Topic 9: Redox Processes IB Chemistry SL Coral Gables Senior High School Ms. Kiely
SPRING BREAK ASSIGNMENT 1) Read Topic 9. 2) Do all worked examples throughout the chapter, and also exercises 1-19. Answers to the exercises are posted on the website as Topic 9 Book Answers. THIS WILL BE GRADED. After Spring Break, we will go over the chapter and the review questions for ONE. DAY. Then, we will take the test the. following class!.
Redox Reactions An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. Oxidation: loss of electrons; Reduction: gain of electrons Redox reactions are common and vital to some of the basic functions of life, including photosynthesis, respiration, combustion, and corrosion or rusting. Redox reactions are also vital for taking selfies, posting your teachers on snapchat, and finding love online! How, you ask? Well, transferring electrons from one substance to another leads to a flow of electrons, which you know also means ELECTRICITY! Redox reactions, thus, generate electricity.
Early attempts to define oxidation and reduction were based on observations made on reactions that involved the gaining and losing or electrons due to the gaining or losing of oxygen or hydrogen; hence the term oxidation which obviously refers to oxygen. 2Mg(s) + O₂(g) 2MgO(s) For instance, in this oxidation reaction magnesium bonds with oxygen to produce magnesium oxide. The magnesium atom loses its electrons since it transfers them to oxygen (ionic). This is what oxidation, or being oxidized, refers to: the loss of electrons in a chemical reaction. However, it is now recognized that this process can occur even without the presence of oxygen. Oxidation-reduction reactions are therefore referred to as any reaction that involves a transfer of electrons, whether complete of partial, and whether oxygen and hydrogen are involved or not.
3Mg(s) + N₂(g) Mg₃N₂(s) Just as in the last slide when we took a look at the synthesis of magnesium oxide, magnesium is losing electrons; however, this time it is losing electrons to nitrogen in the formation of magnesium nitride. This is still an oxidation reaction since a species (magnesium) is losing electrons. We can therefore say Magnesium has been oxidized even though it has not bonded to oxygen. Reduction is also taking place in this reaction. The nitrogen is being reduced since it gains the electrons that have been transferred to it from the magnesium. Oxidation and reduction actually always take place together. This is therefore a redox reaction.
It might help you to remember the following acronym: OIL RIG OIL: Oxidation Is Loss (of electrons) RIG: Reduction Is Gain (of electrons)
In a way, all chemical reactions can be considered redox reactions since all reactions involve some kind of electron exchange. However, for the purposes of IB Chemistry, we will strictly consider redox reactions to occur only when there is an exchange of electrons that results in a change in the oxidation states of a species. For instance, acid-base neutralization reactions are NOT considered to be redox reactions because all the atoms on the reactant side will end up with the same oxidation state on the product side.
Oxidation States Oxidation state: a value we assign to each atom in a compound that is a measure of the electron control or possession it has relative to the atom when it is alone as a pure element. This enables us to keep track of the relative electron density in a compound and how it changes during a reaction. It is the apparent charge that an atom has in a molecule or ion. + sign means the atom has lost electron control - sign means the atom has gained electron control
When working out the oxidation states of atoms in compounds, it is usually best to assign the oxidation state to the atoms that are easy to predict first, then use rules 3 and 4 to find the more predictable elements by subtraction.
Practice: Assign oxidation states to all the elements in H₂SO₄: Rule #5 states that the usual oxidation state for an element is the same as the charge of its most common ion, therefore H = +1 and O = -2 However, since there are two hydrogens, the total oxidation state of H is +2 And since there are four oxygens, the total oxidation state of O is -8 Rule #3 states that the oxidation states of all the atoms in a neutral (uncharged) compound must add up to zero, therefore: H₂ S O₄ 2(+1) + (S) + 4(-2) = 0 (+2) + (+6) + (-8) = 0 Oxidation state of H is +1, S is +6, and O is -2
Practice: Assign oxidation states to all the elements in SO₃² Rule #5 states that the usual oxidation state for an element is the same as the charge of its most common ion, therefore O = -2 However, since there are three oxygens, the total oxidation state of O is -6 Rule #4 states that the oxidation states of all the atoms in a polyatomic ion must add up to the charge on the ion: S O₃ (S) + 3(-2) = -2 (+4) + (-6) = -2 Oxidation state of S is +4 and oxidation state of O is -2
What is the significance behind these numbers? Well, since the oxidation state is a measure of the electron control that an atom has, it follows that the higher the positive number, the more the atom has lost control over electrons, in other words the more oxidized it is. Likewise, the greater the negative number, the more it has gained electron control, so the more reduced it is. H₂S represents sulfur in its most reduced form (lowest oxidation number) SO₃ and H₂SO₄ represent sulfur in its most oxidized form (highest oxidation number)
A redox reaction is a chemical reaction in which changes in the oxidation states of a species occur. Oxidation (loss of electrons, whether full or partial) occurs when there is an increase in oxidation state of an element. Reduction (gain of electrons) occurs when there is a decrease in oxidation state of an element. Hydrogen has been oxidized (oxidation state increased from 0 to +1) Oxygen has been reduced (oxidation state decreased from 0 to -2)
Practice: Use oxidation states to deduce which species is oxidized and which is reduced in the following reactions: a) Ca(s) + Sn²+(aq) Ca²+(aq) + Sn(s) b) 4NH₃(g) + 5O₂(g) 4NO(g) + 6H₂O(l) Step 1: Assign oxidation states to each atom. Step 2: Exercise the following: Oxidation (loss of electrons, whether full or partial) occurs when there is an increase in oxidation state of an element. Reduction (gain of electrons) occurs when there is a decrease in oxidation state of an element.
