On heating a 0.200 g sample of a certain semimetal M in air, the corresponding oxide M2O3 was obtained. When the oxide was dissolved in aqueous acid and titrated with KMnO4, 10.7 mL of 0.100 M MnO4- was required for complete reac-tion. The unbalanced equation is (a) Balance the equation.

Video transcript

everyone. So ask the balance of following redox reaction under acidic conditions and identify the oxidizing and reducing agents. I need to first separate the whole reaction into two half reactions. F. cr 207 two minus. And this goes to see our three plus. Then we have you four us And this goes to you 02 10 plus. Now we need to balance the non hydrogen and non oxygen elements first. So for the first reaction, you need to balance the chromium. This is already balanced. So for the first we actually need to balance the chromium to chromium on the raft inside. But one on the product side we can put a two in front of cr three plus. Now we need to balance the oxygen by adding H20 liquid to the side. That needs oxygen seven auction on the wrapped inside. And the first equation, We need to add 7 H20 liquid to the product side. For the second reaction have to auction on the product side, we need to add to H20 liquid to the reacting side. Now now I need to balance the hydrogen by adding H plus to decide they need hydrogen 14 hydrogen on the product side, they need to add 14 H plus to the acting side on the first reaction. And then for the second reaction we have four hydrogen on the right hand side. So we need to add four H plus to the product side. Now I need to balance the charges and add the electrons to the more positive side, have a look at the charges plus 14 here. I got to hear six here. You're out here. We have a zero here plus four here. Last two here. Last four here on the first reaction we're gonna need to add electrons to the reacting side. We need to add 14. So we need to add six electrons. For the second reaction we need to add electrons to the product side. I need to add two electrons. Now I need to balance electrons on the two half reactions. We have six on the top and two on the bottom and he's most probably three on the bottom. And for the oxidation reactions were gonna lose electrons. We're gonna have electrons on the product side. For the reduction reaction we're gonna gain electrons. We're gonna have electrons on the react inside. For oxidation half reaction We're gonna have three U four plus six H 2 L. It's gonna be a three year 02 two plus plus 12 h plus six electrons. And for reduction half reaction You see our two 72 minus last 14 H plus plus six electrons. Getting to cr three plus seven H 20. For our loss of electrons. This oxidation and this is our reducing agent, It's gonna be you four plus And then for a gain of electrons is reduction is an oxidizing agent. NBC are to go 7 to -. So now I need to get the overall reaction by adding the two reactions, canceling it out. We cancel out any common species. We're gonna get rid of 6 H20. We're gonna be left with 1 H20. Over here. We're gonna have to H plus over here. We're get we're gonna get rid of our electrons as well. So for the balanced reaction, Get three U 4 plus plus cr two plus, seven to minus plus two H plus. And then she was 302 in plus Plus 2, er 3 plus H 20. Thanks for watching my video and I hope it was helpful.