Consider the following voltaic cell:

(c) What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10?

Video transcript

Welcome back everyone calculate the change in cell voltage when the ion concentrations in the half cell of the cathode is increased by a factor of 100 for the following voltaic cell. So what we should first recognize is that because the prompt mentioned that this is a voltaic cell, we're going to recall that our reaction should be spontaneous, meaning that our reduction potential is going to be positive. And we want to recall that for our redox reactions. The half reaction with a more positive reduction potential occurs at our cathode of our voltaic cell. So let's look at this cell, we can see that Beginning with our cadmium two plus caddy on solution here. We have our reaction which produces our solid cadmium metal which is defined by this battery here. And so we're going to write this out as our first half reaction. So we have cadmium two plus this is an ion, so it's Aquarius. And to form our solid cadmium metal, we need to gain those two missing electrons based on that caddy on charge, which we recall means we are missing electrons. So we would gain two electrons here to form our solid stable cadmium metal here. And in our textbooks we're going to see that this reduction of cadmium canyon has a reduction potential value of negative 0.40. And because this is a negative reduction potential value, we see that it's not positive and is therefore occurring at the anodes. So this would be a reaction at the anodes, meaning that we would assume that our copper cat ion reaction to form solid copper is occurring at our cathode. And so writing that half reaction out, we have our copper caddy on to form solid copper metal. We'll have to gain it's one missing electron to form our solid copper. And this reduction has a reduction potential of .52. So this is a positive reduction potential value confirming that this reaction occurs at our cathode of our voltaic cell. Now recall that when we have half reactions we need to come up with an overall reaction. So we need to first ensure that we have a balanced number of electrons for both of these reactions. So to bounce out our electrons here We're going to go ahead and multiply this second reaction by a factor of two which is going to give us two copper cat ions two electrons and two solid copper metal, meaning that we can now add up our reactions which will result in us being able to cancel out those two electrons leaving us with everything else. So we have two of our copper cat ions which react with our one mole of cadmium caddy on To produce on our product side, two moles of solid copper metal and one mole of solid cadmium metal. And actually to be even more clear, we want to recognize that since our reaction with cadmium acadian occurs at the anodes of our voltaic cell. We're going to rewrite it so that we begin with our cadmium metal as our reactant. So we have let's say we have C. D. Medal which produces the products cadmium two plus and Releases two electrons. Which is why this is our anote half reaction occurring as an oxidation because we have electrons released on the product side. And so again canceling things out. We can get rid of the two electrons and rewriting our overall equation. So it's more accurate. We're going to have our two copper cat ions plus our cadmium metal which produces on our product side are solid copper metal plus our cadmium two plus catalon. And based on our electrons that we balanced out, we can say that our electrons transferred N. Is equal to two. And now to calculate our change in cell voltage we want to find our standard cell potential first. So we're going to recall our nerds equation. We're using the equation. We recall that we calculate our standard cell potential by taking our cell potentials that we calculated subtracted from 0.592 divided by our electrons transferred N. Which is two. And then this is going to be multiplied by the log of our concentration of our products over our concentration of our reactant. Raised to any corresponding coefficients X and Y. So plugging in what we know we can say that our standard cell potential is equal to our cell potential subtracted from 0.592 divided by N. Where N. We can plug in as two now multiplied by the log of our concentration of our products. And this is only going to include our acquis substances. So we'll plug in our concentration of cadmium two plus caddy on which has a coefficient of one divided by our concentration of our reactant, which are only a crease reactant. Are are two moles of our copper canyon which have a coefficient of two. Now going back to our prompt, we see that the concentrations of our ion in the half cell of the cathode is increased by a factor of 100. So as we stated, our ion concentration of our cathode is going to be associated with our ion concentration of our copper cat ion. So writing this out in notes, will say if concentration of copper is increased By 100, we'll rewrite our equation so that it says that our standard cell potential is equal to our self potential. Or sorry, standard reduction potential is equal to our reduction potential. We're actually we want to go ahead and subtract these two. So we're going to subtract our two reduction potentials. We're setting them equal to negative one times 0.592 divided by two, Multiplied by the log of our concentration of our cadmium caddy on which will just plug in as a concentration of one Divided by our concentration of our copper caddy on which we are increasing by a factor of 100. And so we'll plug in 100 as our concentration raised to the coefficient of two as the power. And so simplifying this, we can say that the difference between our reduction potentials is going to result in a voltage of .118V. And just to be clear, that is why we formatted our reduction potentials as being subtracted from one another and said our equation In this way here because we need to again calculate our change in cell voltage. And so our final answer is going to be that the .118V is our change in cell voltage. So I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.