Practice: Use oxidation states to deduce which species is oxidized and which is reduced in the following reactions: ANSWERS: a) Ca(s) + Sn²+(aq) Ca²+(aq) + Sn(s) 0 +2 +2 0 Calcium is oxidized, tin is reduced b) 4NH₃(g) + 5O₂(g) 4NO(g) + 6H₂O(l) -3 +1 0 +2-2 +1-2 NH₃ is oxidized because the oxidation state of N increases from -3 to +2. O₂ is reduced because its oxidation state decreases from 0 to -2.
Naming Compounds using Oxidation Numbers Remember the Stock System, which is used to name compounds containing transition metals? Example: copper(ii)bromide The Roman numeral is indicating that this particular copper atom has lost 2 electrons, forming the copper ion, Cu²+. This means the formula unit of this ionic compound is CuBr₂. The naming method is pretty much the same used for naming compounds by their oxidation states.
Practice: Deduce the name of the following compounds using oxidation numbers. a) V₂O₅ b) Ni(OH)₂ c) TiCl₄
Practice: Deduce the name of the following compounds using oxidation numbers. a) V₂O₅ b) Ni(OH)₂ c) TiCl₄ ANSWERS: a) V₂O₅, vanadium(v) oxide O is -2 x 5 = -10, therefore V must be +5 x 2 = +10 b) Ni(OH)₂, nickel(ii) hydroxide OH is -1 x 2 = -2, therefore Ni must be +2 c) TiCl₄, titanium(iv) chloride Cl is -1 x 4 = -4, therefore Ti must be +4
VIDEO Photochromic Lenses - Redox Reactions
Practice: Assign oxidation states to all the elements in the following compounds:
Writing Redox Equations Something called half equations help us write-out redox reactions. Despite oxidation not being able to take place without reduction and vice versa, it is something useful to separate out the two processes from a redox equation and write separate equations for the oxidation and reduction processes. We do this by adding electrons on one side of each of the half equations, (the oxidation half equation and the reduction half equation), to balance the charges. Example: Deduce the two half equations for the following reaction: Zn(s) + Cu²+(aq) Zn²+(aq) + Cu(s)
Writing Redox Equations Example: Deduce the two half equations for the following reaction: Zn(s) + Cu²+(aq) Zn²+(aq) + Cu(s) 0-2 +2 0 Answer: We can see that zinc is being oxidized and copper is being reduced. Oxidation reaction: Zn(s) Zn²+(aq) + 2e- Reduction reaction: Cu²+(aq) + 2e- Cu(s) Electrons are lost, oxidation Electrons are gained, reduction *Note that you include the electrons on the side of the equation that has the ion. **Note that eventually there must be equal numbers of electrons in the two half equations, so that when they are added together the electrons cancel out.
Writing Redox Equations Half equations come in very handy when we need to write a redox equation, but we only have information on which species are involved in the redox reaction and not the overall actual equation. This means we need to put something together and make sure it is balanced for both atoms and charge. A good way to do this is to write the oxidation half equation and then write the reduction half equation. These two are then added together to give the overall reaction. Many of these reactions take place in acidified solutions and we therefore use H₂O and/or H+ ions to balance the half-equations.
EXAMPLE: Write an equation for the reaction in which NO₃ and Cu react together in acidic solution to produce NO and Cu²+.
EXAMPLE: Write an equation for the reaction in which NO₃ and Cu react together in acidic solution to produce NO and Cu²+. Step 1: Assign oxidation numbers and determine who is oxidized and who is reduced. NO₃ (aq) + Cu(s) NO(s) + Cu²+(aq) +5-2 0 +2-2 +2 Cu increased from 0 to +2 so it has been oxidized. Nitrogen decreased from +5 to +2 so it has been reduced. Step 2: Write half equations for oxidation and reduction using rules a) - e) for guidance: Oxidation half: Cu(s) Cu²+(aq) Reduction half: NO₃ (aq) NO(s)
Step 2: Write half equations for oxidation and reduction using rules a) - e) for guidance: Oxidation half: Cu(s) Cu²+(aq) Reduction half: NO₃ (aq) NO(s) Copper is balanced, but oxygen is not. We balance oxygen atoms in particular by introducing water into the equation since we know it is occurring in an acidified solution. Reduction half: NO₃ (aq) NO(s) + 2H₂O(l) Now we balanced the O s, but not the H s. We can balance the H s using H+ ion. Reduction half: NO₃ (aq) + 4H+(aq) NO(s) + 2H₂O(l) Now we balanced the H s. However we now need to balance BOTH of the half equations for charge by adding electrons to the sides of the equation with the more positive charge. Oxidation half: Cu(s) Cu²+(aq) + 2e- Reduction half: NO₃ (aq) + 4H+(aq)+ 3e- NO(s) + 2H₂O(l)
Step 3: Now we need to equalize the number of electrons on each side of the equations by multiplying each appropriately: Oxidation half: Reduction half: 3(Cu(s) Cu²+(aq) + 2e-) 2(NO₃ (aq) + 4H+(aq)+ 3e- NO(s) + 2H₂O(l)) Oxidation half: 3Cu(s) 3Cu²+(aq) + 6e- Reduction half: 2NO₃ (aq) + 8H+(aq)+ 6e- 2NO(s) + 4H₂O(l) Step 4: Put the half equation together and cancel out anything that is the same on both sides. Oxidation half: 3Cu(s) 3Cu²+(aq) + 6e- Reduction half: 2NO₃ (aq) + 8H+(aq)+ 6e- 2NO(s) + 4H₂O(l) FINAL EQUATION: 3Cu(s) + 2NO₃ (aq) + 8H+(aq) 3Cu²+(aq) + 2NO(s) + 4H₂O(l)