Dhamnekar Win,odd Posted August 20, 2022 Posted August 20, 2022 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?
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