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Question 15 of 38SubmitComplete and balance the following redox reaction in basic solution CrzOv? (aq) + Hgl) _ Hg?t(ad) + Crst (aq)HzOHtCrHzoOHHgResetHzO Delete...

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Question 15 of 38SubmitComplete and balance the following redox reaction in basic solution CrzOv? (aq) + Hgl) _ Hg?t(ad) + Crst (aq)HzOHtCrHzoOHHgResetHzO Delete

Question 15 of 38 Submit Complete and balance the following redox reaction in basic solution CrzOv? (aq) + Hgl) _ Hg?t(ad) + Crst (aq) HzO Ht Cr Hzo OH Hg Reset HzO Delete



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Balance the following redox chemical equation. Rewrite the equation in full ionic form, then derive the net ionic equation and balance by the half-reaction method. Give the final answer as it is shown below but with the balancing coefficients.
$\mathrm{KMnO}_{4}(\mathrm{aq})+\mathrm{FeSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow$
$\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+$
$\quad \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)$

This next question is a complete balancing of a single Redox reaction using the half reaction technique. When using the half reaction technique, you either need to balance it in acidic solution or basic solution. Because there is an acid involved in this redox reaction sulfuric acid. We will balance it using the half reaction technique in acidic solution. The chemical reaction we're considering. It's potassium permanganate, plus iron to sulfate plus sulfuric acid goes to iron three sulfate, manganese to sulfate, potassium sulfate and liquid water. So if we look at the reaction, we can see that iron is going from a plus two to a plus three. And the main Janie's is going from a little bit harder to tell to a plus two. If sulfate is two minus than the manganese associated with, it needs to be two plus. But what is it? Over here, where we have well, we have oxygen with a minus two oxidation state. There are four of them that's minus eight. The permanganate has a charge of minus one, so the manganese oxidation state needs to be plus seven. So it goes from plus seven to plus to the permanganate is reduced and the iron is oxidized. So one half reaction then is going to be the permanganate going to the MN two plus. So essentially this entire reaction could be simplified. Just focusing on what's oxidized and what's reduced in one half reaction will look at the manganese and then the other half reaction will look at the irons. The next step, after making sure we have the two correct half reactions, is to balance everything but hydrogen and oxygen. Manganese air already balanced irons are already balanced. We then balance the oxygen's with water. We have four oxygen's on the left hand side, none on the right. So we need four waters on the right hand side. We just introduced eight hydrogen ions. It's an acidic solution, so we can add H eight. H plus is then we balance charges with electrons. We have a two plus charge here. No charge here, So we have a total of two. Plus, here we have a total of eight plus minus 17 plus over here, so we need to add five electrons. So we have a two plus charge on both sides. Then we go to the iron half reaction. We see irons or balance. There's no hydrogen zor water, so we'll balance charges with electrons. We need to add it one electron to the right hand side, so both sides have a plus to charge. Now we sum everything up. But when we sum everything up, we need to make certain that the electrons cancel for the electrons to cancel. This half reaction needs to be multiplied by five. Then the electrons will cancel, and we sum everything up. We get eight. H plus is plus one permanganate plus five iron. Two Pluses goes to five iron three plus is plus mn two plus and four h two os. Then we can take this net ionic reaction and transcribe it into the full entire equation, Recognizing that for every one m n o minus, there is one MN for plus for everyone F e two. Plus. There is one iron to sulfate. We have five of those, so we're going to need five of these. For every hydrogen ion, it's coming from sulfuric acid for every F E three plus, it's coming from iron three sulfate. The MN two plus is coming from manganese to sulfate and then the potassium and the sulfate are spectator ions. But they are up here, so we'll include them and then our waters. So now let's plug in are appropriate coefficients. We needed five here because of this, we need a four here because we have eight hydrogen. We need a five halves here because we need five iron. Three pluses, and we already have two. We need a one half here because we have one potassium here and we've got to Potassium is here. Then we can multiply everything by two. And this will be our complete balanced redox reaction, and we add in the coefficients.

So we'll start with the 2/2 reactions that says the dye chrome eight is going to chromium and that if he two plus is going to be three plus. So to balance the first half reaction, we have to chromium. So we need to put a to coefficient here to get our two chromium. Then we've got seven oxygen's on the left hand side, so it's needs seven waters on the right hand side that introduces 14 hydrogen ion, so at 14 hydrogen ions to the left hand side. Now we have a total of six plus here, and we have a total of 12 plus here. So to make each side six plus you need add six electrons to the left hand side. Then the balance are Charges for the F B two plus F E three plus couple will add one electron to the right hand side so that both sides have a plus to charge that will multiply this bottom, have reaction by six so that the number of electrons will cancel and then some everything up. We sum everything up. We get this balance chemical reaction. However, this textbook prefers to express hydrogen ion as hydro knee. Um, so to do that, we could add 14 waters to both sides, producing 14 hydro knee um, from the 14 hydrogen ions and then 14 more waters to seven gives us 21 waters.

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 are, given this chemical reaction, and we will need to complete it and balance it using the half reactions. So if we look up our half reactions for each of our reactant for the chromium part are half reaction is CR 207 tu minus plus 14 h plus plus six electrons forms to see our three plus plus seven h +20 and those do all have their forms with them. But just to save space. I've left those off and then the second part is for iron and that half reaction is iron two plus forms iron three plus bless on electron. And then one thing to note is that there are six electrons in our blue equation, but there's only one in our red equation. So because of that, we're going to multiply everything in that red equation Bus six. So we're going to give a coefficient of six to all parts of that, Um so we can then go in and we know that those electrons are on opposite sides, so therefore they will cancel out, and then we can write our full equation. So the left side is CR 27 less that 14 hydrogen from the top blue equation. And then our other reactant is that six every two plus from our red equation, and then those will form are three products which were to see our three plus plus seven h 20 plus six f e three plus. So that whole thing in green is now our new balanced.


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