To 100.0 mL of a solution that contains 0.120 M Cr(NO3)2 and 0.50...

Question

To 100.0 mL of a solution that contains 0.120 M Cr(NO3)2 and 0.500 M HNO3 is added to 20.0 mL of 0.250 M K2Cr2O7. The dichromate and chromium(II) ions react to give chromium(III) ions. 
(a) Write a balanced net ionic equation for the reaction.

Accepted Answer

Hello. Everyone in this video we're being asked to find the balance net ionic reaction when this reaction over here occurs. So this reaction is bringing out in english words, we'll put it in terms of just pure chemistry. So what we have is that the overall reaction that we're talking about is M N 04 minus. Just a quiz, reacts with M N two plus again that's acquis and that gives us M N. 02, which is a solid. So bring us down into half reactions and we'll do one at a time. So the first reaction that we deal with is M. N two plus to give us M. N. O. To a solid. So what we can do first is bounce out our oxygen's. You see here on the right that we have two atoms of oxygen on the left, we have none. So how we add our oxygen sources is by adding H20. So we go ahead and add in two moles of H 20. Which of course is a liquid. Unless it's the first step which is balancing out our oxygen. Next we can go ahead and balance out our hydrogen. So now on the left side we can see that we have four atoms of hydrogen but on the right we have none. How can add just purely hydrogen is by adding our hydrogen ions. So again just repeating the same material side. Okay, and then we have M. N. 02. Alright, so again we said we are adding four moles of H plus to give us four hydrogen ions. Alright, and now let's go ahead and balance out our charges. So we see that on the left side we have a two plus overall charge for the entire left side. On the right side we have a four plus charge. We need to add two electrons to the right side so that we have a overall charge of two plus on the right side to match the left side. So we have Mn two plus acquis two moles of H 20 liquid and then repeating all basically at the same on the right side. And of course we're just adding those two negative charges. How we can add those two negative charges is with electrons. So adding two electrons here. Okay, so that's the first half reaction. The second one let's do a different color and both scroll down. All right, so the second half reaction is the second starting material. So, Mn 04 minus which is a quiz to give us the same product because there's only one of M N 02. The first thing we can do again is to bounce our oxygen's. So we see on the left side and starting material side we have four items of auction on the right, we only have two. So we'll just add in two bowls of water to bounce out those heights. Or these oxygen's Next up we can do is balance out our hydrogen. So on the left side we have none but the right side we have four. So to the left we can add those hydrogen ions. And how many will add is four. Alright, now that we balance out the hydrogen, let's go ahead and actually bounce out the charges so we can see on the left side then we have this minus one and four plus. So we have the overall three plus charge on the left side. On the right side here we have just simply neutral. So how to make the left side neutral is by adding negative charges. So let's go ahead and just repeat what we have so far. So again we have a three plus charge right now we need to go ahead and add three negative charges. So three electrons to neutralize everything. And then the right side. So the product side will just remain as is that thing needs to change their Alright, so now we have our two half reactions. Let's actually write those two out. So then we'll have MN two plus which is a glorious with two moles of H20 which is liquid That gives us MNO. to a solid with four age pluses and two electrons. The second half reaction that's balanced so far is M N 04 minus and for H pluses With three electrons to give us MNO two solid and to H 20 s. Alright, so final thing we do here before finding the net reaction is to balance out our electrons for the 1st and 2nd half reaction. So we see here for the first reaction we have two electrons And on the bottom we have three. So how we can balance this out is by multiplying the first half reaction. Well, the all the whole entire verse reaction to be multiplied by three. In the second half reaction will all be multiplied by two. Now, if we do that We'll get three moles of MN two plus, which is Aquarius and then three times two is six H two Os to give us three moles of MNO two, 12 walls of H plus and six electrons. And then for the second half reaction, that would be too M N 04 minus which is Aquarius with eight moles of H plus and six electrons. And that will produce two moles of M N O solid and four moles of H 20. Alright, so just scrolling down here. Okay, so we can now see well, first of course are electrons will cancel out our hydrogen cell counts up on the top. One lb now, four. Our water Hero cuts out. The first fraction will be left with two moles of water and that's about it. So the net reaction then will be three M N two plus with two bowls of H 20 With two SLMN 04 -. And that would give us four moles of MNO two, which is a solid. Let's make this clear, okay, solid. And then we have four moles of age plus, and it's going to be the overall net reaction that we have for this problem. All right. Thank you all so much for watching.

To 100.0 mL of a solution that contains 0.120 M Cr(NO3)2 and 0.500 M HNO3 is added to 20.0 mL of 0.250 M K2Cr2O7. The dichromate and chromium(II) ions react to give chromium(III) ions. (a) Write a balanced net ionic equation for the reaction.

Watch next