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. a. What is the initial cell potential?

Video transcript

welcome everyone. We have to calculate the initial cell potential of a voltaic cell consisting of a 0.25 moller magnesium magnesium carryin half cell and a 1. moller silver, silver carry on half cell. So our first step is to write out our half reactions first showing the formation of the silver solid compound. We have our silver carryin which gains an electron to form solid silver as a product. And then second our second half reaction is showing the oxidation of solid magnesium where we have our magnesium carry on which gains an electron. And so our next step is to make sure that our charges are balanced in the second half reaction notice we have a net charge of plus two on the product side, whereas on the reaction side we have a net neutral charge. So we want to cancel out this net charge of plus two on the product side. And we would do so by adding two electrons. So we would place a coefficient of two in front of our electrons, meaning we would now add two electrons. And our next step is to add these next two is to add these two half reactions together. So we would do so by making sure the electrons are matching in each of our reactions. And in our first half reaction we only have one electron. So we will have to take this entire first half reaction and multiply it by two. And so what this means is that now we would have two moles of silver Catalan plus two electrons producing two moles of solid silver. And then for our second reaction, we still have one mole of solid magnesium which produces magnesium catalon and two electrons. And so now we can add these two equations together noticing that we can cancel out our two electrons on the reacting side as well as the two electrons on the product side here. And then for our final reaction, we're going to have two moles of silver Catalan plus one mole of solid magnesium produces two moles of solid silver plus one mole of magnesium catalon. Now we also want to make note for each of our half reactions which one occurs as an oxidation and which one occurs as a reduction. So recognize that because we have an electron added on the reactant side of our first half reaction. This will occur as a reduction and recall that our reduction half reaction occurs as a or at the cathode half cell, whereas because in our second half reaction, we have electrons added on the product side. We recognize that this occurs as an oxidation. And recall that oxidation occurs at the an ode half cell. Now we want to refer to the standard cell potential reduction table in our textbooks or online and we would see that for our first half reaction, we have a cell potential value for the reduction of silver carry on equal to a value of 0.80 volts and then our cell potential for the second half reaction for the oxidation of solid magnesium is equal to a value of negative 2. volts. And so now we're able to calculate our standard cell potential which is the degree cell and this is equal to we recall our self potential of our cathode subtracted from our cell potential of our A node. And so we would say that this would equal ourself potential of our cathode which above we stated is 0.80 volts subtracted from our self potential of our node which is above as negative 2.37 volts. And so this gives us a standard cell potential equal to a value of 3. volts. Our next step is to determine our reaction quotient Q. And so we would recall that Q. Is going to equal the concentration of our ion products minus the concentration of our ion react ints. And so we would say that that is equal to in our numerator are ion products according to our reaction is one mole of magnesium Catalan. So we have our concentration of magnesium catalon. And then in our denominator our concentration of ion reactant are two moles of silver carry on. So we would say our concentration of silver carry on and we would raise this to an exponent of two since we have that coefficient of two there. So now we would go ahead and plug in the info from the prompt. So according to the prompt for our magnesium carryin have cell we have a concentration of 0.0-50 moller. And then in the denominator for our concentration of our Silver Caddy on half cell. We have a concentration of 1.32 molar which is then squared. And so we would say that our reaction quotient is equal to a value. We're going to plug this quotient into our calculators and we're going to get a value of and I'll just make more room here. So our reaction quotient should equal 0.01435. And so now that we have our value for Q. We want to go into our next equation, recall that are nursed equation is taking our cell potential and setting that equal to our standard cell potential. And this is subtracted from the nerds equation constant. 0.592 volts divided by the electrons transferred. Where that is a value of N. And we would say above that are electrons transferred is unequal to two because as you can see, we canceled two electrons that were transferred in each of our half reactions. And then next we want to recall that this is then multiplied by the log of our reaction quotient Q. And so to solve for our initial cell potential which is what the prompt is asking us to solve for. We want to solve for T cell here. And so we would say Below plugging in what we know that the cell is equal to our standard cell potential, which above we calculated as 3.17V subtracted from our nearest equation constant. 0.0592V divided by the electrons transferred. Which above we stated is two electrons Multiplied by the log of our reaction quotient, which above we determined is 0.01435. And so solving in our calculators, we would get that our initial cell potential e cell is equal to a value of 3.22V. And so this would be our final answer to complete this example as the initial cell potential of our cell. So I hope that everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.