Question

9. In Part 2 of this experiment, you will observe the reaction of aluminum in an aqueous CuCl, solution. a) Given the following half-reactions, predict if aluminum will dissolve in a CuCl, solution. A13+ (aq) + 3e + Al(s) E = - 1.70 V Cu2+ (aq) + 2e - → Cu (s) E = 0.34 V b) The more positive the cell potential the more spontaneous the reaction will be. What precautions should you take in performing this experiment if E is positive and large? 10. In Part 3 of the experiment you will observe the reaction of copper metal in an aqueous AgNO3 solution. Given the following half-reactions, predict if copper will dissolve in AgNO, (aq). Ag+ (aq) + e- → Ag (5) Cu2+ (aq) + 2e- Cu (s) E = 0.80 V E-0.34 V 11. In Part 4 of this experiment, you will construct an electrolytic cell in which copper metal is oxidized to produce Cu2+ ions in solution and H ions are removed from solution to produce hydrogen gas. Cu (s) → Cu2+ (aq) + 2 e- E = - 0.34 V 2 H+ (aq) + 2e → H2 (8) E-0.0 V a) Write the overall reaction and determine the corresponding cell potential. b) Is this reaction spontaneous? Explain why or why not. c) How can we get this reaction to occur? d) To maximize the current, why must the burette be placed about a half inch above the bottom of the beaker? e) What will you use to clean the copper electrodes? Why is it important to use clean copper electrodes in this experiment? nr metal dissolved in the reaction is

Answer 1

Dear Student, solved the first two question, post multiple question to get answer to problem 11 containing multiple sub-parts.

Question 9:

a) Yes,aluminium will dissolve in CuCl2 solution.

E0(reaction) = E0(oxidation) + E0(reduction) = 1.70 + 0.34 = 2.04 V

b) Since the E0(reaction) is very high, hence the reaction will be quite vigorous producing a lot of heat. The liquid may splash out of the container which can cause accidents. Better to add the mixture slowly in order to avoid accidents

Question 10:

Yes, the copper will dissolve, oxidizing himself and reducing the Ag+ to Ag. Since E0(reaction) is greater than zero

E0(reaction) = E0(oxidation of copper) + E0(reduction of silver) = -0.34 + 0.80 = 0.46 V

Note - Post any doubts/queries in comments section.