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For the reaction Cu²⁺ + 2e⁻ → Cu with E° = +0.34 V, what is the cell potential when [Cu²⁺] = 0.01 M at 298 K?

A{'text': '+0.34 V', 'label': 'A'}
B{'text': '+0.399 V', 'label': 'B'}
C{'text': '+0.281 V', 'label': 'C'}
D{'text': '+0.281 V (using log 100)', 'label': 'D'}
Answer & Solution
Correct answer: C. {'text': '+0.281 V', 'label': 'C'}
1. E = E° − (0.059/n) log (1/[Cu²⁺]) = E° + (0.059/n) log [Cu²⁺]. 2. n = 2. Substitute: E = 0.34 + (0.059/2) × log(0.01). 3. log(0.01) = −2. Term = (0.059/2) × (−2) = −0.059. 4. E = 0.34 − 0.059 = +0.281 V. _Source: NCERT Class 12 Chemistry, Unit 3 "Electrochemistry", §3.3_
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