MHT CET · Chemistry · Electrochemistry
\(\mathrm{E}^{\circ}\) cell is \(1 \cdot 049 \mathrm{~V}\) and involves transfer of 2 electrons, calculate equilibrium constant of cell?
- A \(2.75 \times 10^{35}\)
- B \(2.75 \times 10^{10}\)
- C \(0 \cdot 524 \times 10^{35}\)
- D \(2 \cdot 098 \times 10^{10}\)
Answer & Solution
Correct Answer
(A) \(2.75 \times 10^{35}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{\text {cell }}^{0}=1.049 \mathrm{~V}, \quad \mathrm{n}=2, \quad \mathrm{~K}=?\)
\(\mathrm{E}_{\text {cell }}^{0}=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
\(\therefore \log _{10} \mathrm{~K}=\frac{\mathrm{E}_{\text {cell }}^{0} \times \mathrm{n}}{0.0592}=\frac{1.049 \times 2}{0.0592}\)
\(\therefore \log _{10} \mathrm{~K}=35.439\)
\(\therefore \mathrm{K} =\) Antilog \((35.439)\)
\(\therefore \mathrm{K}=2.75 \times 10^{35}\)
\(\mathrm{E}_{\text {cell }}^{0}=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}\)
\(\therefore \log _{10} \mathrm{~K}=\frac{\mathrm{E}_{\text {cell }}^{0} \times \mathrm{n}}{0.0592}=\frac{1.049 \times 2}{0.0592}\)
\(\therefore \log _{10} \mathrm{~K}=35.439\)
\(\therefore \mathrm{K} =\) Antilog \((35.439)\)
\(\therefore \mathrm{K}=2.75 \times 10^{35}\)
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