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
It looks as if chromium gives up electrons to Tin+2, and aluminum gives up electrons to chronium.
So does that mean aluminum is most active, and tin the most able to reduce?
To determine the increasing reductant properties of the metals involved in the spontaneous reactions, we need to consider their reduction potentials. The metal with a more negative reduction potential (E°) will be a stronger reductant.
To solve this, we need to compare the standard reduction potentials (E°) of each metal half-reaction in the given reactions. The greater the magnitude of the reduction potential (in the negative direction), the stronger the reductant.
Let's look at the two reactions:
a) 2Cr + 3Sn2+ -> 2Cr3+ + 3Sn
b) Al + Cr3+ -> Al3+ + Cr
For reaction (a):
- The half-reaction for Cr: Cr3+ + 3e- -> Cr (E° = -0.74 V)
- The half-reaction for Sn: Sn2+ + 2e- -> Sn (E° = -0.14 V)
For reaction (b):
- The half-reaction for Al: Al3+ + 3e- -> Al (E° = -1.66 V)
- The half-reaction for Cr: Cr3+ + 3e- -> Cr (E° = -0.74 V)
To determine the increasing reductant properties, we need to compare the reduction potentials:
From reaction (a):
- Sn has a reduction potential of -0.14 V
- Cr has a reduction potential of -0.74 V
From reaction (b):
- Al has a reduction potential of -1.66 V
- Cr has a reduction potential of -0.74 V
Based on these reduction potentials, we can rank the metals in terms of increasing reductant properties:
Increasing reductant properties: Sn < Cr < Al
So, the metals involved (Cr, Sn, Al) according to increasing reductant properties are Sn < Cr < Al.