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NEET · Chemistry · STD 12 - 2. Electrochemistry
For a cell involving one electron \(E_{cell }^{\ominus} =0.59 \mathrm{V}\) at \(298 \;\mathrm{K}\), the equilibrium constant for the cell reaction is Given that \(\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V} \text { at } \mathrm{T}=298 \mathrm{K}\right]\)
- A \(1.0 \times 10^{2}\)
- B \(1.0 \times 10^{5}\)
- C \(1.0 \times 10^{10}\)
- D \(1.0 \times 10^{30}\)
Answer & Solution
Correct Answer
(C) \(1.0 \times 10^{10}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log _{10} \mathrm{Q}\) at equllbrium \(\mathrm{E}_{\mathrm{cell}}=0, \mathrm{Q}=\mathrm{K}_{\mathrm{eq}}\)
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