MHT CET · Chemistry · Electrochemistry
Which from following equation is correct for relation between strandard cell potential and equilibrium constant?
- A \(\mathrm{E}_{\text {cell }}^0=\log _{10} \mathrm{~K} \frac{\mathrm{n}}{0.0592}\)
- B \(\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
- C \(\mathrm{E}_{\text {cell }}=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
- D \(\mathrm{E}_{\text {cell }}=\log _{10} \mathrm{~K} \frac{\mathrm{n}}{0.0592}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
We know \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{Q}\)
Or at equilibrium \(0=\mathrm{E}_{\text {cell }}^0-\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
Or \(\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
Or at equilibrium \(0=\mathrm{E}_{\text {cell }}^0-\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
Or \(\mathrm{E}_{\text {cell }}^0=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
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