COMEDK · Chemistry · 18. Electrochemistry
For the cell reaction \(4 \mathrm{Br}^{-}+\mathrm{O}_2+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}\) at 298 K , the \(\mathrm{E}^0\) cell \(=0.16 \mathrm{~V}\).
What would be the \(\mathrm{K}_{\mathrm{c}}\) (Equilibrium constant) value if the reverse reaction were to take place?
- A \(8.47 \times 10^{-9}\)
- B \(2.012 \times 10^{-10}\)
- C \(1.422 \times 10^{-11}\)
- D \(7.031 \times 10^{-10}\)
Answer & Solution
Correct Answer
(C) \(1.422 \times 10^{-11}\)
Step-by-step Solution
Detailed explanation
For the reverse reaction, \(E°_{rev} = -0.16\) V and \(n = 4\). Using \(E° = \dfrac{0.059}{n}\log K_c\): \(-0.16 = \dfrac{0.059}{4}\log K_c\) \(\log K_c = \dfrac{-0.16 \times 4}{0.059} = \dfrac{-0.64}{0.059} \approx -10.847\) \(K_c = 10^{-10.847} \approx 1.422 \times 10^{-11}\)
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