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Complete and balance the following redox equation. When properly balanced using the smallest whole-number coefficients, the coefficlent of 5 Is: pt) H28(g) - HNOa(a...

Question

Complete and balance the following redox equation. When properly balanced using the smallest whole-number coefficients, the coefficlent of 5 Is: pt) H28(g) - HNOa(aq) S(s) + NO(g) (acidic solution)A. 15 B.2. C D E.6.

Complete and balance the following redox equation. When properly balanced using the smallest whole-number coefficients, the coefficlent of 5 Is: pt) H28(g) - HNOa(aq) S(s) + NO(g) (acidic solution) A. 15 B.2. C D E.6.



Answers

Balance the following redox equations by the ionelectron method: (a) $\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}$ (in basic solution)
(b) $\operatorname{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}(\mathrm{in}$ basic solution) (c) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}$ (in acidic solution) (d) $\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}$ (in acidic solution)

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Okay, Answer for the question. Here. Balanced the following Redox equation by the half Reaction meter reaction. Balanced rations for what? A To its positive plus hydrogen peroxide is two or two plus to E to positive to e iron to positive class to it 24 and Violence Dictation for part B six. It's positive because to it, in a tree, plus free mhm. See you proper into three. See you to positive plus two in all. Plus four is +20 yeah, Balanced equation for part C is too old plus two in, you know for plus three c n to em in or two plus three cno plus do or H perfect. Balanced off. Reaching MM balance station for party by using a prediction method. Yeah, six or IT plus three b R. Two. We are old. Three plus three is to all plus five b r. What in for part e two s two or three to Plus, are you Dean? Bless ISS for 462 plus two ivy. These are the balanced big vision by using calculation method

