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Calculate the equilibrium constant (Kc) at 25°C for the following redox reaction involving a weak oxidizing agent: H2O2(aq) + 2I^-(aq) + 2H^+(aq) ⇌ I2(aq) + 2H2O(l)

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To calculate the equilibrium constant  Kc  for the given redox reaction, we need to use the Nernst equation. First, we need to determine the half-reactions for the redox reaction:Oxidation half-reaction: 2I aq   I2 aq  + 2eReduction half-reaction: H2O2 aq  + 2H aq  + 2e  2H2O l Now, we need to find the standard reduction potentials  E  for both half-reactions. The standard reduction potential for the reduction half-reaction involving H2O2 is +1.776 V. The standard reduction potential for the oxidation half-reaction involving I is +0.536 V.Next, we need to calculate the standard cell potential  Ecell  for the redox reaction:Ecell = E reduction  - E oxidation Ecell =  +1.776 V  -  +0.536 V Ecell = +1.240 VNow, we can use the Nernst equation to calculate the equilibrium constant  Kc  at 25C:Ecell =  RT/nF  * ln Kc Where:- R is the gas constant  8.314 J/molK - T is the temperature in Kelvin  25C + 273.15 = 298.15 K - n is the number of moles of electrons transferred  2 moles in this case - F is the Faraday constant  96,485 C/mol Rearranging the Nernst equation to solve for Kc:Kc = e^nFEcell/RT Plugging in the values:Kc = e^ 2 * 96,485 C/mol * 1.240 V  /  8.314 J/molK * 298.15 K  Kc = e^239,322.4 / 2474.7 Kc  1.1  10^41So, the equilibrium constant  Kc  for the given redox reaction at 25C is approximately 1.1  10^41.

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