If we assume a vanadium redox battery is only limited by the positive electrode's charge transfer reaction at 25 C, with jo = 2.89 10-4 A cm2 and the symmetry factor is 0.3, what is the active area of the positive electrode if the measured current is -1.59 A for an overpotential of -0.6 V? Answer is 5cm2, please explain how to get this. Thanks