A voltaic cell consists of a PbPb half-cell and a CuCu half- ce...

Question

A voltaic cell consists of a Pb>Pb half-cell and a Cu>Cu half- cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. c. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35 V?

Accepted Answer

Welcome back everyone. We're told that a voltaic cell initially consists of a 0.0 to 50 moller magnesium, magnesium two plus Catalan half cell and a 1.32 molars. Silver, silver caddy on half cell. We need to calculate the cell potential when the silver Catalan has already dropped to 3.18 volts. Our first step is to write out our half reactions. So for our first half reaction we have the reduction of our silver caddy on which forms solid silver by gaining one electron. Our second half reaction is to show the oxidation of solid magnesium which is formed from the magnesium two plus Catalan gaining an electron. Now for each or for each of our half reactions, we need to make sure that the atoms are balanced. In our first half reaction, we have one atom of silver on the react inside and one atom of silver on the product side. So the atoms are balanced. Now looking at our second half reaction, we also have one atom of magnesium on the reactant side and one atom of magnesium on the product side. However, the next step is to make sure the charges are balanced. So for the second half reaction we have a net charge of plus two on the product. Whereas on the reacting side we don't have a net charge at all. We have a neutral charge. So we need to go ahead and cancel out this net charge of plus two on the product side by adding two electrons. So we would place a coefficient of two in front of our electrons, meaning that we would actually have two electrons being added now. And this is going to cancel out that net charge of plus two. And because we added two electrons to the second half reaction, we need to in order to add these two reactions together, we need to make sure that the electrons are matching in the first half reaction. So we're going to also take this first half reaction and multiply it by two so that we also have two electrons here And so what this means is that now we would have to Silver Catalans gaining two electrons to form two moles of solid silver. And then for our second half reaction we still have one mole of magnesium solid formed from a magnesium carry on gaining two electrons. And so now that we have electrons and atoms balanced here, we can go ahead and add up these two half reactions. So we should recognize that we have two electrons on the reactant side in the first half reaction, we can cancel them out with the two electrons on the product side in the second half reaction. And so what we would get when we add these two reactions together is that two moles of our silver caddy on plus one mole of solid magnesium produces two moles of solid silver and one mole of magnesium cat ion. Now the next thing we need to make note of is our reduction and oxidation reactions. So for our first half reaction we have electrons added to the reactive side. And we should recall that this tells us that this first reaction occurs as a reduction and we should recall that our reduction half reaction occurs at the cathode of our half cell here or the catholic half cell. And then for the second half reaction we have electrons added to the product side. Whenever we have electrons added to the product side, we should recall that this means that this half reaction occurs as an oxidation, which we should recall occurs at the anodes half. So So now we can go ahead and make note of our standard cell potential reduction table which is online or in our textbooks. And for our first half reaction, we would see for the reduction of the silver carry on, we have a cell potential value equal to .80V. And then for the oxidation of solid magnesium, we would see that we have a self potential value of negative 2.37 volts. And now that we have these values noted down, we can calculate our standard cell potential E degrees cell and this is going to equal the cell potential of our cathode subtracted from the cell potential of our node. And so what we would say is that for a cell potential of our cathode, we determined that that is 0.80 volts from our textbooks Subtracted from the cell potential over a note, which from our textbooks is negative 2.37V. And so this gives us a standard cell potential equal to a value of 3.17V. So now we have our standard cell potential and now we need to recall our Nerdist equation. So we'll write that below in black here which we should recall is taking our standard cell potential and setting that equal to our or sorry, this is just a cell potential from the prompt. And that is set equal to our standard cell potential E degree cell which is then subtracted from our nancy equation constant 0.592 divided by our electrons transferred N. And then this is going to be multiplied by the log of our reaction quotient. Q. So we want to solve for Q. So plugging in what we know according to the prompt. Our cell potential drops to 3.18 volts. Whereas our standard cell potential above, we calculated as 3.17 volts. This is then subtracted from our quotient where we take our Equation constant 0.0592 divided by N. Which is our electrons transferred. And we could have made note of it over here where we said N is equal to two because we have two electrons that were transferred in both of our reactions. So we would plug in two for N. And then this is then multiplied by the log of Q. Our reaction quotient. So to simplify this we would say that the log of Q. Is equal to negative 0.33784. And we would be able to sulfur Q. By making each side and exponents. And so we would be able to cancel the log term. And so we would say that Q. is equal to 10 to the power here. Where in our calculators we would get that our reaction quotient Q should equal a value of 0.4594. And we should recall that our reaction quotient Q. Is describing the ratio between our ions in our reaction. And so in our numerator we want the ratio of the ion products. So that would be our carry on being the magnesium two plus carry on since it was one of our ion products. And then in our denominator we will plug in the ion reactant which was just our silver canyon where we have a coefficient of two in front of that canyon. And so we would raise it to an exponent of two in our reaction quotient. Now, according to the prompt above we have a concentration of our magnesium two plus catalon equal to 0. to 50 moller. And actually that is for the half cell that it's in because according to the prompt that's for the half cell of magnesium and magnesium two plus carry on. And so we would say that this would be that the concentration of magnesium two plus carry on would be this concentration plus X. Because we need to find the concentration of our magnesium two plus Catalan and it's a product. So we would say plus X. Since it only has a coefficient of one. Now we would take the same process to find the concentration of our silver Catalan And we would say that this is going to equal according to the prompt. We have a half cell concentration of 1.32 molar. So we would say that that is equal to 1.32 moller. And because the silver carry on is a reactant we would say minus and it has a coefficient of two times X. So this is how we will solve for the concentration of each of our ions. And so we're going to go back to our reaction quotient value for Q. Above which we set is 0.4594. And we're going to set that equal to the following equation where we take in our numerator the Equation for solving for the concentration of Magnesium Catalan, so 0.0250 plus x. And then this is in our denominator divided by the Algebra for finding our concentration of Silver Catalan. So that's 1.32 -2 x. And because we know that this should be squared, we're going to square this in the denominator. So in our calculators, solving for X, we can say that X is going to equal 0.264. And now that we have our X values, we can say that therefore our concentration of magnesium Catalan is equal to 0.0 to 50 moller plus our X value of 0.264. And this is going to give us our concentration of magnesium kata and equal to 0.2 89 moller. So this would be our first answer for this example. Our concentration of magnesium two plus carry on. And then for our second final answer, we have our concentration of silver caddy on which is equal to 1. Moeller minus two times our X value of 0.2 which is going to give us a concentration of 0. to moller. And so this would be our second final answer to complete this example. So we've solved for our concentration of our ions and what's highlighted in yellow is our final answers. To complete this example. If you have any questions, leave them down below and I will see everyone in the next practice video

A voltaic cell consists of a Pb>Pb half-cell and a Cu>Cu half- cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. c. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35 V?

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