Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO₃^-(aq) + 5 Hg( $\ell$ ) + 12 H⁺(aq) ? I₂(s) + 5 Hg²⁺(aq) + 6 H₂O( $\ell$ )
The standard reduction potentials are as follows:
IO₃^-(aq) + 6 H⁺(aq) + 5 e^- ? I₂(s) + 3 H₂O( $\ell$ )
E° = +1.20 V
Hg²⁺(aq) + 2 e^- ? Hg( $\ell$ )
E° = +0.86 V

Question

Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO₃^-(aq) + 5 Hg(

\ell

) + 12 H⁺(aq) ? I₂(s) + 5 Hg²⁺(aq) + 6 H₂O(

\ell

)
The standard reduction potentials are as follows:
IO₃^-(aq) + 6 H⁺(aq) + 5 e^- ? I₂(s) + 3 H₂O(

\ell

)
E° = +1.20 V
Hg²⁺(aq) + 2 e^- ? Hg(

\ell

)
E° = +0.86 V

Quizplus · Accepted Answer

The equilibrium constant (K) can be calculated using the Nernst equation: ΔG° = -nFE°cell and ΔG° = -RTlnK. First, calculate E°cell = E°cathode - E°anode = 1.20 V - 0.86 V = 0.34 V. Then, ΔG° = -nFE°cell = -(10)(96485 C/mol)(0.34 V) = -328049 J/mol. Finally, K = e^(-ΔG°/RT) = e^(328049/(8.314*298)) ≈ 3 × 10^57.

Calculate the Equilibrium Constant for the Following Reaction at 25 ℓ\ellℓ

Calculate the Equilibrium Constant for the Following Reaction at 25 $\ell$