A battery relies on the oxidation of magnesium and the reduction ...

Question

A battery relies on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0 * 10-4 M and 1.5 M, respectively, in 1.0-liter half-cells.
b. What is the voltage of the battery after delivering 5.0 A for 8.0 h?

Accepted Answer

everyone in this example, we're told that the reduction of lead and the oxidation of calcium creates a theoretical battery. If there are 0.55 molar of lead two plus and 20. moller of calcium two plus in a one liter half cell. Initially calculate the voltage of the battery after giving three ampules for 4.5 hours. So our first step is to write out our half reactions we have for our first Catalan lead two plus, which should form sorry, which should form solid lead by adding two electrons here and then for our second Catherine, we have calcium two plus which forms our solid calcium. When we gain two electrons. Now we need to go ahead and look up our standard potential value self potential values for these two oxidation reactions. So for our oxidation of solid lead, we would see that on this table which we would find either online or in our textbooks that we have a self potential value equal to negative 0.13 volts. And then for our self potential value for the oxidation of solid calcium, We would see that we have a self potential value equal to negative 2.76V. Now we need to figure out which of these cat ions occur at the cathode and an ode of our cell. So we're going to write out the full reaction where we have Our lead two plus reacting with solid calcium to produce solid lead and the calcium two plus cat. I own Now we should recognize that our lead went from a two plus charge to a neutral charge on the product side, meaning we gained electrons. And so therefore we have a reduction which we would recall occurs at the cathode of our cell. And then in comparison for our calcium, we went from a net charge of zero on the reactant side to a net charge of plus two on the product side, meaning we lost electrons, which would mean we went through an oxidation. And we would recall that an oxidation occurs at the anodes of our cell. And so now we can go ahead and calculate our standard cell potential, which we recall is equal to the cell potential of our cathode, minus the cell potential of our anodes. And so what we would say is that plugging in our values from above our self potential of our cathode which we determined correspond to the oxidation of our lead, Has a value of -0.13V. And this is subtracted from the cell potential of our anodes which we determined occurs at the oxidation of our solid calcium And has a self potential value of -2.76V. And so to get our self potential, what we should get here is a value equal to positive 2.63 faults. Our next step is to calculate our concentration of our lead two plus caddy on as well as our concentration of our calcium two plus caddy on. After the 4.5 hours of the 3.0 ampule charge is added to our reaction. So we're going to go ahead and recall. So I'll just make some more room here actually. So we're going to recall that to get the concentration. After time we first need to calculate the moles of lead that is oxidized. And so we would find that by taking our time from the prompt as 4.5 hours And we want to cancel our hours. So we should recall that in one hour. We have 60 seconds. So we can cancel out ours. Now we're going to move from seconds and correction that should be 60 minutes. So we're going to convert from minutes, one minute, two seconds where we recall that in one minute we have 60 seconds. So now we're able to cancel out minutes and now we're going to convert from seconds, two columns. Because we recognize that according to the prompt, we're told that we have three point oh ampules and we should recall that one. Ampule is equivalent to One column per second. And so we can interpret that three ampules as a conversion factor. Where we would say that for one second we have 3. columns. So now we're able to cancel out seconds. And now from columns, we're going to convert from Columns, two moles of our electron. And so we should recall Faraday's constant, which is an F symbol. And we recall that that has a value of 96, columns for Permal of electron, so for one mole of an electron. And so for our last conversion factor after we've canceled out columns, we're going to move from moles of electrons two moles of our lead. And from above we would agree that we transferred two electrons to form our solid lead as a product. And so we would say that for two moles of electrons transferred we have one mole of our lead two plus caddy on. So now we're able to cancel out moles of electrons and we're left with moles of our lead two plus catalon. And so for our moles of lead oxidized, we're going to get a value equal to 0.2519 moles of the lead two plus that is oxidized. And so to see how much of our concentration we have of lead two plus at the amount of time that passes according to the prompt, we're going to take our initial concentration of lead two plus and