Adding a catalyst to a redox reaction does not affect the equilibrium position of the reaction. A catalyst works by lowering the activation energy of the reaction, which speeds up the rate at which the reaction reaches equilibrium. However, it does not change the equilibrium constant K or the equilibrium concentrations of the reactants and products.To illustrate this, let's consider a hypothetical redox reaction:2A + B 3CSuppose the initial concentrations of A, B, and C are [A] = 1.0 M, [B] = 1.0 M, and [C] = 0 M, respectively. Let's assume the equilibrium constant K for this reaction is 4.0.At equilibrium, the reaction quotient Q is equal to the equilibrium constant K . For this reaction, Q = [C]^3 / [A]^2 * [B] . We can set up an ICE Initial, Change, Equilibrium table to determine the equilibrium concentrations:`` [A] [B] [C]Initial: 1.0 1.0 0Change: -2x -x +3xEquil.: 1.0-2x 1.0-x 3x``Now, we can substitute the equilibrium concentrations into the expression for Q:4.0 = 3x ^3 / 1.0 - 2x ^2 * 1.0 - x Solving this equation for x, we find that x 0.21. Therefore, the equilibrium concentrations are [A] 0.58 M, [B] 0.79 M, and [C] 0.63 M.Now, let's add a catalyst to the reaction. As mentioned earlier, the catalyst will speed up the reaction, but it will not change the equilibrium constant or the equilibrium concentrations. Therefore, even with the catalyst, the equilibrium concentrations will still be [A] 0.58 M, [B] 0.79 M, and [C] 0.63 M.In summary, adding a catalyst to a redox reaction does not affect the equilibrium position of the reaction, regardless of the number of moles of reactants and products. The catalyst only affects the rate at which the reaction reaches equilibrium.