To calculate the equilibrium constant Kc for the reaction between potassium permanganate KMnO and sulfur dioxide SO in aqueous solution, we first need to write the balanced chemical equation for the reaction. The reaction can be written as:2 KMnO + 5 SO + 2 HO 2 MnSO + KSO + 5 HSOHowever, we cannot directly calculate the value of Kc without any experimental data or standard electrode potentials. Instead, we can use the Nernst equation to find the standard cell potential E and then use it to calculate the equilibrium constant Kc .The Nernst equation is given by:E = E - RT/nF * lnQwhere E is the cell potential, E is the standard cell potential, R is the gas constant 8.314 J/molK , T is the temperature in Kelvin 298 K , n is the number of moles of electrons transferred in the reaction, F is the Faraday constant 96,485 C/mol , and Q is the reaction quotient.To find the standard cell potential E , we need the standard electrode potentials for the two half-reactions:1. Reduction of permanganate ion MnO to manganese II ion Mn :MnO + 8 H + 5 e Mn + 4 HO; E = 1.51 V2. Oxidation of sulfur dioxide SO to sulfate ion SO :2 SO + 2 HO SO + 4 H + 2 e; E = -0.17 VNow, we can find the standard cell potential E for the overall reaction by subtracting the standard electrode potential of the oxidation half-reaction from the standard electrode potential of the reduction half-reaction:E = E - E = 1.51 V - -0.17 V = 1.68 VSince the Nernst equation relates the cell potential E to the equilibrium constant Kc , we can rearrange the equation to solve for Kc:E = E - RT/nF * lnQAt equilibrium, E = 0 and Q = Kc, so:0 = E - RT/nF * lnKcNow, we can solve for Kc:lnKc = nFE / RT Kc = e^ nFE / RT For our reaction, n = 5 5 moles of electrons are transferred in the reaction . Plugging in the values, we get:Kc = e^ 5 * 96,485 C/mol * 1.68 V / 8.314 J/molK * 298 K Kc 2.17 10So, the equilibrium constant Kc for the reaction between potassium permanganate KMnO and sulfur dioxide SO in aqueous solution at 298 K is approximately 2.17 10.