+6 -2 +3 -1 -1

1. a) Cr₂O₇^2- + 3C₂H₅OH + 8H⁺ → 2Cr³⁺ + 3CH₃CHO +7H₂O

You first need to determine the oxidation half reaction. The Cr oxidation number decreases from +6 to +3, so it is reduced. The C₂H₅OH must be oxidized as the oxidation number for C increases from –2 to -1. Balance the half reaction in acid.

C₂H₅OH → CH₃CHO + 2H⁺ + 2e^-

b) The reduction in orange colour of Cr₂O₇^2- is proportional to the amount of alcohol C₂H₅OH in the breath sample.

2. Oxidizing Agent Reducing Agent

Au³⁺ + 3e^- → Au

Sn²⁺ + 2e^- → Sn

Zn²⁺ + 2e^- → Zn

Ti²⁺ + 2e^- → Ti

a) Ti Zn Sn Au

b) Ti + Zn²⁺ → Tⁱ²⁺ + Zn

3. 8NO₃^- + 2H⁺ + 3H₂Te → 8NO + 4H₂O + 3TeO₄^2-

a) Oxidation: H₂Te + 4H₂O → TeO₄^2- + 10H⁺ + 8e^-

b) Reduction: NO₃^- + 4H⁺ + 3e^- → NO + 2H₂O

4. Reject 14.21 mL and average the other two to get 12.66 mL Fe²⁺. Balance the redox reaction.

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O x 1

Fe²⁺ → Fe³⁺ + 1e^- x 6

Cr₂O₇^2- + 6Fe²⁺ +14H⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

0.02500 L 0.01266 L

? M 0.100 M

[Cr₂O₇^2-] = 0.01266 L Fe²⁺ x 0.100 moles x 1 mole Cr₂O₇^2-

1 L 6 moles Fe³⁺

0.02500 L

= 0.00844 M

5. a) Ni²⁺ + 2e^- → Ni_(s)

b) 0 v

6. Use the Reduction Chart to determine a spontaneous reaction and the balance it.

The top reaction is written forward and the bottom one is reversed

2Cr³⁺ + 7H₂O → Cr₂O₇^2- + 14H⁺ + 6e^- x 5

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O x 6

6MnO₄^- + 48H⁺ + 10Cr³⁺ + 35H₂O → 5Cr₂O₇^2- + 70H⁺ + 6Mn²⁺ + 24H₂O

6MnO₄^- + 10Cr³⁺ + 11H₂O → 5Cr₂O₇^2- + 22H⁺ + 6Mn²⁺

0.02855 L 0.0100 L

0.0500 M ? M

[Cr³⁺] = 0.02855 L MnO₄^- x 0.0500 moles x 10 mole Cr³⁺

1 L 6 moles MnO₄^-

0.0100 L

= 0.238 M

7. Both Na and Ca oxidize in water and water reduces. See chart for spontaneous reaction.

a) H₂O + 2e^- → H₂ + 2OH^-

b) Na and Ca

c) Ca(OH)_2(s) has low solubility whereas NaOH has high solubility.

8. Ag @ 0.80 v is either the anode or cathode of this cell.

To find out which, add 0.93 v and then subtract 0.93 v and see what you land on.

On the subtraction you get -0.13 v, which is Pb. This means Ag is the cathode (higher) and Pb is the anode.

a) Pb

b) Pb(NO₃)₂

c) NaNO₃ or any other soluble salt that does not react with Pb²⁺.

9. H₂O₂, which is on both sides of the reduction table, must react with one solution and not the other.

From their placement on the chart you can see that H₂O₂ reacts with Cr₂O₇^2- and not Cr³⁺.

Circle them all with a pencil on the chart and look!!!

a) Cr₂O₇^2- turns into Cr³⁺, which is green. The picture on the right tells you that Cr³⁺ is green, hello….

b) O_2(g) is produced in the oxidation of H₂O₂.

c) Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O x 1

H₂O₂ → O₂ + 2H⁺ + 2e^- x 3

Cr₂O₇^2- + 8H⁺ + 3H₂O₂ → 2Cr³⁺ + 3O₂ + 7H₂O

10. From the electrons going from anode to cathode we know that Ga is the anode and Ti is the cathode.

a) Ga → Ga³⁺ + 3e^- -0.56 v

Ti⁺ + 1e^- → Ti x

+0.22

Solving –0.56 + x = 0.22

x = 0.78 v

b) Ti⁺ is the oxidizing agent.

11. Anode: 2I^- → I₂ + 2e^-

Cathode: 2H₂O + 2e^- → H₂ + 2OH^-

12. The reaction of acidified KMnO₄ and SnF₂ needs to be found on the reduction chart.

You basically have four ions to look at, as the typical reaction between the two metals is non-spontaneous in this case. The word acidified should clue you in to the MnO₄^- + 8H⁺ + 5e^- reaction.

Your pencil is your friend! Circle all four ions. There is only one spontaneous reaction- see below.

F₂ + 2e^- → 2F^-

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Sn⁴⁺ + 2e^- → Sn²⁺

K⁺ + 1e^- → K

a) MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O x 2

Sn²⁺ → Sn⁴⁺ + 2e^- x 5

2MnO₄^- + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 8H₂O + 5Sn⁴⁺

b) ? L 0.0350 L

0.0150 M 0.0150 M

0.0350 L Sn²⁺ x 0.0150 mole x 2 moles MnO₄^- x 1 L = 0.0140 L

1 L 5 moles Sn²⁺ 0.0150 moles

2 e^-

2 e^-

13.

14. You can find the spontaneous redox reaction on the reduction chart in acid solution.

Convert it to basic solution and you are done like dinner.

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O x 2

2I^- → I₂ + 2e^- x 5

2MnO₄^- + 16H⁺ + 10I^- → 2Mn²⁺ + 8H₂O + 5I₂

The reducing agent is the species that oxidizes I^-

15. Al³⁺ + 3e^- → Al_(s)

Impure Anode

Cu Zn Au

Pure Cu

Cathode

Power Source

- +

16.

Cathode Reaction Anode Reactions

Cu²⁺ + 2e^- → Cu_(s) Cu → Cu²⁺ + 2e^- -0.34 v

Zn → Zn²⁺ + 2e^- +0.76 v

A voltage is supplied that would be enough to cause the oxidation of Cu.

This would also oxidize Zn, which is a stronger reducing agent, but not the Au, which is a weaker reducing agent.

At the cathode the voltage is kept low enough so that only the Cu²⁺ is reduced, as it is the strongest oxidation agent.

17. Again you need to find the reaction. The reactant possibilities are: Fe²⁺ SO₄^2- O₂ H₂O H⁺

Anytime you have a reactant in air its most likely O₂.

Aqueous tells you that water or H⁺ @ 10^-7 M might be involved.

Find them on the reduction table and look for a spontaneous reaction.

½ O₂ + 2H⁺ + 2e^- → H₂O

Fe³⁺ + 1e^- → Fe²⁺

Write a balanced reaction.

½ O₂ + 2H⁺ + 2e^- → H₂O

Fe²⁺ → Fe³⁺ + 1e^- x 2

½ O₂ + 2H⁺ + 2Fe²⁺ → H₂O + 2Fe³⁺