Chemistry Forum

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, \( E^\circ_{rx} = E^\circ_{red} - E^\circ_{ox} = -0.37 V - (-1.63 V) = + 1.26 V \)

My answer matches with author's answer.

But, Is my logic in answering this question correct?

It can not match, because its negativ , its - 1,26 V

https://www.periodensystem-online.de/in ... 2&id=redox

You have to change both voltages.

E = -0,163 V -(-0,37V) = -1,26 V

Author has given the following answer: is there typographical error in the above answer?

I think your answer is correct. There is a typographical error in the given answer.

Oxidation and Reduction is mixed up. Also for both voltages the same reaction is used. There is no reaction Ti 2+ to Ti 3 +.

Why did you choose ① as reduction half reaction and ② as oxidation half reaction?

Would you explain that?

Author's given answer is wrong.The correct answer is + 1.21 volts using the formula for ∆G°= -nFE°

No also in this case you lost one minus. In tables this gives - 1.21 V

https://www.periodensystem-online.de/in ... 2&id=redox

The table in your link says Ti³⁺ + 3 e⁻ ⇌ Ti⁰ E° = -1.208 V

Reverse it for oxidation of Titanium metal Ti⁰ ⇌ Ti³⁺ + 3 e⁻ E° = +1.208 V

The redox potential of a half reaction is measured to the hydrogen electrode.
So it doesn't matter if you look Ti 3+ + 3 e- => Ti or Ti => Ti 3+ + 3 e- . The voltage of 2 H+ +2 e- => H₂ or H₂ => 2 H+ +2 e- is 0 V
and the difference between both reaction is - 1,26 V.

Also in your postings it's given with the ⇌ sign. Also back and forward is the same regarding the potential.