When the concentration of the oxidizing agent is increased by a factor of 5, the reaction will no longer be at equilibrium. To determine the effect on the position of the equilibrium and the cell potential, we can use the Nernst equation:E = E - RT/nF * ln Q where 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 25C = 298 K , n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant 96,485 C/mol , and Q is the reaction quotient.Since the concentration of the oxidizing agent is increased by a factor of 5, the reaction quotient Q will also change. Let's assume the initial reaction quotient is Q1 and the new reaction quotient after increasing the oxidizing agent concentration is Q2. Then:Q2 = 5 * Q1Now, we can compare the cell potentials before and after the change in concentration:E1 = E - RT/nF * ln Q1 E2 = E - RT/nF * ln Q2 Since Q2 > Q1, ln Q2 > ln Q1 , and therefore E2 < E1. This means that the cell potential will decrease when the concentration of the oxidizing agent is increased by a factor of 5.As for the position of the equilibrium, since the cell potential is decreasing, the reaction will shift to counteract the change in concentration. In this case, the reaction will shift towards the side of the reducing agent to consume some of the excess oxidizing agent and re-establish equilibrium.