Multiply our oxidation potential by two because voltage is an intensive property. Us two Al three plus, and six electrons, two times three gives us six electrons. So we would have two aluminum, so two aluminum, and then two, this is going to give Our next half-reaction, so our oxidation half-reaction. So we're multiplying our half-reaction, but remember, we do not multiply our standard reduction potential by three because voltage is an intensive property. So we have three Pb two plus, so for reduction, we would have three Pb two plus, plus three times two electrons, gives us six electrons,Īnd this would give us three solid Pb. Our second half-reaction by two, because two times three also gives us six electrons. Needs to be the same, so we need to multiply So three times two gives us six electrons. Multiplying our first half-reaction here by three, because that gives us six electrons. The overall reaction, we need to balance everything. So the standard oxidation potential is positive 1.66. So the standard reduction potential is negative 1.66 volts, all we have to do is change the sign, so Oxidation half-reaction which is what we've done down here, you just change the sign on the standard reduction potential. When you reverse a reduction half-reaction and turn it into an Standard oxidation potential for this half-reaction now. Lose three electrons, so loss of electrons is oxidation. So we write solid aluminum, right, is going to aluminum three plus. Next, we're going to write the oxidation half-reaction for aluminum, so here is aluminum, so we're going to write an oxidation half-reaction, so we need to reverse this And so the standard reduction potential for this half-reaction So a reduction half-reaction is lead two plus, plus two electrons, going to solid lead. So if lead two plus is oxidizing, we're going to write the lead two plus, this half-reaction here, we're going to leave it how it is asĪ reduction half-reaction. So we'll start with lead two plus oxidizing solid aluminum. Let's go ahead andĬalculate these standard cell potentials forĮach of these reactions to just confirm our predictions. So no, lead two plus cannot oxidize copper under standard state conditions. And so this should not work, so our prediction is So this reducing agent, copper, is above lead two plus. Potential tables, so if I draw a line from lead two plus to solid copper, all Oxidize a reducing agent that appears above it on our standard reduction Next, let's predict whether lead two plus can oxidize solid copper. So sometimes you'll see thisĬalled the diagonal rule. So we can actually draw a diagonal line from lead two plus to aluminum, and we predict this will work. So we would predict that lead two plus can oxidize aluminum, so our prediction is yes for aluminum. An an oxidizing agentĬan oxidize any reducing agent below it on our table, so if I find aluminum here, aluminum is below lead two plus. In general, an oxidizing agent can oxidize any reducing agent that lies below it on our standard reduction potential table, so here is lead two plus, all right, so we have our stronger oxidizing agents going up on the left side. So therefore, lead two plus is functioning as an oxidizing agent, so it itself must be being reduced,īut it is functioning as an oxidizing agent. Know whether lead two plus can oxidize these solid metals. Reduction in potential table, a very shortened version of it where we have our half-reactions written as reduction half-reactions on the left, and the standard reduction potentials for those half-reactions on the right, measured in volts. Standard cell potentials E zero for each reaction at 25 degrees C. Our goal is to predict whether or not lead two plus can oxidize solid aluminum or solid copper under standard state conditions.
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