SOMEONE PLSS HELP
. 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
To classify the metals involved (Cr, Sn, Al) according to increasing reductant properties, we need to examine their ability to lose electrons in a redox reaction. The metal that can lose electrons more easily is a stronger reductant.
To determine the reductant properties, we can consider the reduction potentials of each metal. Reduction potential is a measure of the tendency of a species to gain electrons.
In this case, we can compare the reduction potentials of each metal involved. If the reduction potential is higher, it means that the metal is stronger in terms of reductant properties.
Let's analyze the given reactions:
a) 2Cr + 3Sn2+ -> 2Cr3+ + 3Sn
Here, Cr is being oxidized from Cr(0) to Cr(III) and Sn2+ is being reduced to Sn(0). Since Cr is undergoing oxidation, it will act as a reductant, and Sn is undergoing reduction, indicating that it is acting as an oxidant.
b) Al + Cr3+ -> Al3+ + Cr
In this reaction, Al is being oxidized from Al(0) to Al(III) and Cr3+ is being reduced to Cr. Similar to the previous reaction, Al is acting as a reductant, and Cr3+ is acting as an oxidant.
Comparing the given reactions, we can see that Al is acting as a reductant in both reactions, whereas Cr and Sn are acting as oxidants. Therefore, Al has the highest reductant property among the three metals.
From highest to lowest reductant properties, the classification would be: Al > Cr > Sn.