Q29 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry

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
    $2.012 \times 10^{-10}$
  • Correct answer
    C
    $1.422 \times 10^{-11}$
  • D
    $7.031 \times 10^{-10}$

Solution

For the reverse reaction, $E°_{rev} = -0.16$ V and $n = 4$.

Using $E° = \dfrac{0.059}{n}\log K_c$:

$-0.16 = \dfrac{0.059}{4}\log K_c$

$\log K_c = \dfrac{-0.16 \times 4}{0.059} = \dfrac{-0.64}{0.059} \approx -10.847$

$K_c = 10^{-10.847} \approx 1.422 \times 10^{-11}$

Hence, the correct answer is option: 'C'.

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