Determine the optimum mass ratio of Zn to MnO2 in an alka- line battery.

Video transcript

Hey everyone in this example, we're told that for a nickel cadmium battery to calculate the optimal mass ratio of cadmium to nickel oxide hydroxide. So what we should first recognize is that because this is a redox reaction in a discharging nickel cadmium battery, we're going to need to write out half reactions. So right now, what we would understand is that we have cadmium reacting with nickel oxide hydroxide and we would predict that we have cadmium hydroxide as a product as well as nickel hydroxide. However, like I said, we need to go ahead and write out our half reactions. So for our first half reaction, we would have cadmium producing our cadmium hydroxide. And what we should recognize is that we have because cadmium is in its elemental form. It has an oxidation number of zero. And then on the product side we don't know the oxidation number of cadmium since it is a transition metal in a compound here. So we would just place X. But we should recall that oxygen will have an oxidation number of -2. Whereas hydrogen we recall has an oxidation number of plus one. Now we have the subscript of two here. So we need to multiply that by our oxidation numbers And what we would have is X - Plus two is equal to zero since our product here is a neutral compound. And so solving for X, we would have X is equal to positive too. And so that would be the oxidation number for X. Or cadmium. So we can say that this compound cadmium hydroxide has an oxidation number of plus two here. Now. Our next step is to balance each of our atoms on both sides of the equation. We have one atom of cadmium on both sides, so that's taken care of. So now our next step is to balance out our oxygen. So we recognize that in our product cadmium hydroxide, we have two oxygen's because we have that coefficient of two. So we would balance this out by adding water to our react inside here. So we would have cadmium plus we would have two moles of water added because we need to have two moles of oxygen to balance out our oxygen producing our cadmium hydroxide. Sorry. So now we have our oxygen's balanced. However, now we need to focus on balancing our hydrogen. We would see that we have two moles of hydrogen on the product side. And here we have four moles of hydrogen. So we would balance out our hydrogen by using H plus, so we would have cadmium plus two moles of water yields cadmium hydroxide plus two moles of h plus, however, adding these two moles of H plus would create water on our product side. So what we really have now is cadmium plus two moles of water yields cadmium hydroxide Plus two moles of water. So we want to get rid of this excess water and so to do. So we are going to go ahead and add two Molds of hydroxide so that we are able to cancel out our two moles of water on both sides of our equation. So this is going to give us now cadmium plus two moles of hydroxide yields our cadmium hydroxide. However, we would see that because we now have two moles of hydroxide on the reactant side, we have a net charge of -2 -2 Because we have that coefficient of two there. So we have a net charge of -2 on the reactant side. And then on our product side we have this neutral compound. So we need to balance the charges by adding two electrons here. So we would say plus two electrons to bounce out the charge. So we have a -2 charge over here now and now we're going to complete this and move on to our second half reaction. So I'm just going to create some more room here. So, for our second half reaction, we're going to be starting out with our nickel oxide hydroxide which produces our nickel hydroxide on the product side. Now, as before, we don't know the oxidation number of nickel here in this compound because it is a transition metal, meaning that we would label it as X. We would recall that and we'll actually use this in red. So we would recall that oxygen has an oxidation number of negative two. So that would count for both of our oxygen's and then hydrogen. We recall has an oxidation of plus one and so we would say that overall we have for this compound a charge of plus three when we solve for our oxidation number of of nickel being X. And this would be our oxidation number for our compound plus three. Now for our product we would follow the same steps and we would get that X as equal to an oxidation number of plus two for our compound here. And because we see that we went from a plus three two a plus two charge for our product here, we would agree that this is going to be the reduction half reaction. Whereas for our first half reaction, we would agree that we have a net charge of plus two for our product here. So this was our oxidation half reaction where we lost electrons. So now moving forward with balancing our half reaction with nickel oxide hydroxide. We would see that our nickels are already balanced. There's one atom of each on both sides of our equation. So now we're going to first balance our oxygen's. So we recognize that we have two oxygen's on our product side and we have two oxygen's on the reacting side. So our oxygen's are already balanced as well as our nickel. However, we can balance our hydrogen out because we have two on the product side and just one here on the react inside. So we're going to bounce out hydrogen. We recall by using our proton H plus. So what we should have is nickel oxide hydroxide plus age plus yields nickel hydroxide. And we should recall that when we add H plus we're really creating water. So now we have nickel oxide hydroxide Plus H 20. Is going to yield nickel hydroxide and we would go ahead and cancel out this excess water. And our H plus by adding one mole of hydroxide. And now we've created a net charge in adding this hydroxide. We've created a net charge of -1 on our product side. However, we have a charge of zero on our oxygen. Or sorry, or react inside. So we would go ahead and add an electron here to our reactant side. So now we would have nickel oxide hydroxide plus water plus one electron. And this will give us balance charges for our product side where we have again that -1 charge coming from that hydroxide. Now we need to combine our half reactions. However, before we do so, we need to make sure that the electrons match for both reactions. Now, for our first half reaction, we have two electrons on the product side. And for our second half reaction we have one electron on the reactant side. And this means we would need to multiply this entire equation by two to get two electrons for this second half reaction. And this would give us a new equation where we have two moles of nickel oxide hydroxide Plus two moles of water Plus two electrons yields two moles of nickel hydroxide Plus two moles of hydroxide. And so we're going to take these two half reactions. So we have Cadmium plus two moles of hydroxide Yields one mole of cadmium hydroxide plus our two electrons from that half reaction added to our second half reaction. Two moles of nickel oxide hydroxide and I'll just make more room here, Nickel oxide hydroxide plus two moles of water Plus the two electrons yields two moles of nickel hydroxide Plus two Mnolds of Hydroxide. So we're going to cancel out the electrons because they're on opposite sides. We have it on the reactant side here and on the product side here. And we can also cancel out the hydroxide because they're also on opposite sides in both half reactions. And this is going to give us an overall reaction for our balanced redox reaction, which equals cadmium plus two molds of our nickel oxide hydroxide, Plus two moles of water yields cadmium hydroxide plus two moles of our nickel hydroxide. And now that we have this full redox reaction, we can accurately calculate our mass ratio of cadmium to nickel oxide hydroxide. And looking at the ratio between cadmium to nickel oxide hydroxide. According to our equation above, We would say that we have a coefficient of one in front of cadmium and a coefficient of two in front of nickel oxide hydroxide. So this would be a 1 to 2 molar ratio. So we're going to start off our calculation by having our one mole of cadmium To our two moles of Nickel oxide hydroxide. And then from our periodic tables, we are going to go ahead and get the molar mass of cadmium. And we would see that for one mole of cadmium, We have a molar mass from our periodic tables of 112. g of cadmium. So this would allow us to cancel out moles of cadmium. And now we want to cancel out moles of nickel oxide hydroxide. So we're going to go ahead and multiply by our next conversion factor. Where from our periodic tables, we see That for one mole of nickel oxide hydroxide, We have a molar mass of 91.7 g of nickel oxide hydroxide. So this allows us to cancel out moles of nickel oxide hydroxide, leaving us with grams. And so for our mass ratio of cadmium to nickel oxide hydroxide, we're going to get a value from our calculators, a value of 0.613 g of cadmium to our grams of or we can say per our grams of nickel oxide hydroxide. And so this here would be our final answer of our mass ratio of cadmium to nickel oxide hydroxide from this equation. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video