This is a problem when of Chapter 19 redox reactions and electro chemistry. So in discussion, balanced the following redox reaction by half reaction method. Okay, so in the first part of the question, we have given H2 or two plus every two plus I own, that can form every three plus ion plus water. So we balance uh this uh redox reaction in the acidic solution. So the redox reaction is the reaction in which both oxidation and reduction can take this simultaneously. So this type of reaction is called the redox reaction. So oxidation means loss of electron and reduction means to gain of electrons. Or in other words, oxidation were defined as the increase in oxidation number that is called the oxidation or decreasing and decreasing oxidation number. That is called the reduction. So the first step is we can write the oxidation number of age element and to give an equation. So then uh you can see that uh this uh effort to plus two form every three plus. So that means this undergo the process of the oxidation. So this is the process. And to go the observation and uh this is the oxidation and this uh this is undergo the reduction because oxygen in this case that undergo the reduction. So this is the reduction. So our represents the reduction or represents the observation. So we divide the body equation into two parts, oxidation half part, and the reduction half part. Now. So first we have to balance this oxidation half part. So in order to balance the oxidation half reaction, we can see that the one electron is lost. So uh this is the balanced oxidation half reaction. So now in order to balance the reduction half reaction. Uh First we can see the number of atoms, so the hydrogen atoms are balanced, but oxygen are not balanced. So in order to balance the oxygen atom, we multiplied by the stock geometry coefficient to in the right hand side of the reduction half reaction. But now hydrogen is unbalanced. So we just add two H plus iron on the opposite side. So as we know this is the addiction and uh mm. And we know we see that from the situation that the two electrons are again in this equation so that we can add to electron in the left hand side of the reduction half reaction. So now in order to cancel the electrons, what we can do, we multiply this oxidation half reaction by it too, so that the body electrons are cancel out on observation half reaction and reduction half reaction. Now see that this becomes to a free two plus ions that can form too, F E three plus ion plus two electron. And when we add this uh That's two or 2, So H two or 2 plus two H plus ion plus two electrons that can form to which tour. So if we add that, we can see that these two electron and electron on both side cancel out. So what we're left with the so we are left with the twice off feet two plus ion plus H two or 2 plus to which plus I own that can form to F E three plus ion plus two H two. So this is the balanced redox reaction. So this is the answer of the party of the question. And uh in the part of the question that we have to use the similar approach. So in the second part of question, uh we have to balance distribution in the acidic solution. So first we write the oxidation number of each element. Now clearly you can see that this copper Androgel oxidation because its oxidation number increased from zero to plus two. And this nitrogen Androgel the reduction so I can try it in this way. So this is the reduction Because the oxidation number of nitrogen degrees from plus 5 to plus two Plus 2 Plus two. Mhm. Yeah. So now they were the equation into two parts that is solution half part and her reduction half part. So just to balance the oxidation half part. So now the oxidation half balance. And in order to balance the reduction half part we can see that the oxygen is not balance and hydrogen also not balance. So in order to balance the oxygen we just add the water molecule, those side at which the oxygen's are deficient. So that's why we add to which to her. But now this hydrogen it's not balanced. So in order to balance the hydrogen, we add three H plus iron on the opposite side. So now just are the electron because the reduction half that when they gain of electrons can take place. So we just add three electrons. And now in order to balance the electrons, what we can do we just multiply the reduction half reaction equation by the two and the oxidation half reaction equation by three so that the electrons are cancel out each other and add both the question. So on adding we get the final answer. So on adding we get three copper plus place off at january plus six hydrogen ion that can form price off Copper two plus ions plus twice of you know and four H door. So this is the balanced redox reaction in acidic solution. All right. So in the third part of the question we have to balance the situation by the basic solution. So first of all we write the observation number of each element and then divide the both oxidation half reaction and a reduction half reaction. Now you can clearly see that this nitrogen that Andrew Cody observation and this magnus that undergo the reduction so divide. Okay, so in order to balance the oxygen in the oxidation half reaction, we add water molecule on the deficient side and uh to which plus iron on the opposite side. So after that, ah we can see the change. So the change is for the two electrons. So that's why oxidation wins loss of electrons so that the two electrons we you can see. So in the reduction half reaction, the magnus is balance but oxygen is not balanced. So in order to balance the oxygen, we just add water molecule on the deficient side. So just add to which to in order to balance the oxygen but hydrogen is not balanced. So that's why we at four H plus iron on the opposite side. And uh then the electrons in order to balance this whole occasion. So now you can see that the electron sarna uh in order to cancel the electrons on both sides, we just multiply this observation half reaction by the By three and a reduction half reaction by two so that the electrons are cancel out each other. So on adding and on adding about the equation, we are left with distribution. And because we have to balance this uh equation in the basic solution, this redox reaction, the basic solution. So that's why we just add a hydroxide ion. So at 20 H. I. N. On the both side, because we have two H plus uh I own so that's why we had to watch minus on both sides. So just add and just rearrange the water molecules and what we are left with the final equation. Okay, so this is the balanced equation in the basic solution. So three cyanide minus two amino four minus plus water molecule that can form +30 minus plus two women or two plus uh to hydroxide ions. So in the part of the ocean that we have to balance distribution in the basic solution. So again right? The oxidation number of uh each element. And then you can see this uh undergo the oxidation and this undergo the reduction. And this type of reaction, we can also call it as the disproportionate reaction. Um Okay, so now the oxidation half reaction that is uh B. R. Two to form B. R. 03 minus. So first we balance the brahmins by using the documentary coefficient oh in B. R. 03 miners. And then we can balance the oxygen on adding water molecule on those areas which have and efficient oxygen's and on the opposite side 12 h. Plus iron. And in order to balance the A charge, we had 10 electron. So in the reduction half reaction again. First of all we just balance the brahmin by using the stock America fish into in the right hand side of the reduction half reaction. And then in a reduction half reaction had two electrons. So in order to balance The situation and cancel the electrons, we multiply this reduction half reaction equation by five so that 10 electrons are on both sides are cancel out each other. Okay so all right so what we're left with we left 26 br two plus six waters that can form to borrow three minus plus 12 H plus plus 10 br minus. So adding 12 H minus on both side because we have a large plus, so 12 h plus plus 12 oh H minus that. Uh We know that this is called Watch tour and just rearrange this a public tuition. And yeah finally get the balanced redox reaction in the basic solution. So we get six BR two plus 10 hydroxide down that conform to B. R. 03 minus plus six water plus 10 br minus. And finally in the last part of the equation again uh the same processes applicable. So we have to balance the situation in the aesthetic solution. And the first we write the oxidation number of each element and then divide the oxidation half reaction and reduction half reaction. So you can see that this Eileen undergo the reduction part and the cell phone and your body oxidation. But so and a similar approach we have to use in order to balance the situation. So here or represents the oxidation. These are represents the reduction. This represents the oxidation half reaction, and this r h. R. Represents the production half election.

