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We have been discussing proton transfer reactions...acids donate protons and bases accept protons. In this chapter we are going to be discussing electron transfer reactions. The following questions are preparation to our discussion of oxidation-reduction chemistry. You may want to have your book available as you answer these questions. NOTE: Charge is very important in electron transfer reactions because an electron has a -1 charge!

1. Consider the reaction the following reactions

S(s) + O₂(g) ----> SO₂(g)

2SO₂(g) + O₂(g) ----> SO₃(g)

CH₄(g) + 2O₂(g) ----> CO₂(g) + 2H₂O(g)

2HgO(s) ---> 2Hg(l) + O₂(g)

a) There are four reactions shown above, what is common to the first three reaction? Explain

b) What is different about the fourth reaction compared to the first three?

2. In Table 4.3 in Chapter 4 on page 155 are a set of rules for assigning oxidation numbers. (See Sample Problem 4.9 to see how to use oxidation numbers.) Those rules are;

A. Any atom in a pure element or molecule such as O₂, H₂, N₂, P₄, or Zn has an oxidation number of zero.

B. For ions consisting of a single atom, the oxidation number is equal to the charge on the ion; Cl^- has an oxidation number of -1 and Zn²⁺ has an oxidation number of +2.

C. Fluorine is always -1.

D. Chlorine, bromine and iodine are -1 except in compounds with fluorine or oxygen (both of which have higher electronegativities).

E. The oxidation number of hydrogen is +1 except when bound to a metal (hydrogen is more electronegative).

F. The oxidation number of oxygen is usually -2 (except in H₂O₂ when oxygen has an oxidation number of -1)..

G. The sum of the oxidation numbers in a neutral compound must be zero.

H. The sum of the oxidation numbers for a polyatomic ion must equal the ion charge.

Assign the oxidation numbers to the following elements in the following compounds:

a) H₂O

b) SO₂

c) NH₃

d) CaCl₂

e) KMnO₄

3. Knowing the oxidation number for an element in a substance allows us to determine the change in the number of electrons. For example in the reaction

Zn(s) + Cl₂(g) ----> ZnCl₂(s)

Zinc has an oxidation state of 0 (zero) in the reactants and an oxidation state of +2 in the products. So zinc has lost two electrons when zinc reacts with chlorine to form zinc chloride. Also chlorine, Cl₂, has an oxidation state of 0 in the reactants and an oxidation state of -1 in the products. Each chlorine has gained one electron when it reacts with zinc. Notice in the balanced chemical equation the number of electrons lost by zinc is exactly equal to the number of electrons gained by chlorine.

How many electrons are lost or gained by each of the elements in the following reactions.

a) 2H₂(g) + O₂(g) ----> 2H₂O(l)

b) Mg(s) + 2HCl(aq) ----> MgCl₂(aq) + H₂(g)

c) Zn(s) + Cu²⁺(aq) ---> Zn²⁺(aq) + Cu(s)

4. In the last equation in Q3 (Zn(s) + Cu²⁺(aq) ---> Zn²⁺(aq) + Cu(s)) the overall reaction can be broken into two half-reactions that reflect which element lost electrons and which element gained electrons. If we divide the overall reaction into two half-reactions based on the elements we would have

first half-reaction : Zn(s) ---> Zn²⁺(aq)

second half-reaction : Cu²⁺(aq) ----> Cu(s)

Notice that the number of elements is balanced for each of these half-reactions, but the charge is not. For the first half-reaction the charge is 0 (zero) on the reactants side and +2 on the products side. To balance the charge in the half-reaction we must add two electrons (2e^- to the right side of the half-reaction). When we look at the second half-reaction it is clear two electrons must be added to the left side of the half-reaction.

first half-reaction : Zn(s) ---> Zn²⁺(aq) + 2e^-

second half-reaction : 2e^- + Cu²⁺(aq) ----> Cu(s)

Check out that when we add the first and second half-reactions together the electrons cancel and the overall reaction is the same as we had when we started. Being able to recognize the half-reactions given the overall reaction is important in Chapter 21. Lets try a few examples;

a) Fe(s) + 2Ag ⁺(aq) ---> Fe²⁺(aq) + 2Ag(s)

first half-reaction (iron)

second half-reaction

b) 2Al³⁺(aq) + 3Ca(s) ---> 3Ca²⁺(aq) + 2Al(s)

first half-reaction (aluminum)

second half-reaction

c) 10Br^-(aq) + 2MnO₄^-(aq) + 16H⁺(aq)--> 2Mn²⁺(aq) + 8H₂O(l) + 5Br₂(l)

first half-reaction

second half-reaction

In class we'll discuss more of these interesting aspects of electron transfer reactions. The important idea to notice at this point, is that being able to separate an overall reaction into half-reactions, and that there are electrons being transferred from one half-reaction to the other, that it might be possible to have those electrons flow through a circuit to do some work. For example, we could get the electrons to power a calculator, a watch, or a pacemaker.

4. Is there anything about the questions that you feel you do not understand? List your concerns/questions.

5. If there is one question you would like to have answered in lecture, what would that question be?