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Spontaneity and redox reactions

Using standard cell potential to predict whether a redox reaction will be spontaneous under standard conditions.  Created by Jay.

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  • female robot ada style avatar for user TheBitWall
    What is the reasoning behind Copper not working with Lead? I still don't understand.
    (10 votes)
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  • mr pants teal style avatar for user adenike.adedapo
    Could Pb^2+ be the oxidizing agent of Cu if one added energy to it?
    (9 votes)
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    • old spice man green style avatar for user Matt B
      Sure, you can make your own electrical circuit with an additional battery and then choose your cathode/anode but then this would not be a "spontaneous" redox reaction, which is the title of the video, because "spontaneous" does not require additional energy ;-)
      (15 votes)
  • hopper jumping style avatar for user Yuya Fujikawa
    If a reaction is not spontaneous, it can't happen?
    (4 votes)
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    • orange juice squid orange style avatar for user awemond
      If a reaction is not spontaneous, it can't happen under the given conditions. However, it could happen under different conditions. For example, changing the temperature or the concentrations of the molecules in your reaction might make the reaction spontaneous.

      As an example, consider water freezing into ice at 20 °C. This is not spontaneous. However, if you lowered the temperature to -10 °C, then suddenly water would spontaneously freeze into ice.
      (12 votes)
  • piceratops seedling style avatar for user Sheqi Zhang
    Does it means that is the overall voltage is greater than 0, then the reaction can happen?
    (3 votes)
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    • duskpin ultimate style avatar for user rosafiarose
      If the overall electric potential (voltage) is greater than 0, then the reaction is spontaneous. A spontaneous reaction is one that can occur by itself without any outside addition of energy. A non-spontaneous reaction can still occur, but it will need an addition of energy (the activation energy) before it will start occurring.
      (3 votes)
  • spunky sam blue style avatar for user lizy210210
    at , Jay rewrites the two half reactions so that the electrons can cancel, but because making the electron number equal by multiplying by a number will not affect the voltage value (as intensive property) is it even necessary to find the equation of the overall reaction? Because the question does not expricitly ask for the overall equation.
    (just trying to save time on the AP exam :) )
    Can I just look at the voltage values and add them right away?
    (1 vote)
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  • blobby green style avatar for user abdulrharris
    I thought that because our standard cell potential for Cu was higher, that means it MUST be reduced (+0.34 > -0.13). Thus I got the answer as being +0.47. Why, if Pb^2+, cant oxidize copper, would I then show lead being reduced?
    (2 votes)
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  • starky tree style avatar for user Ryan
    When calculating cell potentials, in my tutorial in university they said to subtract the more positive potential from the smaller or more negative potential and not add them?
    Eg question was:
    Rechargeable nickel-cadmium batteries normally operate (discharge) with
    the following oxidation and reduction half-cell reactions.

    Cd(s) + 2OH-(aq) ⇌ Cd(OH)2(s) + 2e- (-0.82v)

    NiO(OH)(s) + H2O(l) + e- ⇌ Ni(OH)2(s) + OH-(aq) (0.60v)
    And the overall cell potential answer was: 0.6-(-0.82)=1.42

    Is this the same thing or is uni wrong XD?
    (2 votes)
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  • starky seedling style avatar for user Tanya Henry
    When you are figuring out if lead can oxidize copper at , why are we reversing the equation for copper?
    (1 vote)
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  • blobby green style avatar for user 😊
    Is it true that we have to reverse the reaction which the lowest reduction potentials?
    (1 vote)
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  • piceratops sapling style avatar for user akiilessh
    why is that it doesnt work with the metals above it?? the reason apart from saying that the meatals above has higher reduction potential.
    (1 vote)
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Video transcript

