A voltaic cell is constructed that uses the following half-cell reactions: Cu+1aq2 + e- ¡ Cu1s2 I21s2 + 2 e- ¡ 2 I-1aq2 The cell is operated at 298 K with 3Cu+4 = 0.25 M and 3I-4 = 0.035 M. (a) Determine E for the cell at these concentrations.

Video transcript

welcome back everyone. The following half cell reactions are used in a certain voltaic cell. We have liquid romaine gaining two electrons to form two moles of our bromide carry on. And this reduction occurs under the cell potential. Standard cell potential of 1.7 volts. Then we have silver carry on gaining one electron to form solid silver. And this reduction occurs with the standard cell potential of 10.80 volts. If the concentrations of bromide and silver caddy on at 2 98 kelvin are 0.15 moller and 0.10 molar respectively, calculate the E M. F. Electric magnetic force generated by the cell at these concentrations. So what we should recognize is that we are not at standard conditions. We want to recall that standard conditions are represented by a self potential or a standard cell potential at 25 degrees Celsius one moller and a ph of seven And a pressure of 1 80 M. So right now because our concentrations are .015 and .10, we know we're not under standard conditions. So we want to recall our first equation Where we take our self potential. That's under nonstandard conditions. Set that equal to our self potential under standard conditions. So that's a degree sell subtracted from 0.059, two. Multiplied by or sorry, the multiplication symbol should be the unit volts actually and then we will divide by our number of electrons transferred, which we recall is represented by N. And so this is then multiplied by the log of our reacting quotient Q. And because we need to use our first equation, we need to figure out our electrons transferred by identifying which of our half reactions here, given in the prompt occur at the anodes or at the cathode. So we want to recall also that the more positive our standard cell potential that will correlate to occurring at the cathode, meaning the reaction is a reduction. And so if we look at our given values for our standard cell potentials of these two half reactions, we see the more positive value associated with reaction one. So we'll say this is more positive standard cell potential. So therefore this reaction occurs as a reduction and this means that this will occur at the cathode of our voltaic cell. And that means that we can label the second one this is a lower self potential value. Standard cell potential value. And so therefore this means that this will occur as an oxidation, meaning that this will occur at the anodes of our voltaic cell. And if we know that the second given half reaction is occurring as an oxidation. That means we should rewrite it so that we have solid silver on the reactant side forming silver catalon as a product as well as one electron as a product. Recall that whenever we have electrons on the product side, we know that our reaction is an oxidation. So now we need to combine these two reactions. But before we can do so, we need to balance out the electrons on both equations. So because we see that we have two electrons in our first half reaction, we need two electrons for the second half reaction. So we're going to take this entire half reaction and we're going to multiply it by two. So this is now going to give us two moles of our solid silver, producing two moles. Let's fix that. Two moles of our silver cat ion Plus two electrons that it's releasing. And then we can add this finally to our first half reaction where we have our one mole of remain liquid, Gaining two electrons to form the product two moles of the bromide Catalon. And so now we can add these two half reactions together to come up with an overall equation here where we will first go ahead and cancel out the matching pairs of electrons on both equations while carrying down everything that's left over. So what we're left with is one mole of our roaming liquid. Then we have two moles of solid silver, both on the reactant side. And then we have our product arrow where we have our first product which is our two moles of bromide and ion Plus our two moles of our silver cat ion. And because we canceled out two electrons from both equations were going to say that n is equal to two. So now that we have this overall equation, we need to find our standard cell potential for this entire reaction. So we're going to recall that we take our standard cell potential degrees cell and we calculate this by taking the standard cell potential of our cathode subtracted from the standard cell potential of our a node. So plugging in what we know from our prompt, we would say that R S L. Is equal to the standard cell potential of our cathode, which is given as 1.7 volts. So we plug that in 1.7 volts subtracted from the cell potential of our and not given as 0.80 volts. So 0.80 volts. And from this difference we would get a standard cell potential for the overall reaction being 0.27 volts. And now to get to our final answer for the prompt, we need to calculate our self potential under nonstandard conditions since we have concentrations under one molar for both of our ions. And so let's go ahead and plug in. What we know, we would say that our standards or are self potential under nonstandard conditions is equal to our self potential Under standard conditions. 8° sell subtracted from. And actually let's go ahead and plug in the 0.27 volts. Since we know that value. So 0.27 volts is subtracted from Our quotient where we have 0.0592V divided by N which is our electrons transferred. Which above we stated is equal to two. And then this is multiplied by the log of our reaction quotient, where we should recall that our reaction quotient Q is represented by a concentration of our products, divided by our concentration of our reactant. And so for our products and reactant we're referring to only our overall equation here. So we would focus on these two products. So let's go ahead and plug in those concentrations which were given to us from the prompt. So we have our cell potential under nonstandard conditions is equal to point 27V subtracted from 0.0296. When we divide our quotation out and then we're multiplying by the log of our concentration of our product where our product is our two moles of our bromide, an eye on here, which is our first product. So we plug that concentration given from the prompt which was 0.015. And we're going to raise this to a power of two. Since we have a coefficient of two in our overall reaction. Next we have a second acquis product. So we're going to multiply it by the concentration of r 0.10 moller of our silver caddy on which we will also raise to a power of two since it has a coefficient of two in the overall reaction. Now we don't include our reactant because if we look at our overall our overall reaction, our reactant have solid and liquid phase. We only want to include the Aquarius face. And so that is why we are disregarding reactant and only including our products here. So now to simplify this calculation, we're going to get zero point and sorry about that. So 0.27 volts subtracted from 0.296. So let's make that six a bit neater. This is multiplied by the log and we're going to focus on taking the product of these two quantities here squared before doing the log operation. And so when doing so we're going to get a value of 2.25 times 10 to the negative six power. And so now we can take the log there where in our next line will have 0.27 volts subtracted from 0. multiplied by, we'll have negative 5.64782 as the result of that long term. And now, following our order of operations, we're going to do the multiplication here between these two terms which will now give us zero point and sorry, let's keep our pen color consistent. So 0.27 volts subtracted from our product here which is now -0.167175. And now taking the difference between .27V and this value here, we get a result of 0.437 which we can round to about 0.44V. And this would be our final answer for our self potential under nonstandard conditions. A k a. R. Electric magnetic force based on our concentrations of our products being, are two moles of our bromide and two moles of our silver cat lion, so it's highlighted in yellow is our final answer. I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.