[Win,odd Dhamnekar]

It is sometimes necessary to compute E° for a given half reaction from other half reaction of known E°.

For example, the standard electrode potential for the oxidation of Titanium metal to Ti³⁺ can be obtained from the following half reaction:
Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts ①

Ti³⁺ + e⁻ ⇌ Ti²⁺ E° = -0.37 Volts ②

Calculate E° for Ti⁰ ⇌ Ti³⁺ + 3 e⁻

My answer: ② is more spontaneous reaction than ①.

Reduction: Ti³⁺ + e⁻ ⇌ Ti²⁺ E° = -0.37 Volts ;

Oxidation: Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts

Hence,
 E°_rx= E°_red - E°_ox = -0.37 V - (-1.63 V) = + 1.26 V

As the E°_rx is +ve, the reaction Ti⁰ Ti³⁺  + 3 e^- is spontaneous.  Is that correct?


My answer matches with author's answer.

But, Is my logic in answering this question correct?

[Babcock_Hall]

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.

[mjc123]

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!

[Hunter2]

Should the value - 1,21 V

https://www.periodensystem-online.de/index.php?el=22&id=redox

[mjc123]

No, those values are reduction potentials (even though the half-reactions are written as oxidations).

[Hunter2]

If I solve the equation 3E3 = 2E1 + E2

E3  = (2E1+ E2)/3

E3 = (2* (-1,63V) +(-0,37 V))/3 =
(-3,26-0,37V)/3  = - 1,21 V

How do you get the positiv value.

And Titanium  is a inoble element to hydrogen electrode,  what has everytime negative  reduction Potentials.

[mjc123]

OP asked about the oxidation of Ti to Ti³⁺. So the reactions I've written as 1, 2 and 3 are oxidations [I should have said "sequential oxidation" by the way, not "sequential reduction" - apologies] and the potentials are the negative of the reduction potentials.

[Hunter2]

I understand what are you saying.

But in my opinion it is like this.

Ti²⁺ + 2 e⁻ ⇌ Ti⁰ E° = -1.63 Volts ①

Ti³⁺ + e⁻ ⇌ Ti²⁺ E° = -0.37 Volts ②

These are reduction But with the ⇌ sign I can also write

 Ti⁰  ⇌Ti²⁺ + 2 e⁻ E° = -1.63 Volts ①

 Ti²⁺⇌Ti³⁺ + e⁻    E° = -0.37 Volts  ②

These are Oxidation.

In my opinion the voltage will not change to positiv , because measured to H+/H2 value 0V and ⇌ is used and not => sign.

See two tables

As oxidation

http://www.chemgapedia.de/vsengine/media/vsc/de/ch/11/aac/vorlesung/kap_11/kap11_2/grafik/spannungsreihe.gif

As reduction

https://mmsphyschem.com/wp-content/uploads/2021/12/standard-reduction-potential-table_193-3076027.jpg

No change  from plus to minus or opposit.

So Ti⁰  ⇌Ti³⁺ + 3 e⁻  and Ti³⁺  + 3 e⁻   ⇌Ti⁰ with  E° = -1,21 V

[Win,odd Dhamnekar]

No, those values are reduction potentials (even though the half-reactions are written as oxidations).

I agree with what you said. In my opinion, In computing oxidation potential of a half reaction using reduction potentials table, sign should be reversed.   

[Aldebaran]

In data tables for standard electrode potential equilibria it is customary to write them as reductions and attribute the sign accordingly. If the equilibrium is written in reverse (as an oxidation) the sign is reversed.
When written as reductions the more positive value indicates the more favourable direction.

[Hunter2]

My question is still if I have H+/H2 = 0 V and another metal let say Zn/ Zn2+ has a potential  of  -0,76 V . How can it have + 0,76 V in other direction. The same question was asked for Titanium in this exercise.
Copper Cu2+ has + 0,34 V. How it can have - 0,34 V.

Reaction takes place Zn + Cu2+ => Zn2+ + Cu not the opposit direction, because zinc has lower potential as copper. If the voltages would change the polarity then reaction would run in opposit,but this don't happen.

[Win,odd Dhamnekar]

 F₂(g) + 2 e^-  2F^-(aq)  E°(V) = +2.87 means this reduction reaction is reduction favourable and oxidation unfavourable.

If we write this reaction in reverse order 2F^-(aq) F₂(g) + 2 e^-,then  E°(V) = -2.87 indicating that this reaction is oxidation unfavourable and reduction favourable.

I hope this explanation will clear the confusion.

[Hunter2]

No, because the value is measured to H2/H+.

And it's +2,87 V Doesn't matter Oxidation or Reduction. Show me one table where it is opposit.
How to Deal with this number drives  the Nernst equation.

[Aldebaran]

Although it's more common to see these equilibria written as reductions here is a table showing both: https://www.eesemi.com/ox_potential.htm

[Hunter2]

Ok thankyou for digging. Have never seen it before. Then forget all what I said.