quantitative nernst

Question

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:

RankGuru · Accepted Answer

1. **Nernst Application**: $E = E^\circ - \frac{0.0591}{1} \log \frac{[Red]}{[Ox]}$.
2. **Values**: $E^\circ = 0.40$ V, $[Red] = 1.0$ M, $[Ox] = 0.01$ M.
3. **Log term**: $\log \frac{1.0}{0.01} = \log(100) = 2$.
4. **Calculation**: 
   - $E = 0.40 - 0.0591 	imes 2$
   - $E = 0.40 - 0.1182$
   - $E = 0.2818$ V.
5. **Result**: Approximately **$0.28$ V**.

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quantitative nernst

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