CH 117 Spring 2015 Worksheet 19

Chapter 16

1. Define each of the following. These are all phase transitions that are worth memorizing! a). Freezing – liquid to solidb). Melting – solid to liquidc). Boiling (aka evaporation) – liquid to gasd). Condensation – gas to liquide). Sublimation – solid to gasf). Deposition – gas to solid

2. How does changing either the temperature or the volume affect entropy?Entropy increases with increasing temperature and with increasing volume direct relationship!

3. Describe how molecular complexity and molecular weight affect entropy. Increasing molecular complexity (the number of bonds in a compound) will increase the entropy).Increasing the molecular weight will increase entropy.

4. Calculate K for the following reaction given that G = -174.8 kJ/mol: CO (g) + 2 H2 (g) CH3OH (g). Is this reaction spontaneous or non-spontaneous? Product-favored or reactant-favored?The formula that relates K to G: G = -RTlnKRearrange formula to solve for K: K = e-G/RT

Plug and chug! G needs to be in J to use the formula, so multiply the given value by 1000 before starting.K = e- (-174800 J)/(8.314 * 298 K) = 4.37 x 1030

G is negative, so reaction is spontaneous. K is a lot bigger than 1, so reaction is product-favored.

Chapter 17

5. List a few rules for determining the oxidation number of an atom.Anything in its elemental state has an oxidation number of 0.For any monatomic ion, its charge is equal to its oxidation number.Oxygen has an oxidation state of -2 unless in H2O2.Hydrogen has an oxidation state of +1. All the oxidation states in a compound must add up to the overall charge on that compound.

6. Define each of the following terms. OIL RIG or LEO says GERa). Oxidation – loss of electronsb). Reduction – gain of electronsc). Reducing agent – what was oxidized, reduces something elsed). Oxidizing agent – what was reduced, oxidizes something else

CH 117 Spring 2015 Worksheet 19

7. Identify which element was oxidized, which was reduced, the reducing and oxidizing agents, and the half-reaction for the following.a). 2 Cr+ + Sn4+ Cr3+ + Sn2+

First step is to assign oxidation states to all elements.Cr+: +1Sn4+: +4Cr3+: +3Sn2+: +2Cr+ is oxidized (lost electrons) acts as reducing agentSn4+ is reduced (gained electrons) acts as oxidizing agentOxidation half-reaction: Cr+ Cr3+ + 2e-

Reduction half-reaction: Sn4+ + 2e- Sn2+

b). 3 Hg2+ + 2 Fe (s) 3 Hg2 + 2 Fe3+

Hg2+: +2Fe (s): 0Hg2: 0Fe3+: +3Hg2+ is reduced acts as oxidizing agentFe (s) is oxidized acts as reducing agentOxidation half-reaction: 2 Fe (s) 2 Fe3+ + 6e-

Reduction half-reaction: 3 Hg2+ + 6e- 3 Hg2

c). 2 As (s) + 3 Cl2 (g) 2 AsCl3

As (s): 0Cl2 (g): 0As in AsCl3: +3Cl in AsCl3: -1As (s) is oxidized acts as reducing agentCl2 (g) is reduced acts as oxidizing agentOxidation half-reaction: 2 As (s) 2 As3+ + 6e-

Reduction half-reaction: 3 Cl2 (g) + 6e- 6 Cl-

8. Distinguish between an electrolytic and a voltaic or galvanic cell. An electrolytic cell is one that is nonspontaneous needs the addition of electricity to make the reaction go forward, G is positive, Ecell is negative.A voltaic/galvanic cell is one that occurs spontaneously without the input of electricity, G is negative, Ecell is positive.

9. Draw a typical galvanic cell using the following half-reactions and cell potentials. Write the overall reaction, calculate the Ecell for the reaction, and label all of the necessary parts of the cell.

Al3+ + 3e- Al (s) E = -1.66 oxidationPb2+ + 2e- Pb (s) E = -0.36 reduction

To write the overall reaction the number of electrons must be balanced and cancel each other out. We have to multiply the top reaction by 2 and the bottom one by 3

CH 117 Spring 2015 Worksheet 19

for this to happen. We also need to flip the first reaction around since it is given as a reduction but will only occur as an oxidation.

2 Al (s) 2 Al3+ + 6e- E = 1.663 Pb2+ + 6e- 3 Pb (s) E = -0.36

Now, add the two half-reactions together to get the overall reaction: 2 Al (s) + 3 Pb2+

2 Al3+ + 3 Pb (s) Ecell = Eox + Ered = 1.66 – 0.36 = 1.30 V