. Using the following spontaneous reactions, classify the 3 metals involved (Cr, Sn, Al) according to increasing reductant properties.
a) 2Cr + 3Sn2+ ---> 2Cr3+ + 3Sn
b) Al + Cr3+ ---> Al3+ + Cr
Look at the activity series. Which metal is above other metals.
Frankly I don't know what the question means. I will assume you want the strongest reducing agent.
I do that this way. There are much shorter ways of doing it but I like this one.
Cr is above Sn because Cr replaces Sn^2+.
Al is above Cr because Al replaces Cr^3+.
So the activity series will be
Al ==> Al^3+ + 3e
Cr ==> Cr3+ + 3e
Sn ==> Sn^2++ 2e
Notice that these are oxidations which means Al is easiest to oxidize. That makes Al the strongest reducing agent. If I have assumed incorrectly I'll try again if you rephrase the question.
To classify the 3 metals (Cr, Sn, Al) according to increasing reductant properties based on the given spontaneous reactions, we need to compare their ability to act as reducing agents.
In reaction (a):
2Cr + 3Sn2+ ---> 2Cr3+ + 3Sn
We can see that chromium (Cr) is oxidized from a oxidation state of +2 to +3, while tin (Sn) is reduced from a oxidation state of +2 to 0. Therefore, tin (Sn) is acting as the reducing agent in this reaction.
In reaction (b):
Al + Cr3+ ---> Al3+ + Cr
In this reaction, aluminum (Al) is oxidized from an oxidation state of 0 to +3, while chromium (Cr) is reduced from an oxidation state of +3 to 0. Therefore, aluminum (Al) is acting as the reducing agent in this reaction.
Based on the reactions, we can conclude that:
1. Sn is the strongest reducing agent since it can reduce Cr3+ to Cr (reaction a).
2. Al is a weaker reducing agent than Sn, but stronger than Cr, since it can reduce Cr3+ to Cr (reaction b).
3. Cr is the weakest reducing agent since it cannot reduce Sn2+ or Al (based on the given reactions).
Therefore, the classification of the metals according to increasing reductant properties is:
Cr < Al < Sn
To classify the metals according to increasing reductant properties, we need to determine the order in which they are oxidized in the given spontaneous reactions. The metal that gets oxidized is the reductant or reducing agent, while the metal that gets reduced is the oxidant or oxidizing agent.
In reaction (a): 2Cr + 3Sn2+ ---> 2Cr3+ + 3Sn
Here, chromium (Cr) is being reduced from Cr to Cr3+, and tin (Sn) is being oxidized from Sn2+ to Sn.
In reaction (b): Al + Cr3+ ---> Al3+ + Cr
Here, aluminum (Al) is being oxidized from Al to Al3+, and chromium (Cr) is being reduced from Cr3+ to Cr.
To determine the order of reductant properties, we compare the metals based on their ability to be oxidized. A metal that can be easily oxidized is a stronger reductant (reducing agent), while a metal that is less likely to be oxidized is a weaker reductant.
In this case, comparing reaction (a) and (b), we see that Sn (tin) gets oxidized in reaction (a) and Cr (chromium) gets oxidized in reaction (b). Since chromium is less likely to be oxidized than tin, chromium (Cr) is the weaker reductant compared to tin (Sn). Therefore, tin (Sn) has the highest reductant properties, followed by aluminum (Al), and then chromium (Cr) has the lowest reductant properties.
So, the three metals involved (Cr, Sn, Al) can be classified according to increasing reductant properties as follows:
1. Chromium (Cr) - lowest reductant properties
2. Aluminum (Al)
3. Tin (Sn) - highest reductant properties.