The equilibrium constant, K, for the following reaction is 2.90x10-2 at 1.15x103K. 2503(g) =2502(g) + O2(g) If an equilibrium mixture of the three gases in a 15.3 L container at 1.15*10PK contains 0.326 mol of S03(g) and 0.332 mol of SO2, the equilibrium concentration of O2 is M.​

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Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.