we're going to subtract that from our moles of lead two plus that is reduced or oxidized, sorry, reduced. So we should actually correct this to oxidize because above we stated that this is reduced and this is for one liter of solution. And so what we would calculate is according to the prompt, we have an initial concentration of lead two plus as 20. molar. And we're going to subtract this from our lead two plus that is reduced that we calculated above as 0. moles of lead that is reduced and this is for one liter of solution. So what we're going to get here is a value equal to 0.2981 Molar. And this would be our concentration of lead two plus after 4.5 hours according to the prompt. So we're going to take the same steps and apply this to our calcium. So right now we need to first calculate our moles of calcium that is oxidized. And for another correction here, this should say reduced for lead. So going back to our moles of calcium oxidized. We're going to calculate that by taking our moles of lead that was reduced, which we determined above as 0.2159 moles of lead. And we're going to multiply this by a conversion factor. Where we're going to go from moles of lead two moles of calcium that is oxidized. So we can be more specific and say moles of lead that is reduced to moles of calcium that is oxidized. And from our equation above, we would see that for one mole of lead reduced, we have one mole of calcium that is oxidized because we have coefficients of one in our equation here, in front of calcium and in front of our lead as well as in front of our ions here, we have coefficients of one in front of everything. So we use that as a conversion factor where now we can cancel out moles of lead and we're left with moles of our calcium that is oxidized. Sorry. So this is going to give us a value equal to zero point 2519 moles of our calcium that is oxidized. And so now we can go ahead and calculate our concentration of the calcium two plus Catalan after the 35 hours. And that is going to equal our concentration of calcium two plus initially according to the prompt added to our moles of calcium that is oxidized per liter of solution. And so what we would get is from the prompt, we have an initial concentration of calcium two plus as 0.15 moller which we will now add to the amount of calcium two plus oxidized, Which we above stated is equal to 0.2519 moles of calcium two plus per leader of solution. And what we're going to get here is a value equal to. And I'll just actually write this below because we're running short on room so we would get a value equal to 0.2669 Moller. And so now that we have these concentrations after the amount of time we can go ahead and continue on our solution to calculate our standard cell potential. So we should recall that our standard cell potential is equal to and rather this is for the cell potential of our battery. So this is going to equal our self potential. Our standard cell potential subtracted from our new ERnst equation constant which has a value of 0.592 volts multiplied by N. Which is our electrons transferred at our reduction. Multiplied by the log of our equilibrium constant. Q. Where we would recognize that Q. is representative of our concentration of the calcium two plus Catalan Divided by the concentration of our lead two plus Catalon. And so in outlining these facts, we can say that our self potential or our battery ourself potential of our battery rather which would give us our voltage is equal to Our standard cell potential which we calculated earlier on in our solution and we said that that was equal to a value of 2.63V. And this is subtracted from our new Ernst equation constant which again we state it has a value of 0.0592V. We would divide this by two of our electrons that are transferred at a reduction. So this part of our equation should actually be corrected two divided by two. And we would still multiply by the log of Q. And so just to make more room here We're gonna go ahead and multiply this by the log of Q. Where we have in our numerator our concentration of our calcium two plus carryin. Which above we stated is equal to 0.2669 moller. Where in our denominator we have the concentration of the lead two plus Catalan which above we calculated as 0.2981 Moeller. So we would cancel out our units of polarity and we're left with volts units in our numerator. And what we're going to get is that our voltage of our battery When we add this or when we apply the math in our calculators, we're going to get a value equal to 2.631 volts. And so this here would be our final answer as the voltage of our battery. After 35 hours when we have a charge of 3. ampules given off from the battery. So what's highlighted in yellow represents our final answer. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

A battery relies on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0 * 10-4 M and 1.5 M, respectively, in 1.0-liter half-cells. b. What is the voltage of the battery after delivering 5.0 A for 8.0 h?

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