What is the E° for the reaction Ti⁰ ⇌ Ti³⁺ + 3 e⁻ ?

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=−0.37V−(−1.63V)=+1.26VErx∘=Ered∘−Eox∘=−0.37V−(−1.63V)=+1.26V

My answer matches with author's answer.

But, Is my logic in answering this question correct?