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 Ti

^{0} Ti

^{3+} + 3 e

^{-} is spontaneous. Is that correct?

My answer matches with author's answer.

But, Is my logic in answering this question correct?