Quantitative Nernst
Hardchemistry
At 298 K, for a one-electron reduction with [Ox]=0.010 M, [Red]=1.0 M, E = E° − (0.0591 V) log([Red]/[Ox]). If E° = 0.40 V, E is approximately:
Select the correct option:
Solution
Incorrect! Answer:
0.28 V
- Nernst Application: E=E∘−10.0591log[Ox][Red].
- Values: E∘=0.40 V, [Red]=1.0 M, [Ox]=0.01 M.
- Log term: log0.011.0=log(100)=2.
- Calculation:
- E=0.40−0.0591×2
- E=0.40−0.1182
- E=0.2818 V.
- Result: Approximately 0.28 V.
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About This Question
- Subject
- chemistry
- Chapter
- electrochemistry
- Topic
- quantitative nernst
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.28 V
- Nernst Application: E=E∘−10.0591log[Ox][Red].
- Values: E∘=0.40 V, [Red]=1.0 M, [Ox]=0.01 M.
- Log term: log0.011.0=log(100)=2.
- Calculation:
- E=0.40−0.0591×2
- E=0.40−0.1182
- E=0.2818 V.
- Result: Approximately 0.28 V.
This hard difficulty chemistry question is from the chapter electrochemistry, covering the topic of quantitative nernst. It appeared in the 2025 exam.
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