To balance to redox reaction shown in the screen, which take place in in ECO CCD conditions. So let's start with the first reaction oxidation of tin two plus between four plus by higher date and iron. So let's start with uh oxidation reaction first. So Tin two plus loses two electrons And becomes 10 4 plus. And this reaction doesn't need any additional balancing. So additionally, um how to date is reduced to idea. Mhm. Then we have to balance this reaction in the media. Okay, so first let's balance Aydin and Now we have to add three water molecules to the right side of the equation to balance, sorry, six water molecules to the right side of the equation to balance oxygen. So now there are six oxygen atoms on its side. So now we have to balance hydrogen and church. So on the right there are 12 hydrogen. Therefore we need to add 12 hydrogen. Okay, to the left side of the equation. Now let's calculate the total charge on the left, So its 12th -2. So it's 10 and On the right side and that charges zero. So therefore we have to have 10 electrons to the left side of the equation. So now it's balance and basically 10 electrons means that each idea and five plus gets five electrons and is reduced to eyes and zero. Which makes sense. So now we can balance oxidation reaction. We just simply to which we've already done. And now we can calculate the overall net reaction. So overall it's tera electron reaction. So therefore we have to multiply teen by five. Um 2nd reaction by one. Yeah, So now let's write down the total reaction. So 5:15 to plus. Okay reacts. We is too Mhm violate an ions 12 broughton's Uh huh. And it forms um five teen four plus One idea in and six water molecules. So let's check for the church balance is plus 20 on the right and on the left, head stem minus to plus 12, which makes 20. So there are 12 hydrogen is on the left, 12 hydrogen is on the right, 5 15 items on the left, five teen items on the right to I. D. In atoms on the right to I jin atoms on the left, so in six oxygen's on this side. So therefore it's balanced. Now let's move on to the next equation, which is the oxidation of arsenic oh 33 minus two, arsenic oh 43 minus. So this is oxidation because oxidation number of arsenic changes from last 32 plus five and it's oxidized by brahman which is reduced in turn. Yeah. Now let's balance that equation in the echoes a cd media. So first we'll start with oxidation reaction. So in the oxidation reaction mhm uh are saying oh 33 minus is oxidized to are saying oh 43 minus. So we have to balance oxygen first. So therefore we add one water molecule to the left side of the equation. So now we have four oxygen atoms on the each side. Now we need to balance hydrogen so we add to protons to the right side of the equation. So and now we need to balance charge. The net charge on the left side of the equation is negative three on the right side, it's negative one. Therefore we have to add two elections. Which makes perfect sense because arsenic is oxidized from plus 32 plus life. So it's a two electron process. So in the second in the reduction reaction brahman, Mhm who is reduced to to brahman negative bromides by getting two electrons. So therefore overall it's a two different process. And we have to multiply each equation by one. So let's write down the overall reaction. It's our Singapore three. Three minus. Yeah. Glass one Water molecule plus for I mean, yeah, forming uh one arson 89 RC vehicle for three minus last two broad mites plus two products. So let's again calculate the charge on the right hand, left side of the equation. So on the raft it's three minus on the right. It's uh three minus as well because it's three, it's negative three plus negative two plus two. So it's negative three in total. To prominence on the left. Two province on the right. Uh two proteins on the left to proteins on the right, for oxygen on the left, for oxygen on the right and one arsenic on each side. So therefore this reaction is balanced. Thank you very much for watching this video. I hope I helped you


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