Use the oxidation-number method to balance these redox equations. a. PbS+O2→PbO+SO2

step 1 Question number 62 is the balancing of redox reactions. There are three different ways in which

Use the oxidation-number method to balance these redox...

Key Concepts

Conservation of Mass and Charge

In any balanced chemical equation, particularly redox reactions, the conservation of mass and charge is fundamental. This principle asserts that the number of atoms of each element and the overall charge must be identical on both sides of the equation, ensuring that no electrons are lost or gained in the aggregate during the reaction.

Oxidation-Number Method

The oxidation-number method is a systematic approach used to balance redox equations. It involves determining the oxidation states of the elements in the reactants and products, identifying the changes in oxidation numbers, and then ensuring that the total increase in oxidation numbers (representing oxidation) equals the total decrease (representing reduction), thereby conserving both mass and charge.

Redox Reaction

A redox reaction is a chemical process involving the transfer of electrons between two substances. In these reactions, one species undergoes oxidation (losing electrons) while another undergoes reduction (gaining electrons), making it essential to account for electron transfer when balancing the overall reaction.

Oxidation State

The oxidation state (or oxidation number) is an assigned value that represents the effective charge of an atom in a compound. It indicates the degree of oxidation or reduction of the atom, serving as a tool to track how electrons are distributed in a molecule or ion during chemical reactions.

Transcript

00:01 Question number 62 is the balancing of redox reactions.

00:07 There are three different ways in which you can balance redox reactions.

00:11 You can balance them by inspection, which means you just inspect the chemical equation, and after inspection, determine where appropriate coefficients go in order to balance the atoms of each element.

00:23 Or you can balance it using the oxidation number method, which is what is requested in this problem, or you can use the half -reaction method.

00:34 So for this one, to balance using the oxidation number method, we assign oxidation states to each element in all the reactants and all the products in order to determine the number of electrons that need to be transferred in order for the reaction to be balanced.

00:51 So for this first one, we have lead to sulfide, reacting with oxygen gas, producing lead to oxide, and sulfur dioxide.

01:00 Side.

01:02 We assign oxidation numbers.

01:04 We have sulfur with an oxidation number of minus two because it has a charge of minus two.

01:11 Therefore the lead, which could be 4 plus or 2 plus, is 2 plus because we only have one lead and one sulfur.

01:18 The oxidation state of elements in their elemental form, elemental form of oxygen is 02, is 0.

01:27 We then go to oxygen in lead 2 oxide and the oxygen except in the case of peroxides and super oxides has an oxidation state of minus 2.

01:39 So the lead in order to be a sum of oxidation states equal to 0 would be plus 2.

01:45 We then go to oxygen which is minus 2 there are two of them so that's total minus 4 so the 1 sulfur needs to be plus 4.

01:53 So then we look at the change in oxidation states of the elements.

01:57 Lead stays the same.

01:58 Sulfur goes from minus 2 to plus 4, so that's a total of 6 electrons that were transferred.

02:06 The oxygen goes from 0 to minus 2.

02:12 So that means we need to have 3 times as many oxygens as we have sulfurs.

02:20 How do we know that? because one sulfur needs to...

02:26 Now one sulfur releases six electrons, therefore there needs to be three oxygens each taking on two electrons to accommodate all of the six electrons that were released.

02:40 So to get three oxygens per one sulfur with o2, we need to multiply it by three halves.

02:50 Then when we multiply it by three halves, we have a fraction in our balanced chemical reaction, which we can fix by multiply.

02:58 Everything by two.

03:01 Then we will verify that it is balanced.

03:04 We have two leads and two leads, two sulphurs, two solfers, six oxygens and two and four, six oxygens.

03:15 The next chemical reaction is n -aw -o -3 plus n -a -o -h plus oxygen goes to naw -o -4 plus water.

03:27 We assign our oxidation states.

03:30 This is a this is an ionic compound, so sodium will have an oxidation state equal to its charge of plus one.

03:37 Oxygen is pretty much always minus two.

03:39 There are three of them, so that's minus six, but the w -o -3 needs to have just minus one charge to neutralize the one plus coming from sodium.

03:52 So the oxidation state of tungsten will be plus five.

03:57 Sodium is plus 1 because that's its charge.

04:00 Oxygen is minus 2, requiring hydrogen to be plus 1 to have a neutral species.

04:07 Oxygen in its elemental form is 0.

04:10 Sodium here again is plus 1.

04:13 Tongsten is minus 2.

04:14 There are 4 of them, so that's minus 8.

04:20 But we need just minus 2 to accommodate the 2 pluses right here, so that means tungsten needs to be plus 6.

04:27 We then go to water, oxygens minus two, hydrogens plus one.

04:33 So we see that it's tungsten that has a change in oxidation state from plus 5 to plus 6, so one electron is released per tungsten.

04:44 Oxygen goes from zero to minus 2, but not all the oxygens, because we already have some oxygens that are at minus 2.

05:02 So it's this one oxygen that was zero here.

05:06 One of these oxygen atoms becomes a minus two here.

05:11 But the oh right here gets incorporated into the water over here, still with the oxygen at a minus two.

05:21 So because we have one ohh with one hydrogen and we have two hydrogens here, we need to make sure our hydrogens are equal.

05:32 Then we recognize that we have tungsten that needs to take on just one electron.

05:42 I'm sorry, he's going to release one electron...

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