Balance the following net ionic reactions, and identify which elements are oxidized and which ar

Balance the following net ionic reactions, and identify...

Key Concepts

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between chemical species. In these reactions, one species loses electrons (is oxidized) while another gains electrons (is reduced), making them fundamental to understanding reaction mechanisms in chemistry.

Oxidation

Oxidation refers to the process in which a chemical species loses electrons. This loss results in an increase in the oxidation state of the species and plays a key role in redox reactions, helping identify which elements undergo electron loss.

Reduction

Reduction is the gain of electrons by a chemical species, leading to a decrease in its oxidation state. Recognizing reduction allows one to determine which species are accepting electrons during redox reactions.

Oxidation States

Oxidation states (or numbers) are numerical values assigned to atoms in a compound that indicate the degree of oxidation or reduction. They are essential for tracking electron transfer during redox processes, enabling the identification of which atoms are oxidized and which are reduced.

Half-Reaction Method

The half-reaction method is a systematic approach used to balance redox reactions. It involves splitting the overall reaction into two parts—one for oxidation and one for reduction—and separately balancing mass and charge before combining them to form the balanced overall reaction.

Net Ionic Equations

Net ionic equations show the actual chemical species involved in the reaction by excluding spectator ions. They focus on the fundamental redox processes occurring in a solution, emphasizing the changes in oxidation state and electron transfer.

Transcript

00:01 For this next group of redox reactions, i'm going to do the same as i did before, using the half -reaction technique, which is a little bit more comprehensive, and allows for us to add in hydrogen ions and water when the redox reaction is occurring in aqueous solution.

00:17 So for the first reaction, we have mno4 -1 plus s2 -1 -2, goes to m -n -o -2 plus s.

00:24 So one -half reaction is m -n -o -4 -1, goes to m -n -o -2.

00:27 The second half -reaction is s -2 -2 -2 -2.

00:30 Minus goes to s.

00:31 We'll balance everything but the hydrogen and the oxygen atoms and they're balanced.

00:36 So we next balance all of the oxygens with water.

00:42 We have four oxygens here.

00:44 We have two oxygens here, so we need two more oxygens by adding two waters.

00:48 We now have four hydrogen, so we're going to need four hydrogen ions on the left hand side.

00:53 Then we can begin to balance charges.

00:55 We have no charge over here.

00:57 We have a three plus charge here, so we need three electrons on the left -hand side.

01:03 And then in the case of the sulfur half reaction, we have s2 minus and s.

01:07 So we need two minus on both sides.

01:10 We will add two electrons to the right -hand side.

01:13 Now we need to make sure the electrons are equal, so that when we add them up, the electrons will cancel.

01:18 We'll multiply the first reaction by two, the second reaction by three, so that we get a total of six electrons, which will then cancel.

01:27 We then sum what is left over, canceling anything that is common to both sides.

01:32 We have four multiplied by two.

01:38 That gives us eight hydrogen ions, which we have here, and then 2mn 04 minus.

01:47 We've got the 3 s2 minus, and then we've got the three sulfurs, and we've got 2mno2, and then four waters.

01:58 If we then look at everything, we've got four hydrogens, i'm sorry, we've got eight hydrogens and eight hydrogens, two manganese, two manganese, three sulfurs, three sulfers, and then eight oxygen.

02:12 We've got four and four more oxygen charges.

02:15 We've got eight plus minus two, that's six plus, minus six more, that's zero, and we've got zero charge over here.

02:25 For the next one, we've got the complete reaction, i -o -3 -1 plus i -minus goes to i -2, which seems a little strange because there's only one product, but the iodine from i -o -3 -1 goes to i -o -2, and the iodide from i -minus goes to i -o -2.

02:45 So half -reactions can be written as the i -o -3 -minus goes to half of the product, i -2, and then the iodide goes to the other half of the product, i -2.

02:59 Recognized this and you didn't put in the half, that's okay, you'd still be able to balance the equation.

03:05 In the end, you just may need to divide everything by two in order to get the smallest whole number coefficients.

03:16 So we'll balance the iodines first.

03:20 We've got one and one, one and one.

03:24 We'll then balance the oxygens with water, so we need three waters on the right hand side.

03:29 We'll then balance hydrogens.

03:31 We need six hydrogen ions on the left hand side...