In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries. The overall cell reaction for this relatively new battery is: 2 H2O1l2 + 2 NiO1OH21s2 + Zn1s2 ¡ 2 Ni1OH221s2 + Zn1OH221s2 (b)What is the anode half-reaction?

Video transcript

welcome back everyone, nickel iron batteries have less impact on the environment because nickel iron batteries don't contain cadmium, which is considered a hazardous material. The overall cell reaction for a nickel iron cell is two moles of nickel oxide hydroxide solid reacting with two moles of liquid water and one mole of iron solid to produce two moles of nickel to hydroxide solid and one mole of iron to hydroxide solid provide the an ode half reaction. So we need to first assume standard conditions because we're not given conditions for our reaction. So we're called that that means we have a temperature of 25 degrees Celsius, a pressure of 1 80 M APH equal to about seven. And we have concentrations of one molar. So next we want to write out our half reactions based on our overall reaction. The first beginning with our first solid reactant, which we do not carry coefficient. So we have just nickel oxide hydroxide which produces nickel to hydroxide as a product, which is a solid. Our second half reaction is our second solid reactant being our solid iron. And sorry, this is an S. Here which produces as a product, our iron to hydroxide solid. Our next step is to balance out the atoms in both half reactions. Beginning with the first half reaction, we recognize that on our product side we have two moles of oxygen and hydroxide, whereas on the reacting side we have just two moles of oxygen and one mole of hydrogen. So we want to recall that we balance hydrogen using hydride and so on our react inside, we're going to expand it by adding one mole of hydride to the reactant side. Now moving on to our second half reaction, we recognize that we have two moles of oxygen and hydrogen on the product side, but none on the reactant side. So now we're going to recall that we balance oxygen using water and so we're going to expand our reactant side here by adding two moles of water. This is going to now give us two moles of oxygen and now four moles of hydrogen where we need to fix that on the product side because we now only have two moles of hydrogen. So we're going to add two moles of hydride to the product side, which will now give us four moles of hydride to the product side, and four on the reactant side bouncing out our atoms. We see that iron and nickel are balanced in both half reactions. So we don't need to worry about those atoms. Now we need to balance out the charges. So we recognize in our first half reaction that we have an overall charge of plus one on the reactant side and an overall charge of zero on the product side. So we need to cancel out the charge on the more positive side, which is our reacting side by adding one electron. Next we look at our second half reaction and we see that we have a net charge of zero on the reactive side. But a net charge of plus two on the product side from that coefficient of two, meaning we're going to cancel out that plus two net charge by adding two electrons so that we have a neutral charge on both sides of our reaction. Next we want to recognize that because we have hydride present in both of our half reactions, we need to add hydroxide. So we always add hydroxide when we have hydride, meaning we're going to add one mole of hydroxide to the first half reaction and two moles of hydroxide to the second half reaction based on the amount of hydride we have in each reaction. And in doing so for our first half reaction, we're now going to have nickel oxide hydroxide solid. When we have hydroxide added to hydride, this will form water. So we would form one mole of water plus our single electron to produce our nickel two hydroxide product for our second half reaction, when we add those two moles of hydroxide, we're now going to have And sorry, just to be clear. So before we go too far ahead, I just want to correct a mistake here. So because we added hydroxide because of the presence of hydride in both of these half reactions, we need to add hydroxide to both sides of our equation so that things balance out. So we're going to add hydroxide. Two moles of hydroxide here for the second half reaction and we're also going to add one mole of hydroxide to the first half reaction. Since we added just one on the reactive side here. And so now what we will have for our first half reaction, it's not going to change to nickel oxide hydroxide solid. Again, when we combine hydride with hydroxide, we form water so we'll form one mole of water since we only added one mole of hydroxide to our one mole of hydride plus our single electron on our product side, we're going to have nickel to hydroxide Solid plus are one mole of hydroxide. Moving on to our second half reaction, adding those two moles of hydroxide to both sides of our half reaction, we're now going to come up with one mole of solid iron And let's make this clear. So one mole of solid iron plus, we would have two moles of water Plus two moles of hydroxide on the reactant side forms are one mole of iron, two hydroxide solid Plus when we again combine two moles of hydride, +22 moles of hydroxide, we would form two moles of water plus our two electrons. And so now we would recognize that we have two moles of water on both sides of this second half reaction, meaning we can now cancel it out on the reactant and product side. And so the second half reaction is going to simplify to iron solid Plus two moles of hydroxide. And this should have had a queue here produces one mole of iron to hydroxide plus our two electrons as a product. And so now we need to determine which of these half reactions, which we're comparing between this first half reaction, which is now fully balanced versus the second half reaction, which is also now fully balanced. We want to decide between these two circled half reactions, which is are an ode half reaction. And we would recognize that because we have electrons on the react inside here. This tells us that this reaction occurs as a reduction, which we would recall occurs at the cathode of our voltaic cell and because we see electrons are added on the product side here of the second half reaction, this reaction occurs as an oxidation, which we would recall occurs at the anodes of our voltaic cell, meaning for our final answer, it would be this half reaction that we have written out fully balanced as are an ode half reaction. This would be our final answer. To complete this example. I hope 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