Write unbalanced oxidation and reduction half-reactions for the following processes. . (a) Te(s) + NO₃^-(aq) → TeO₂(s) + NO(g)

Video transcript

Welcome back everyone. What are the unbalanced reduction and oxidation half reactions for manganese carry on reacting with chlorine dioxide gas to produce manganese dioxide solid and our chloride an ion. So let's begin by writing out our two half reactions. Beginning with our re agents, we have our manganese to Pluskat Ian which should form as a product manganese dioxide solid. And for our second half reaction we have our chlorine dioxide gas which forms as a product our chloride an ion. Our first step is to focus on our first half reaction, where we need to determine whether it's a oxidation or reduction happening. And so we're going to focus on the oxidation number of manganese on our reactant and product side on our reactant side, it's going to equal the ion charge. So we would have an oxidation state of plus two on the reactant side. Whereas on the product side, we need to solve for the oxidation state of manganese. So for the oxidation state of Manganese here, we don't know it. We're going to solve for it by recalling that our oxidation number of oxygen is -2. So just to be clear, oxidation state is minus two for oxygen. So two sulfur oxidation state of manganese, we would have X plus Our subscript of two, multiplied by the oxidation state of oxygen being - and this is going to equal zero since the overall compound here is neutral. So we're going to sulfur X and say that X is going to equal positive four. And so therefore our oxidation state of manganese on the product side is equal to plus four. And so what we can observe is that our oxidation state went from plus two, two plus four, citing an increased oxidation number. And because our oxidation state of manganese increases, we would say therefore the reaction is an oxidation. So now we want to follow the same steps to determine whether the second half reaction is a reduction or oxidation. So looking at the oxidation state on our reactant side for chlorine, we need to solve for that. So we'll put a question mark here and we know our oxidation state for oxygen is going to be minus two. So solving in the same way, we would have X plus R. Subscript of two on oxygen times its oxidation state being minus two Equal to zero. Since the overall compound of chlorine dioxide is neutral. Then solving for X, we would have X is equal to positive for And so our oxidation state of chlorine is going to be equal to positive for on the reactant side on the product side. We need to solve again for the oxidation state of chlorine. We don't know it. So writing out an equation, we would have X plus R. Subscript of two on oxygen times the oxidation state of oxygen being minus two equal to our overall charge of our product. Uh chlorine, sorry, chloride, which is a minus one charge. So our equation is equal to negative one until solving for X. We would find that X is equal to negative four -1. So that we can say X is equal to positive three. I'm sorry correction. We have X is just equal to negative one minus negative four and so X is equal to positive three. And so the oxidation state of chlorine on the product side is equal to positive three. And so we would observe in this case, a change going from plus four two plus three on the product side. And so we would say are Oxidation State decreased by one unit. And so therefore, since the oxidation state decreases for glory in our reaction occurs as a reduction. So for our final answer, we have our unbalanced oxidation reaction and we have our unbalanced reduction half reaction. So what's highlighted in yellow are our two final answers. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.