Question

Aluminum, `"Al"`, produces `"H"_2` bubbles when immersed in `"HCl"`, mercury, `"Hg"` does not. Which is the strongest reducer?

A) Al3+
B) Al
C) Hg
D)Hg+
E H2

Answer 1

Aluminium metal.

Explanation:

Basically, the fact that aluminium produces bubbles of hydrogen gas when immersed in hydrochloric acid tells you that it can reduce hydrochloric acid to hydrogen gas.

`2"Al"_ ((s)) + 6"HCl"_ ((aq)) -> 2"AlCl"_ (3(aq)) + 3"H"_ (2(g)) uarr`

The oxidation half-reaction looks like this

`"Al"_ ((s)) -> "Al"_ ((aq))^(3+) + 3"e"^(-)`

So aluminium metal acts as a reducing agent here because each atom of aluminium loses `3` electrons.

The reduction half-reaction looks like this

`2"H"_ ((aq))^(+) + 2"e"^(-) -> "H"_ (2(g))`

Here, each atom of hydrogen gains `1` electron, which is equivalent to saying that two atoms of hydrogen will gain `2` electrons.

You can balance the number of electrons lost and gained in the two half-reactions to get the overall balanced chemical equation that describes this redox reaction.

So aluminium is the reducing agent because it reduces hydrogen cations, `"H"^(+)`, to hydrogen gas, `"H"_2`. So the metal can reduce hydrochloric acid to hydrogen gas, but the aluminium cation, `"Al"^(3+)`, cannot do the same because it's electron-deficient, i.e. it has no more electrons to donate.

`color(white)(a)`

By comparison, mercury does not produce bubbles of hydrogen gas when immersed in hydrochloric acid, so it cannot reduce hydrochloric acid to hydrogen gas.

So if the metal cannot reduce hydrochloric acid to hydrogen gas, the mercury(I) cation, `"Hg"^(+)`, can't reduce it, either `->` think about the fact that the mercury(I) cation is missing an electron, to begin with.