Home › JEE Advanced › Chemistry › Electrochemical Series › For the reaction Cu²⁺ + 2e⁻ → Cu with E° = +0.34…
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_
Related questions
The electrolytes of the two half-cells are joined by a:The two portions of a galvanic cell are also called:In the Daniell cell, zinc dissolves at the anode and copper:A negative standard potential marks a reducing agent that is:Going from top to bottom of the table, the electrode potential:The weakest oxidising agent in the table of potentials is:The most powerful reducing agent in aqueous solution is:The weakest reducing agent in the table of potentials is: