Question #04d43
1 Answer
Explanation:
Before jumping in, it's worth taking a moment to make sure that you understand what you're dealing with here.
The problem wants you to pick the strongest reducing agent by looking at what should actually be four redox equilibria.
So, what does it mean for a chemical species to be a reducing agent?
As you know, oxidation and reduction reactions involve the transfer of electrons from one element to another. More specifically, you know that
- the element that donates electrons is being oxidized
- the element that accepts electrons is being reduced
So, in order for a chemical species to be a reducing agent, it must reduce another chemical species, i.e. it must donate electrons to that species.
The strongest reducing agent will thus be the chemical species that donates electrons with the most ease.
Now, take a look at the four redox equilibria given to you
#"M"_text((aq])^(2+) + 2"e"^(-) rightleftharpoons "M"_text((s])" " " "E^@ = +"1.21 V"#
#"Q"_text((aq[)^(+) + "e"^(-) rightleftharpoons "Q"_text((s])" " " " color(white)(a) E^@ = +"1.03 V"#
#"Z"_text((aq])^(3+) + 3"e"^(-) rightleftharpoons "Z"_text((s])" " " " color(white)(a)E^@ = - "0.21 V"#
#"X"_text(2(aq]) + 2"e"^(-) rightleftharpoons 2"X"_text((s])^(-) " " " "E^@ = -"1.23 V"#
The idea here is that the
Simply put, a positive
Likewise, a negative
You're looking for the chemical species that is most willing to donate electrons. This means that its redox equilibrium must lie furthest to the left.
This in turn means that you're looking for more negative
This means that
#" " " " "X"_text(2(aq]) + 2"e"^(-) rightleftharpoons 2"X"_text((s])^(-) " " " "E^@ = -"1.23 V"#