What is the oxidizing agent in the reaction Zn + 2 H⁺

Question

What is the oxidizing agent in the reaction Zn + 2 H⁺ → Zn²⁺ + H₂?

Accepted Answer

Hey, y'all, let's get started on this next question. So we want to balance the redox reaction in acidic conditions. So, um and then what we wanna do is we can identify which species are oxidized, which are reduced. And then we're going to want to identify the oxidizing agent and the reducing agent. So let's actually first balance out this reaction. And what we're going to do in order to do that is to write out our half reactions. So let's write out our half reactions. We have T C 04 minus Fields TC two Plus. And then our second half reaction is going to be our Each and no two yields nitrate. Alright, so the first step that we're going to want to do is we're gonna want to balance the oxygen's on each side. So here we have four oxygen's. And in order to balance the oxygen's, we're gonna want to add waters. So here we're gonna be adding four waters to the product side. Now, because we added four waters to the product side, we're actually adding eight hydrogen is to the product side. Now, in order to balance that, we're going to want to add eight protons or eight H plus ions to the react inside. Now let's actually calculate our charge for both sides. So we have a negative one charge from right here and we have a positive eight charge from the eight protons. So that means that we're going to have a positive seven charge on this side. And then for this side we have a positive to charge. Now in order to balance that we want to actually add electrons to our react inside in order to get a positive seven to a positive to charge. And so we'd be adding five electrons to that side. Now let's do the same thing for our second half reaction. So once again we're going to be um balancing our oxygen's first. So here we have two oxygen's while on this side we have three. So we're gonna add one water on the side now because we added a water here, we're adding more hydrogen. So we have three hydrogen on the reactant. So that means we're gonna add three protons to the products. Now let's actually calculate our charge. So we have a zero charge for the reactant and we have a positive to charge from the products because the nitrate gives one negative charge and the protons give three positive charge charges. So in order to balance that we're going to actually add two electrons to the products. So now we have our balanced equations. And essentially what we're going to want to do is we're going to actually want to ensure that the electrons from both equations are balanced. So we have five electrons here and we have two electrons here. So now we have to find a common number between five and 2 um so that we can balance these electrons out And that's going to be 10. So we're going to multiply this half reaction by five. We're gonna multiply this half reaction by two. And so here Let's write out our half reactions. We have 10 electrons plus protons plus two TC 04 yields to TC two plus plus eight waters. So that's gonna be for our first half reaction. Then moving on to our second, we have five waters plus five nitric acids, yields five nitrates Plus 15 hydrogen ions plus electrons. Now, what we need to do is we actually have to cancel out um what is similar on both equations. So what I mean by that is we have 10 electrons on this side. So we have 10 electrons on that side. So that's going to cancel out. We have 16 protons on this side and 15 on this side. So we're gonna actually minus this by 15 to get one, not a proton. And then we have five waters on this side and eight on this side. So we're going to actually -5 here to get three waters overall. And so are balanced overall equation. It's going to look like this. So we have and h plus plus five HNO two plus to T c 04 minus yields three waters plus five nitrates, plus two of these. So this is going to be our overall balanced equation. Now we can move forward to the second part of the question and this is asking us to identify which species are oxidized and which are reduced. So in order to do that, let's actually rewrite our initial equation. We have T. C. 04 minus plus nitric acid. Just yields nitrate. And we have All right. So um let's actually um label these. We have A B, C. And D. And we're going to actually calculate the oxidation states to see which is oxidized and which is reduced. So um we have to follow a specific equation. So for a which is our TCO for minus. We're trying to find the oxidation state for this element. So we have X plus -2 which is the oxidation state for oxygen. But because we have four of those, we're gonna multiply that by four and that's going to equal an overall charge of -1. Which is given right here. So if we were to solve for that, we have a positive seven charge for this next. Let's go on to this. So we have um We have basically are zero. Um so we have my apologies. We have a positive one for hydrogen plus X. Which is what we're trying to find for nitrogen Plus a -2 for oxygen. And we're going to multiply that by two. And so essentially we're going to get an ex of three for nitrogen. All right. Um give me one moment. Yes. So three for nitrogen and I messed up here is actually going to be a I know two. So 3 for nitrogen moving forward we have C. So we have our nitrate. So essentially what's going to happen here is gonna have our X. Which is our nitrogen value um Plus negative two. Which is the value oxidation state for oxygen going to multiply that by three and that's going to equal negative one. And so our x value for this is going to be a positive five for nitrogen. And then we have um d which is our TC and that's going to be positive too. So for looking so if we're looking at our nitrogen Going from our reactant to our products, we go from a positive three To a positive five. And so essentially what's happening here is we're losing electrons And that means that this is being oxidized Now for our other elements, we're going from a positive seven to a positive too. And because we're reducing our positive charge, we're gaining electrons. And so that means that this is going to be reduced. So this is going to be our answer for our second part of the question. Now, moving forward, we're gonna do our 3rd part And that's asking which is the oxidizing agent and which is the reducing agent. So are oxidizing agent is going to be the reactant that's getting reduced. So that's going to be this species right here, our reducing agent is going to be the reactant that's being oxidized. So that's going to be this species right here. And so that's going to be your final answer um for the third part of the question. Alright, everyone, I really hope this video helped, and I will see you in the next one.

General Chemistry 때문에 힘들어하고 계신가요?

What is the oxidizing agent in the reaction Zn + 2 H⁺ → Zn²⁺ + H₂?

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Balancing Redox Reactions: Acidic Solutions 연습 세트

General Chemistry 때문에 힘들어하고 계신가요?

What is the oxidizing agent in the reaction Zn + 2 H+ → Zn2+ + H2?

다음 시청

Balancing Redox Reactions: Acidic Solutions 연습 세트

What is the oxidizing agent in the reaction Zn + 2 H⁺ → Zn²⁺ + H₂?