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Balance the following redox equation in acidic solution: Cu + NO 3" Cu 2+ + NO...

Question

Balance the following redox equation in acidic solution: Cu + NO 3" Cu 2+ + NO

Balance the following redox equation in acidic solution: Cu + NO 3" Cu 2+ + NO



Answers

Balance this redox reaction (in acidic solution). $$ \mathrm{Cu}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{NO}_{2}(\mathrm{~g}) $$

So iron, magnesium and gold. These are all examples of metal elements. And so metals do have many properties in common which define them as metals. So for example, they're shiny, they're very good conductors of heat and electricity and they are considered to be malleable. So here we're looking at the oxidation reduction reactions. So we have species losing and gaining electrons. So in this fast example, what we have is copper going to carpet surplus. So that is the oxidation part of the reaction. And then we have N. 03 miners going to N. 02 That is the reduction part of the reaction. So now we can look at the balanced equation. So we start with C. U. For copper Up to and 03-, Add four H. Plus. And that gives us see you two plus. Like we've already discussed at two N. O. To another product we looked at and finally to H 202 waters. So that's the first example and we do have one more to look at. And so again, we'll start off with the identifying the oxidation and reduction components. So firstly what we have is cr oh H three That gives the CR 04 to minus that is the oxidation portion of the reaction because we're losing electrons. And secondly, we have cl oh minus. That gives us see our minus, that is the reduction part. And so overall we have the defining equation as to see our oh H 33 c l o minus at 40 H minus where the minus what's on top of the oxygen that is in equilibrium with two C r 04 two minus at three C l minus, and finally five waters H 20

So here we're essentially given a specific chemical reaction where we have iron metal reacting with nitrate, leading to the formation of F. E three plus and nitrogen dioxide. And were given that the reaction takes place in the civic solution so we can use H plus in our reactant. So the easiest to write step is essentially for half reaction is essentially the oxidation of iron in its ground state To FE three plus. And then we have to have the reduction of nitrate to nitrogen dioxide. So we can add essentially H2 hour to balance the number of options which will require adding to H plus on this side. And we need an additional electron for charge balance. We need the electrons to cancel out. So we multiply the second equation by three and as a result, we see that the net reaction is the falling and this gives our final answer.

