The first statement says that k1/k1 = 1, where k1 is the rate constant for the forward reaction, and k1 is the rate constant for the reverse reaction. This might at first appear to be true, since, when a reaction is at equilibrium, the rates of the forward and reverse reactions are equal. but keep in mind that the reaction constant is only part of the reaction rate. In fact, statement I equals the equilibrium constant. So k1/k1 will not equal one at equilibrium, unless it happens that the concentrations of the reactants and products are exactly equal at equilibrium. Not all reactions meet this condition, so statement I is not true.
Statement II says that E = E°, where E is the reaction potential and E° is the standard potential for the reaction. There is a condition under which this is true, but that condition isnt equilibrium. The standard potential is defined as the reaction potential under standard conditions. Look at Equation 2 in the passage. E = E°tot (RT/nF)ln([C]c[D]d/[A]ab). Under standard conditions, the concentrations of all the reactants and products are equal to one, so everything cancels out. The natural log (ln) of 1 is 0, so the complicated expression drops out, leaving the reaction potential equal to the standard potential. So Statement II is true under standard conditions, but is not true at equilibriumunless equilibrium happens to occur under standard conditions. So Statement II is not true either.
Statement III, on the other hand, is true at equilibrium. At equilibrium, the voltage is zero, since the forward and reverse reactions are taking place at equal rates. We can rearrange the equation to find that the standard potential equals (RT/nF)ln([C]c[D]d/[A]ab). And if we rearrange this
equation, were left with Statement III. So III alone is true, and the answer is B.
