Home >
Community >
What is the E° for reaction Ti⁰ ⇌ Ti³⁺ + 3 e⁻?
Upvote
VOTE
Downvote
+ Chemistry
Posted by
Nick Parks
What is the E° for reaction Ti⁰ ⇌ Ti³⁺ + 3 e⁻?
One, the reaction that you labeled an oxidation is actually a reduction. Two, a source that I checked gives the standard reduction potential of Ti(III)/Ti(II) as -0.9 volts, and I am not sure which value is correct. Three, I have reservations about the additivity of reduction potentials under certain circumstances, but I will defer to someone who is more familiar with such questions than I am.
One, the reaction that you labeled an oxidation is actually a reduction. Two, a source that I checked gives the standard reduction potential of Ti(III)/Ti(II) as -0.9 volts, and I am not sure which value is correct. Three, I have reservations about the additivity of reduction potentials under certain circumstances, but I will defer to someone who is more familiar with such questions than I am.
No, your logic is not correct. This is not a redox couple - there isn't an oxidation coupled to a reduction - so you can't add the "cathodic" and "anodic" potentials. it is a sequential reduction, as you will see if you write it out explicitly (always a good idea):
1. Ti Ti2+ + 2e 2. Ti2+ Ti3+ + e 3. Ti Ti3+ + 3e
3 = 1 + 2
Here we use Hess's law - what is additive is ΔG, not E. ΔG3 = ΔG1 + ΔG2 -3FE3 = -2FE1 - FE2 3E3 = 2E1 +E2
This can be seen on a Frost diagram (oxidation state diagram), where G is plotted vs. oxidation number, and E is the slope of the line joining two states.
In this case, using your values, I calculate E3 (for the oxidation of Ti to Ti3+) as +1.21V, which by coincidence is not far from your value calculated by wrong logic!
No, your logic is not correct. This is not a redox couple - there isn't an oxidation coupled to a reduction - so you can't add the "cathodic" and "anodic" potentials. it is a sequential reduction, as you will see if you write it out explicitly (always a good idea):
1. Ti Ti2+ + 2e 2. Ti2+ Ti3+ + e 3. Ti Ti3+ + 3e
3 = 1 + 2
Here we use Hess's law - what is additive is ΔG, not E. ΔG3 = ΔG1 + ΔG2 -3FE3 = -2FE1 - FE2 3E3 = 2E1 +E2
This can be seen on a Frost diagram (oxidation state diagram), where G is plotted vs. oxidation number, and E is the slope of the line joining two states.
In this case, using your values, I calculate E3 (for the oxidation of Ti to Ti3+) as +1.21V, which by coincidence is not far from your value calculated by wrong logic!
More
VOTE
https://www.periodensystem-online.de/index.php?el=22&id=redox
https://www.periodensystem-online.de/index.php?el=22&id=redox
More
VOTE
1. Ti
2. Ti2+
3. Ti
3 = 1 + 2
Here we use Hess's law - what is additive is ΔG, not E.
ΔG3 = ΔG1 + ΔG2
-3FE3 = -2FE1 - FE2
3E3 = 2E1 +E2
This can be seen on a Frost diagram (oxidation state diagram), where G is plotted vs. oxidation number, and E is the slope of the line joining two states.
In this case, using your values, I calculate E3 (for the oxidation of Ti to Ti3+) as +1.21V, which by coincidence is not far from your value calculated by wrong logic!
1. Ti
2. Ti2+
3. Ti
3 = 1 + 2
Here we use Hess's law - what is additive is ΔG, not E.
ΔG3 = ΔG1 + ΔG2
-3FE3 = -2FE1 - FE2
3E3 = 2E1 +E2
This can be seen on a Frost diagram (oxidation state diagram), where G is plotted vs. oxidation number, and E is the slope of the line joining two states.
In this case, using your values, I calculate E3 (for the oxidation of Ti to Ti3+) as +1.21V, which by coincidence is not far from your value calculated by wrong logic!
More
VOTE