Use the data in Appendix D to predict whether the following reactions can occur under standard-state conditions. (c) Oxidation of Ag(s) by Pb²⁺(aq)

Video transcript

Welcome back, everyone. Using the given reduction potentials below identify whether the oxidation of platinum solid by zirconium for plus catalon will occur spontaneously at standard state conditions were given the reduction potential of platinum two plus two platinum as 1.18 volts and the reduction potential of zirconium four plus two zirconium as negative 1 45 or 1.45 volts. So from the prompt, we have the oxidation of solid platinum to show its oxidation, it's going to form as a product, our platinum two plus catalon and release two electrons. Now, for our zirconium four plus, recognize that the prompt tells us that zirconium four plus is our oxidizing agent. And Zirconium four plus is therefore going to be reduced. So it will undergo a reduction. So showing its reduction, it's going to gain four electrons and form as a product solid zirconium. For our first reaction with solid platinum showing its oxidation, we have a Reduction potential for its oxidation equal to negative 1.18V. Since in the prompt, the reduction of platinum cap platinum cat ion is 1.18 volts which is positive. So we're just going to flip its magnitude as negative for the oxidation. And now for the standard cell potential of our reduction of zirconium four plus that is given in our pumped as negative 1.45 volts. Now, with these two half reactions written out, we need to bounce out the electrons. Notice that in our second half reaction, we have four electrons, but in our first, we only have two. So we're going to take our entire first half reaction and multiply it by a factor of two, which is now going to give us two moles of solid platinum platinum, which produces two moles of platinum cat cat ion Plus four electrons. So now adding up these two reactions were going to come up with an overall reaction in which we have, we can cancel out the four electrons from our first half reaction with the four electrons from our second half reaction and carry down everything else that we're left with. Our overall reaction is going to now be one mole of solid platinum or rather two moles of solid platinum. Since we multiplied by two plus one mole of zirconium, four plus catalon plus. Actually, we have our arrow sign now yields two moles of platinum, two plus catalon Plus one mole of solid zirconium. Now we need to calculate our cell potential e cell under standard conditions recall that we can calculate this by taking the standard cell potential for the reduction at our cathode or added to the standard cell potential of our oxidation at our node. So, plugging in our values for the reduction, standard cell potential, we have the value negative 1.45 volts given above in our prompt. And then plus our standard cell potential for the oxidation at are an ode. We determine the value to be negative 1.18 volts. So taking the sum of these two terms, we get a standard cell potential equal to negative 2.63 volts. We should now recall that when we have a standard cell potential that is negative or less than zero. This means that our reaction is going to be non spontaneous in the forward direction. But when we have a standard cell potential that is positive or greater than zero, our reaction is going to be spontaneous in the forward direction. And because our calculated self potential was negative 2.63V, we would say that therefore, our reaction is going to be non spontaneous. And so that means that the reduction, we're sorry, that means that the oxidation. So our oxidation of solid platinum, bye zirconium catalon will not occur and it will not occur specifically spontaneously. So this statement is going to be our final answer to complete this example corresponding to choice be in the multiple choice as the correct answer. So I hope this made sense and let us know if you have any questions