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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?

  1. A \(8.47 \times 10^{-9}\)
  2. B \(2.012 \times 10^{-10}\)
  3. C \(1.422 \times 10^{-11}\)
  4. D \(7.031 \times 10^{-10}\)
Verified Solution

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 \(E° = \dfrac{0.059}{n}\log K_c\): \(-0.16 = \dfrac{0.059}{4}\log K_c\) \(\log K_c = \dfrac{-0.16 \times 4}{0.059} = \dfrac{-0.64}{0.059} \approx -10.847\) \(K_c = 10^{-10.847} \approx 1.422 \times 10^{-11}\)