Calculate the equilibrium constant for each of the reactions in Problem 65.

Video transcript

hey everyone in this example, we need to consider the below reaction and calculate the equilibrium constant. Using reduction potentials. So we should recognize that this is a redox reaction. And we need to write out our half reactions for our first half reaction that comes from our first reactant, which is solid lead, Which produces lead, two plus Catalan as a product. We should recognize that we have one mole of lead on both sides of the equation. So the atom is balanced. And our next step is to recognize the charges. So the net charge on the reaction side is zero because we have a neutral atom of lead. However, the net charge on the product side is plus two because we have that too plus charge here. And so we should cancel out this plus two charge that we have a net neutral charge of zero on the product side. And so that would mean that we need to gain two electrons here on our product side. And because we have electrons here on our product side, we would recall that this half reaction is going to be our oxidation half reaction. Now, our next step is to recognize the second half reaction, which comes from our second reactant. We have two moles of the A. G plus one carry on Which forms as a product, two moles of silver. And so now we need to go ahead and recognize that because we have two atoms of silver on the product side, as well as two atoms of silver Catalan on the react inside are atoms of silver are balanced. And so the next step is to focus on calculating the same amount of charge. So we should recognize that we have a net Plus two charge on the reactant side because we have that coefficient of two, whereas on the product side we have a neutral atom. So we have just a net charge of zero. So we're gonna need to add two electrons here to the react inside to go ahead and get a net neutral charge here for both sides of our reaction. So we would add two electrons here and because we added the two electrons on our reactant side here, this is going to be the reduction half reaction. We also can ensure that our electrons here are the same in both of our half reactions. So we don't need to worry about balancing out electrons. And now our next step is to recall but for our oxidation half reaction that occurs in our voltaic cell at the cathode. Or sorry, at the anodes. We also want to recall that in a voltaic cell. The reduction half reaction occurs at the cathode. And so we want to refer to our standard reduction potential table. We would find this either online or in our textbooks and we would see that for our oxidation half reaction which involves our atom led, we would find lead on this table and see its oxidation which has a value, a self potential value to be specific degree. So Equal to negative 0.13V for our a node. And then for our reduction half reaction, which occurs at the cathode. We're going to find silver on our standard reduction potential table and we'll see that it has a cell potential value equal to 0.80 volts for our cathode. And so now we want to calculate our standard cell potential, which is just E. And we would say that this is equal to a value of our self potential for our cathode added or sorry, subtracted from our self potential of our A node. And so plugging in from what we know from above from our table, we get that our standard cell potential is equal to we have for our cathode a value of 0.80 volts For the oxidation of silver. And then this is going to be subtracted from our cell potential of lead from our table, which has a value of its oxidation as negative 0.13V. So because we're subtracting a negative, we're really adding here and we're going to get a value for our standard cell potential equal to zero 0.93 volts. So next we want to recall the two following equations. Our first equation being our equation for Gibbs free energy which can be found by taking negative one times r gas constant R times the temperature in kelvin times the Ln of our equilibrium constant K. Our second equation we want to recall is going to be our second equation for Gibbs free energy, which can be calculated by taking negative one times N, which would represent our electrons transferred at our redox half reaction times F which is our Faraday's constant. And then times the standard cell potential of our cell. Now we're going to combine these two equations to calculate our equilibrium constant. And in doing so, we would form a new equation where we can say that the natural log of our equilibrium constant is equal to our electrons transferred at our redox half reaction N times Faraday's constant, F times the standard cell potential of our cell divided by r gas constant, R times the temperature in kelvin. So we're gonna go ahead and plug in what we know into this formula. So what we would have is that our natural log of our equilibrium constant K. Is equal to our electrons transferred at our redox half reaction. We would see our reduction half reaction occurred with our silver caravan and we transferred two electrons here. So our N value here is equal to two. So we can say two times faraday's constant. So we should recall that Faraday's constant is equal to a value of 96,485 jewels per mole of electron. And so just to be clear for a value of N which is our electrons transferred. We can clarify this by giving it units to say that two moles of electrons are transferred and this is multiplied still by Faraday's constant. So then we're going to multiply this by our standard cell potential value which above we calculated as 0.93 volts. However, we want to recall that vaults are equivalent to jules divided by columns. So we're going to substitute jewels divided by columns in place of volts here and then now we're going to divide by R gas constant R. So we should recall that our gas constant R is equal to a value of 8. joules divided by moles times kelvin, which is then multiplied by our temperature in kelvin. And so according to the prompt, we have a temperature Which we will we were not given actually. So we should assume that we have a standard temperature, which in Kelvin would be to 98.15 Kelvin. So we recall that we assume standard temperature since we were not given temperature and now we want to go ahead and cancel out our units so we can get rid of kelvin, we can get rid of columns. Sorry, actually we would not cancel out columns yet. So we can actually just focus on get ridding of jewels. We can also get rid of moles. So it appears that the only unit we're left with this columns. However, we should recall that the equilibrium constant does not have units. So we're just going to disregard columns and also cancel it out. So in doing so, in our calculators we should get a value equal to and actually we'll just write this down below. So it's clear. So we say that the Ln of our equilibrium constant is equal to a value in our calculators of 72.9 or sorry, . 9822. And so our last step is to isolate our equilibrium constant by taking Jeweler's number on both sides to cancel out the Ln term. So we would have us number two, an exponents of this value here. So it would cancel out the Ln term on the left hand side and we would have equilibrium constant isolated so that now we get a value equal to 2. times 10 to the Positive 31st power. And this would be our final answer here for equilibrium constant that we were able to find using reduction potentials. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video