A voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn1s2 + Ni2+1aq2 ¡ Zn2+1aq2 + Ni1s2 (c) What is the emf of the cell when 3Ni2+4 = 0.200 M and 3Zn2+4 = 0.900 M?

Video transcript

welcome back everyone. A voltaic cells built and operates at 2 98 kelvin Using the following reaction, solid iron reacting with cobalt two plus caddy on to produce solid cobalt as a product and iron two plus Catalan as a second product. When the iron two plus Catalan concentration is 20.750 moller. And our cobalt two plus caffeine concentrations. 20.350 moller. What is the E M. F. Of the cell where E M F we want to recognize is our electric magnetic force otherwise known as our self potential Under nonstandard conditions being a self. We know that these are nonstandard conditions due to the fact that we have concentrations that are under one molar, where one Moeller would be our concentrations under standard conditions. Now because we need to calculate E M f. We want to recall our nursed equation where we have our cell potential under nonstandard conditions sl equal to our cell potential under standard conditions. E degree sell subtracted from 0.5 92 volts divided by our electrons transferred being N. And let's just make some more room here. So this is then multiplied by the log of our reaction quotient Q. Where we would recall that Q. Is our concentration of our Aquarius or gaseous products. Over our concentration of Aquarius or gaseous react mints. So our first step is to write out two half reactions because we should recognize that this is redox. So for our first half reaction according to what were given in the prompt, we have our solid iron which produces on the product side, our iron two plus catty on where we would have two electrons on the product side Now for our second half reaction according to what were given in the prompt, we have cobalt two plus catty on as our reactant here where on our product side we have solid cobalt that is formed and that is formed from our cobalt caddy on gaining its two electrons that it was originally missing. So that would be our second reactant being the two electrons there. Our next step is to identify which of these half reactions occurs at the cathode or a note of our voltaic cell. Now because we recognize that electrons are being released as a product here. In our first half reaction we would recognize this half reaction as an oxidation. And because it's an oxidation, it occurs at the node of our voltaic cell. And our second half reaction we have electrons that are on our reactant side. And so this would signal to us that this half reaction occurs as a reduction. And recall that our reduction occurs at the cathode of our voltaic cell. Now, next we want to refer to our textbooks to find our standard cell potential for the reduction of our cobalt two plus carry on. And we would see that our standard cell potential is equal to a value of negative 20.28 volts. Where for the oxidation of solid iron we have a standard cell potential in our textbooks equal to a value of negative 0.45 volts. So now going back to our first equation, we can calculate our standard cell potential for our overall reaction given in the prompt, Where we're going to recall that we can calculate that 80° sells the standard cell potential equal to the difference between our standard cell potential of our cathode subtracted from our standard cell potential of our node. And this difference is going to give us our standard cell potential of our overall reaction equal to a value of negative 0.28 subtracted from negative 0.45 which will equal 0.17V. And sorry, so our units of volts should be included as they're given to us in our definitions here from our textbooks. So now we have our standard cell potential for our overall reaction given in the prompt. And now we can go into our nearest equation to calculate our non standard cell potential for the reaction. So going into our first equation will have that are Non standard cell potential is equal to our standard cell potential being .17V which is subtracted from 0.0592V divided by our electrons transferred. Where we will see that in both reactions we have two electrons. So N is equal to two. So we would plug in two in the denominator here. And then this is going to be multiplied by the log of our reaction quotient Q. Where we have our concentration of our Aquarius products. So that would be Our Iron two Plus Catalon concentration divided by our concentration of our Aquarius reactant, is being our cobalt two plus catalon. And this is referring to our overall reaction Above here. And so plugging in our known concentrations which we are given both of these ions according to our prompt. We're going to plug those into our first equation. So what we'll have is SSL equal 2.17V subtracted from. And let's actually take the value of this question here. So when we divide this question out we get a value of 0. volts. This is multiplied by the log of in our numerator. Our concentration of our iron cotton is given as 0.750 moller. Where we're dividing by our concentration of our cobalt two plus Catalan given in the prompt as 0.350 moller. And we don't include those moller units because this is a reaction quotient here. So we just plug them in as numbers. And so simplifying this, we're going to get that our cell potential under nonstandard conditions is equal to 0.17V. And we can actually Yeah 0.17V subtracted from. 0.02 96V multiplied by our log of our quotient is going to yield the results of 0.330993. Next we want to take that product there so we'll have our cell potential under nonstandard conditions is equal to 0.17V subtracted from our product, which will get a result of 0.009797V. And now taking that difference, we're going to have that are electric magnetic force or cell potential under nonstandard conditions is equal to 0.1602, which we will round 2366 to about 0.16V. And this would be our final answer as our electric magnetic force or EMF a k a R cell potential under nonstandard conditions for of our voltaic cell based on our given overall reaction. In the prompt. I hope everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.