Calculate the equilibrium constant for the reaction between Fe2+(aq) and Zn(s) (at 25 °C).

Video transcript

hey everyone in this example, we need to consider the reaction of a two plus Catalan of copper and solid nickel at 25 degrees Celsius. We need to calculate the equilibrium constant and we're going to use reduction potentials to do so. So our first step is to write out our reaction, we have copper two plus reacting with solid nickel to form solid copper plus the nickel two plus Catalan. We should recognize that this is a redox reaction, meaning we need to write out half reactions. So for our first half reaction that comes from our first reactant being the copper two plus carry on which we can see forms solid copper as a product. We should recognize that we went from a net plus two charge to a net neutral charge to form our neutral atom of copper, meaning we must have gained electrons. And so we can say that we gained two electrons here to form our neutral atom of copper. And this would mean because we gained these two electrons that this occurs or this half reaction occurs as a reduction. Now moving on to our second half reaction coming from our second atom which is our second reactant, which is solid nickel we produce as a product, the nickel two plus catalon. And so we should recognize that we must have lost two electrons. However, to form our neutral atom of nickel, we would have needed to have gained two electrons here on the product side. And so because we lost two electrons to form our nickel to plus carryin on the product side, this is going to occur as an oxidation. Now we should recall next that the oxidation half of our reaction occurs at the an ode of our voltaic cell. Whereas our reduction half of our reaction we should recall occurs at the cathode of our voltaic cell. Now making note of that, we want to go ahead and refer to the standard reduction potential table online or in our textbooks and we would see that for our first half reaction with the copper two plus carry on. We would have at our cathode For copper. A value or value a voltage where the degree of our cell is equal to negative 0.23V correction. That would be positive. Sorry, 0.34 volts for our cath out of copper. And then for our anote where nickel half reaction is occurring. We would see according to our tables that we have a voltage Equal to a value of a -0.23V. So this is for our an ode. And so now we can go ahead and calculate our self potentially degree of ourselves. And recall that this is calculated by taking the cell potential of our an ode. Or sorry, are self potential of our cathode subtracted from our self potential of our a node. And so as we saw above our self potential of our cathode had a voltage of .34 which was a more positive and higher value. So what we would have is .34 subtracted from our cell potential of our an ode which above was negative 0.23V. And so we can see that subtracting a negative means well really just add here. And so we would get that are self potential Is going to equal a value of 0.57V. And so now we can go ahead and continue to solve for our equilibrium constant. So we're going to recall that. We're going to use the following form of our cell potential formula where we get our standard cell potential E degree cell or E zero cell is equal to our standard voltage which is 0. volts divided by N. Where we should recall that N is going to be our number of electrons transferred in our redox reaction. And then this entire quotient here is going to be multiplied by the log of our equilibrium constant. K. Now the prompt wants us to solve for equilibrium constant. So we're going to reformat this form of our first equation so that we can solve for our log of K equal to we would say we would take the value N. And multiply it by our standard cell potential E zero cell Which is now going to be divided by that standard value of 0.05916V. Now we should make note of the fact that according to our redox reaction which we wrote above, which occurred at our cathode. We stated that we had a transfer of two electrons here and so our value and here is going to be too so we can go ahead and plug in what we know. So what we have is our log of our equilibrium constant. K is equal to two for n, multiplied by the standard cell potential, which we calculated above here as .57V. So we would plug that in Here below as 0.57V as our standard cell potential divided by our standard value of 0.05916V which came from our formula. And so simplifying this, we would get that the log of our equilibrium constant is equal to a value of 19.2698 when we simplify the right hand side. So now we want to cancel out this log term so that we can isolate for our equilibrium constant. So we're gonna take our right hand side and set it as an exponent here to 10 so that we can cancel our log term. And what we will get now is that our equilibrium constant K is equal to a value Of 1.86 times 10 to the positive 19th power. And so this would be our value for equilibrium constant calculated by using standard 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.