This is Problem 29 of Chapter 18 Electric mystery. So in this question balance each of the following Redox reactions in acid solution. So in the party of the question we have given, then bless Ferric ion, which is a free three plus that can form then ions plus fairest ions which is every two plus. So in the first step, in order to balance these Redox reaction, we have to find the oxidation number. So the oxidation number of pin in this case it is zero. And for every three plus ion, this is plus three for tin two plus iron. This is plus two. And for every two plus iron, this is plus two. So these are the oxidation numbers. Oxidation number is basically charge assigned of an element according to some set of rule, and that rule are called the oxidation number rules. So now we can clearly see that the tin undergo oxidation because the increasing oxidation number from zero to plus two. So that means this is the case of oxidation. And this is the case of reduction because the charge on after three plus from a free two plus that plus three, two plus two that went degrees in oxidation number. So this is called the A reduction. Yeah. So in the first step, we have to divide this equation into two half. Part one is oxidation half reaction, and the other one is reduction. Half reaction. Okay, so in the oxidation half reaction, we have 10, 2, 10, 2 plus ions, and in case of reduction, half reaction. We have a f e three plus two f e two plus iron, so as we can see that, So that means, in case of oxidation, half reaction. In order to balance this oxidation half reaction, we know that in case of oxidation, loss of electron can take place. So in this case, two electrons are lost, so we can see these, uh, two electrons and in case of reduction, half reaction, a reduction means gain of electrons. So in this case, one electron again so that this can form a few two plus ions. So that means here we can add one electron like this, but this equation. But this situation is not balance until we have to balance the electron on both the side of the equation. So, in order to balance the electron. What we have Do we just multiply this whole equation by two? Okay, so now what we get, we have tend to tend to plus ion plus two electrons. So this is the oxidation half reaction. And in case of reduction, half we have to f e three plus ion plus two electrons that can form to Effy two plus island. So now we just add all of these equations in order to cancel the electrons. So the two electrons no electrons are canceled out and on adding the equation, we get our balanced chemical equation. So which is finally 10 plus two f e three plus ions that can form Tyne ions plus mhm twice off F E two plus ions. So this is the balanced equation. Mhm. Mhm. Yeah. And in the second part of the question. Okay, so similar approach. We can apply in the second part of the question. So we have given this type of equation mhm. So in this situation, first we find out the oxidation number of each element, So the oxidation number of hydrogen, his plus one arsenic is plus three. Oxygen is minus two Irene zero. Oh, this hydrogen in this case, plus one arsenic. In this case, it is plus five oxygen in this case, minus two, and I'll died. Island, this is minus one. Okay, Right. So now so you can easily calculate the oxidation number. Like, for example, if we consider this example like h two, Yes. Or for minors. So if we consider that for arsenic, it is X, and we know hydrogen. The oxidation number of hydrogen, that is, uh, plus one. So we can try to in two plus one plus X plus four and two minus two. This is for the oxygen. Mhm. So this is four into minus two, which is equal to minus one. So we get two plus X minus eight minus one. So the value of, uh, X minus six at its minus one. So the value of X is classified. So this is the just rough calculation in order to find the oxidation number. Okay, so now we we know about the oxidation number. So now first we can see the oxidation half part and a reduction half part. So now this arsenic undergo the observation because, uh, increasing observation number. And this I wouldn't undergo the reduction because decrease in oxidation number. That is called the reduction. Mhm. Okay, so now we can divide this equation into two half part. That means the oxidation half part and reduction. Half part. Yeah. So the oxidation half reaction, and here we write reduction, half reaction. Mhm. Okay, so the oxidation half reactions comes out to be We have edge. Yes. Arsenic. Mhm. Oh, three to minus. That can form H two s or four minus. So the oxidation number that is plus three for arsenic and here, plus five for arsenic. Now we can, in order to balance the oxidation half reaction, we can see that the hydrogen and oxygen are not balanced. But arsenic is balanced. So in order to balance the oxygen atom first. So we add water molecule on that side in which oxygen are deficient. So that means in the left hand side of the oxidation, half reaction so that we can add one water molecule in order to balance the oxygen. So now oxygen is balanced, but hydrogen is not balance. So in order to balance the hydrogen, we can see their total three hydrogen involved in the left hand side and two hydrogen involved in the right hand side. So we just add one. Uh huh. H plus iron in the right hand, side of direction. So now finally, we can balance the electron so as we know the oxidation that when loss of electrons or how many electrons are lost, so see, the change is only for two so that maybe two electrons are lost in this. Okay, so we can add two electrons in the right hand side of the equation. So Okay, so So this is the balance, the equation balanced reduction, oxidation, half reaction. So now we can reduction half reaction. That is I two to I minus. So here, oxidation. Number of Children zero. And for all right. And this is minus one. So first we can balance the island. So we just multiplied by this documentary coefficient to in the right hand side of the reduction half reaction. So as we know, reduction. So that means the loss in case of reduction, gain of electrons. So how many electrons are so here? Two electrons are again in this case because we're not two Children are involved in this case. That's okay. So now both oxidation, half and reduction. Half reaction are better. So now just we can do just add both this situation in order to get the final balance Paradox the action. So simply we can, right? Yeah, mhm. And the other regulation I Odin plus two locked on that can form too ru minus. So just simply add both the equation in order to get the final balanced Redox reaction. So the two electrons are canceled out on what they said and what we get. So we get finally. Yeah, on adding both the equation, we get balanced Redox reaction, and this is in acidic solution. Okay, And now, in the third part of the question, in the third part of the ocean, we have Cooper plus silver ion that can form called part two. Plus, I own Let's silver. So again, the oxidation number of copper zero. So, in the native form, free form the oxidation number of the element that is zero according to the oxidation number rules in case of Silvera I own that is possible in case of, uh, corporate two. Plus, I own the oxidation number plus two and for silver at +80 So, now see, this is the oxidation because observation number increased from zero to plus two. And this is the reduction because the oxidation number decrees from plus 1 to 0 so that cooper and rocky oxidation and silver plus and silver ion undergo reduction. Okay, so now again, same battered follow divide the population into cool parts. One is oxidation, half reaction. And another one is the reduction half reaction. So the oxidation half reaction, that is copper too copper, two plus iron. Mhm. So, in order to balance this oxidation half reaction, we can see the change in changing the oxidation number, which is plus two. So that means the two electrons are lost in this case because oxidation wins philosophy electrons. So we just lost these two electrons and in case of reduction, half reaction. We have silver plus I own two from silver. So here oxidation number of silver iron plus one. And here it is zero. So now you can see that the change in oxidation number and in case of reduction that when the gain of electrons So how many electrons are gained? So it is only one electron are gained in this case. Okay, so now this occasion both the equation oxidation, half and reduction half reaction are balanced. But you can see the electrons are not balanced until now. So in order to balance and cancel the electrons, what we can do, we just multiply this whole equation by two so that the electrons are cancel out on the side of the equation. So in the final equation, we don't need to write the electrons. So we get copper to copper, two plus ion plus two electrons. And here we get twice of silver, plus ion plus two electron that can form twice of Sullivan. Okay, so if we add about this situation, yeah, two electrons are canceled out on both side of the equation and what we are left with. So we are left with copper, plus those silver ions that can form copper ions less the very soft silver. So this is the balance paradox reactions. So I hope this answer your question. So thank you so much for watching this video

Problem. 21 asks us to use the oxidation number method to balance the following net Ionic Redox reactions where it gives us the reaction for the equation as Ian plus energy minus yields the end to plus plus, you know to So the first part of the oxidation number method is to assign oxidation numbers for all the atoms I saw. Just break those above the chemical equation Mercy and zero uh, N 03 minus the oxygen is gonna be a minus two there, three of them, for a total of minus six by the whole molecule. Needs to be a minus one charge. And so that tells us that the nitrogen is a plus five. Does the end to class is gonna be a plus two again for N o to the oxygen is gonna be a minus two. There's two of them for my sport charged, So our nitrogen has to be plus for in order to give this in a charge of zero. So the nitrogen in this reaction went from a plus five to a plus four, indicating it was reduced on its delta or change. I was minus one, the one from plus five plus four The hi CNN went from zero two plus two, meaning itwas oxidized. And it's Delta they cost to the one from zero to cost you. Then if I rewrite out, this chemical equation will need Thio incorporate this delta so that we multiply The nitrogen is by two in order to even out this differential. So we have the end plus and the three minus yields began to us what? And no. Two, where we're gonna multiply our Nigerians by two to here to hear. And this makes sense because we need for the number of magicians on both sides products and reactions to be the same. So now we have the same two nitrogen zahn react inside to nitrogen is on the product side and we have, uh, one z n on one's yen. But we're gonna then need Thio Ad tells us this is an acidic solution. So we're gonna need to add hydrogen ions and water in order to balance out the number of oxygen's not. And hydrogen is in this equation. And so on the react. Inside we have six oxygen's of three agents to a six on. We only have four on this side two times two. And so we're gonna need to add two waters. So plus two h 20 in order to balance out the number of oxygen's on the product side to the react inside. This gives us four hydrogen sze on the product side, so we're gonna need to add plus four age plus, uh, on the react inside for a balanced chemical equation.


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