- [Voiceover] Our goal is to predict whether or not lead two plus can oxidize solid aluminum or solid copper under standard state conditions. And also, to calculate these standard cell potentials E zero for each reaction at 25 degrees C. So we have a standard 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. Our question wants to know whether lead two plus can oxidize these solid metals. So therefore, lead two plus is functioning as an oxidizing agent, so it itself must be being reduced, but it is functioning as an oxidizing agent. 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. An an oxidizing agent can oxidize any reducing agent below it on our table, so if I find aluminum here, aluminum is below lead two plus. So we would predict that lead two plus can oxidize aluminum, so our prediction is yes for aluminum. So we can actually draw a diagonal line from lead two plus to aluminum, and we predict this will work. This reaction will work. So sometimes you'll see this called the diagonal rule. Next, let's predict whether lead two plus can oxidize solid copper. An oxidizing agent can't oxidize a reducing agent that appears above it on our standard reduction potential tables, so if I draw a line from lead two plus to solid copper, all right, we're going up here. So this reducing agent, copper, is above lead two plus. And so this should not work, so our prediction is no, this will not work. So no, lead two plus cannot oxidize copper under standard state conditions. Let's go ahead and calculate these standard cell potentials for each of these reactions to just confirm our predictions. So we'll start with lead two plus oxidizing solid aluminum. 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 as a reduction half-reaction. So a reduction half-reaction is lead two plus, plus two electrons, going to solid lead. And so the standard reduction potential for this half-reaction is negative .13 volts. So this is negative .13 volts. 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 reduction half-reaction. So we write solid aluminum, right, is going to aluminum three plus. To do that, it needs to lose three electrons, so loss of electrons is oxidation. So we need to find the standard oxidation potential for this half-reaction now. And we've done this several times. When you reverse a reduction half-reaction and turn it into an oxidation half-reaction which is what we've done down here, you just change the sign on the standard reduction potential. So the standard reduction potential is negative 1.66 volts, all we have to do is change the sign, so it's positive 1.66 volts. So the standard oxidation potential is positive 1.66. If we wanted to write the overall reaction, we need to balance everything. So we need to get our number of electrons equal. So we could do that by multiplying our first half-reaction here by three, because that gives us six electrons. So three times two gives us six electrons. The number of electrons needs to be the same, so we need to multiply our second half-reaction by two, because two times three also gives us six electrons. So let's do that calculation here. 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, and this would give us three solid Pb. So three Pb solid. 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 leave our standard reduction potential as negative .13 volts. All right, let's do our next half-reaction, so our oxidation half-reaction. So we would have two aluminum, so two aluminum, and then two, this is going to give us two Al three plus, and six electrons, two times three gives us six electrons. And once again, we do not multiply our oxidation potential by two because voltage is an intensive property. So our standard oxidation potential remains positive 1.66 volts. To get our overall reaction, we just add together our two half reactions, right, so let's add these together, so we're going to add these together to get our overall. All right, so the electrons would cancel out, so six electrons, so let's write our reactants, which are right here, so we would have three Pb two plus... Plus two Al, so plus two Al, giving us, so for our products, we would have three Pb, so three Pb, plus two Al three plus. So we have two Al three plus. So is this reaction spontaneous? So is our overall reaction spontaneous? We could figure that out by calculating the standard cell potential, which is what our problem asked us to do. So what is the standard cell potential? We know we just have to add up our standard reduction potential and our standard oxidation potential. So just like we add our half-reactions, we add our potentials together. So we're going to add our standard reduction potential for that half-reaction, and we're going to add our standard oxidation potential for that half reaction, to get the standard cell potential for the cell. So that's negative .13, right, so it's negative .13, plus 1.66, so plus positive 1.66. So that's equal to positive 1.53, so the standard cell potential, is positive 1.53 volts. And remember, when your cell potential is positive, that means a spontaneous reaction. So this is a spontaneous reaction, which is what we predicted. We predicted that lead two plus could oxidize aluminum, all right, so let's go back up here to our problem again, so remember, our problem asked us, can lead two plus oxidize aluminum? And we predicted yes, by using the diagonal rule here, by drawing this arrow down here, and then we just calculated the standard cell potential, and we saw that it confirmed our prediction. All right, next, let's do copper. So we said that lead two plus will not be able to oxidize copper. Let's find the standard cell potential to confirm our prediction. So we're going to keep this half-reaction, right? So we're going to keep this half-reaction for lead, and then for copper, we need to reverse this half reaction as it's written. So we need to write the reverse of this reaction, and when you're writing the reverse, remember, you need to change the sign, this is positive .34 for the reduction of Cu two plus, so for the oxidation of copper, it's negative .34, the standard oxidation potential. So let's go back down here, where we have some more room, and let's figure out the standard cell potential. So we're going to write our reduction half-reaction first, which was lead two plus, plus two electrons, to give us solid lead here, so the standard reduction potential for this half reaction, is negative .13 volts. And the oxidation half reaction, this was to oxidize solid copper, so copper turns into copper two plus, and we need to lose two electrons, we're oxidizing here, so we're losing electrons, and the standard oxidation potential for this half reaction, we just talked about it above, it's negative .34. Negative .34 because we're writing this as an oxidation half-reaction. So to find our overall reaction, we just add up our half reactions here. We already have the number of electrons the same, so those electrons will cancel out, and we have our reactants, so let's write those in here. So we have lead two plus, plus solid copper, and our products, so our products would be lead and Cu two plus. So this would be lead, and Cu two plus. All right, let's calculate the standard cell potential. We know, to find the standard cell potential, all we have to do is add the standard reduction potential, and the standard oxidation potential, so what is negative .13 plus negative .34? That's equal to negative .47 volts. So the standard cell potential, right, for this reaction, that we've just written out here, for this overall reaction, is negative, and since the standard cell potential is negative, we know this is not a spontaneous reaction. So let me write that in red here. So this is not a spontaneous reaction. So the standard cell potential is negative. So if you calculated the standard change in free energy, since the standard cell potential is negative, you get a positive value for the standard change in free energy. So this is not a spontaneous reaction. So let's go back up to here, let's go back up to the beginning. All right, so we made our predictions using our standard reduction potential. We just confirmed that lead two plus will not oxidize solid copper, this is not a spontaneous reaction. So aluminum, aluminum is more easily oxidized than copper, right, you could figure that out just by thinking about this example, or you could actually look at the standard reduction potentials. You could see that aluminum down here has a more negative value for the standard reduction potential, and therefore, it's more easily oxidized. So aluminum is more easily oxidized than copper. We say aluminum is more active than copper, so let me write that down. Aluminum is more active, it's more active than copper, so if you remember the activity series, you can explain the activity series in general chemistry, by looking at standard reduction potentials. So this is just another way to think about something you probably already know.