiply to remove the electrons.....`3xx(i) + 4xx(ii)`

`"3H"_3"CCH"_2"OH" +4MnO_4^(-) +4H^+ rarr 3"H"_3"CC(=O)OH" +4MnO_2(s) + 5H_2O`

But basic conditions were specified, so we add `4xxHO^-` TO EACH SIDE......

`"3H"_3"CCH"_2"OH" +4MnO_4^(-) rarr 3"H"_3"CC(=O)O"^(-) +4MnO_2(s) + H_2O +4HO^-`

And this (I think) is balanced with respect to mass and charge...

Are all these balanced with respect to mass and charge....? If they are not then they cannot be accepted as a model of reality.