Lab Electrochemistry (Parts II and IV)

Lab Assignment Experiment 10: Electrochemistry (Parts II and IV) Part II: Electrolysis: Qualitative Observations Watch the video Electrochemistry Part II, which is posted on Courseweb, The aim of this part of the report is to identify the halfreactions occurring, based on the observed color changes and appearance of gases or solid deposits. Remember that phenolphthalein is pink in strongly basic solution and colorless in neutral and acidic solution. Consult a table of standard electrode potentials as part of the analysis. Part II - 1 1. Report observations (a)–(c) (see the procedure for Part II - 1). State for which electrode (anode or cathode) each observation applies. (a)Cathode has a little more bubbling than anode. It is bubbling twice as fast. (b) After the phenolphthalein indicator is added, only the cathode section turned pink. The pink is coming from the cathode when it is electrolyzed. (c) The pink disappeared after some stirring. Solution becomes colorless. 2. On the basis of your observations, suggest the half-reaction occurring at the cathode (-). Show clearly that each of your observations is consistent with this half-reaction. 1 The bubbling from the cathode comes from the release of hydrogen gas. When phenolphthalein is added, the cathode section turns pink because of the presence of OH-. And when the solution is stirred, the pink disappears because the ions become more dispersed in the solution (instead of being concentrated). 3. Similarly, suggest the half-reaction occurring at the anode (+). Show clearly that each of your observations is consistent with this half-reaction. The bubbling from the anode comes from the release of oxygen. There was less bubbling in the anode compared to the cathode because oxygen is heavier than hydrogen gas. The anode section didn’t turn pink when phenolphthalein was added because the region is more acidic due to the H+ present. Phenolphthalein only turns pink when the solution is basic. After stirring, the solution becomes colorless again because the ions were more dispersed. 2 Part II - 2 4. Report observations Some bubbling coming off of anode (graphite) and some coming off of cathode (copper). Bubbling in cathode is twice as fast compared to anode. When phenolphthalein is added, the solution turns pink. But after some stirring, the pink disappears. The pink actually comes from the cathode. Part II - 3 5. Report observations (a)–(c). (a) Bubbling coming from cathode (graphite) but not really from the anode. Bottom is turning blue. The top has some bubbling, but it is also getting a little cloudy. (b) Entire solution turns light blue. Formation of precipitate. (c) Solution turns purple-pink after the addition of phenolphthalein. 3 Observation (c) from Part II-3 isn’t really the same observation as (c) from Part II-1 because the solution became more of a purple-pink after the addition of phenolphthalein. And this is because the solution in Part II-3 was slightly blue before. 6. On the basis of your observations, suggest the half-reaction occurring at the cathode (-). Show clearly that each of your observations is consistent with this half-reaction. There was bubbling coming from the cathode and the bubbling was actually caused by the hydrogen gas. And when phenolphthalein was added to the solution, it did turn pink because of the OH - present in the solution. Phenolphthalein only turns pink when the solution is basic. 7. Similarly, suggest the half-reaction occurring at the anode (+). Show clearly that each of your observations is consistent with this half-reaction. 4 Ther e wasn’t really any bubbling coming out of the anode. But during the electrolysis, the bottom of the solution started to turn light blue, which was coming from the anode (copper). A precipitate was also formed because some of the Cu2+ probably reacted with OH(released from the cathode) to form Cu(OH) 2 which is insoluble in water. Summary 8. Explain why there is a change in chemistry when Cu replaces graphite as the anode but not when Cu replaces graphite as the cathode. When Cu replaces graphite as the anode, the metal becomes oxidized during the electrolysis. Therefore, Cu2+ would be released from the metal, which causes the solution to turn blue. But when Cu is the cathode, the metal is being reduced, so it is gaining electrons and no Cu2+ is being released. 5 Part IV: A Practical Battery: The Zinc-Carbon“Dry Cell”: A Galvanic Cell The dry cell was developed in the late 19th century and was one of the first commercially available single use batteries. The anode and cathode reactions for the dry cell are given below. (anode) (cathode) Zn(s) → Zn2+(aq) + 2 e2 NH4+(aq) + 2 MnO2(s) + 2e- → Mn2O3(s) + H2O + 2NH3(aq) One disadvantage of the dry cell is that the ammonia produced at the cathode corrosive to the zinc can that serves as the anode, which leads to leaks. The alkaline battery, which was introduced in the 1950s alleviates this problem by using less corrosive materials. 1. Based on the given half reactions for the dry cell: a. Write the overall cell reaction 2 NH4+(aq) + 2 MnO2(s) + Zn(s) → Zn2+(aq) + Mn2O3(s) + H2O + 2NH3(aq) b. Calculate the standard cell potential for the dry cell. 6 2. A CHALLENGE. You have an opportunity to improve the voltage of the dry cell by replacing the zinc anode with another metal. You have the following metals available to you: Al, Cu, Mg, and Pb. Determine which metal anode would result in the highest voltage for a dry cell. Support you choice with relevant standard cell potential calculations. The metal anode that would result in the highest voltage for a dry cell is Mg because the Eº cell value is 3.88 V, which is greater than all the other Eº cell. Grading rubric Score Assigned Aspect of Report Excellent Good Fair Poor Quality and completeness of Observations 2 1.5 1 0.5 Quality of Conclusions (Including Accuracy of HalfReactions) 3 2.25 1.5 0.75 1 0.75 0.5 0.25 Part II Part IV Question IV - 1 Question IV - 2 2 1.5 1 0.5 Overall Report Quality 1 0.75 0.5 0.25 Total Report /9 7