# Calculating half-cell potential from two reduction potentials [duplicate]

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.

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

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

Hence, $$E^{\circ}_{rx} = E^{\circ}_{red} - E^{\circ}_{ox} = 1.63 V - 0.37 V = + 1.26 Volts$$

Would any member of chemistry stack exchange explain me which $$E^\circ$$ of half reaction be chosen as reduction $$E^\circ$$ and which $$E^\circ$$ of half reaction be chosen as oxidation $$E^\circ$$ in this question and why?
• @orthocresol, So $E^\circ$ for the reaction $Ti^0 \rightleftharpoons Ti^{3+} + 3 e^- = + 1.21 volts$ that means author's answer is wrong. Commented Aug 20, 2022 at 14:53