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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 $

My answer matches with author's answer. But I am doubtful about the correctness of the answer?

But, Is my logic in answering this question correct?

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
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Win_odd Dhamnekar
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    Answer: neither. Look at the link above for a full example of how to do this. Sure, the numbers are slightly different, but otherwise it's entirely the same problem you're trying to solve. – orthocresol Aug 20 '22 at 14:07
  • @orthocresol, So $E^\circ$ for the reaction $Ti^0 \rightleftharpoons Ti^{3+} + 3 e^- = + 1.21 volts$ that means author's answer is wrong. – Win_odd Dhamnekar Aug 20 '22 at 14:53
  • That’s what I’m getting, too. – orthocresol Aug 20 '22 at 14:55

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