124f chemistry

Experiment 9: Electrochemistry – Voltaic Cells Name : JiHun Kim Date : 07/18/2018 Source : Dr. M. Ansari’s lab manual posted on Canvas Goal of Experiment : Conduct an experiment using metal and check the results obtained through oxidation-reduction. Introduction : Each half-reaction has a standard potential (E°, in volts) associated with it, which is a measure of how likely the reaction is to occur spontaneously under standard conditions. By convention, these half-reactions are expressed as reductions and the standard reduction potentials tabulated to compare the relative oxidizing strength of various species. The copper(II) causes the oxidation of the zinc and is therefore, referred to as the oxidizing agent. The equation of net redox reaction is like this: Oxidation Half-Reaction: Zn (s) à Zn2+ (aq) + 2 e 28574281305Reduction Half-Reaction: Cu2+ (aq) + 2 e- à Cu (s) Net Redox Reaction: Zn (s) + Cu2+ (aq) à Zn2+ (aq) + Cu (s) The standard reduction potentials of the zinc and copper half-reactions are shown below: Cu2+ (aq) + 2 e- D Cu (s) E° = + 0.34 V Zn2+ (aq) + 2 e- D Zn (s) E° = - 0.76 V A voltaic (or galvanic) cell uses a spontaneous redox reaction to produce an electric current. In a voltaic cell, the half-reactions are separated in different compartments called half-cell, Figure 1. Zn/Cu voltaic cell schematic: And student will use equation of : E°cell = E°cathode - E°anode , Ecell= E°cell-RTnFlnQ , (R: gas constant, 8.314 J mol-1 K-1, T: temperature in kelvin, n: number of electrons transferred, F: Faraday constant, 9.65 x 104 J V-1 (mole e- )-1). Using the Zn/Cu redox reaction as an example: Zn (s) + Cu2+ (aq) à Zn2+ (aq) + Cu (s), Q = [Zn2+][Cu2+] and lnQ = ln[Zn2+] - ln[Cu2+]. If [Zn2+] is held constant at 1.0 M while [Cu2+] is varied, the Nernst Equation reduces to: Ecell= RTnFln[Cu2+] + E°cell Chemicals and Equipment : Procedure : Part I Place 24-well plate on a paper towel and label it like this figure: Obtain three strips of each metal and put in the well-plate. Watch how it changes and record. Part II i. Zn (s) | Zn(NO3)2 (aq, 1.0 M) || Pb(NO3)2 (aq, 1.0 M) | Pb (s) ii. Zn (s) | Zn(NO3)2 (aq, 1.0 M) || FeSO4 (aq, 1.0 M) | Fe (s) iii. Zn (s) | Zn(NO3)2 (aq, 1.0 M) || AgNO3 (aq, 1.0 M) | Ag (s) iv. Zn (s) | Zn(NO3)2 (aq, 1.0 M) || Cu(NO3)2 (aq, 1.0 M) | Cu (s) This voltaic diagrams will perform each time. Place 24 well-plate and prepare each metals and solutions in it. Put labquest and electric wire, and put each metal into solutions. And record each voltage. Results : Part I: Redox Reactions Cu(NO3)2 Pb(NO3)2 Zn(NO3)2 MgCl2 Cu X Changed to light pink color Changed to light pink color N/R Pb Changed to black color, Pb perfectly crashed X N/R N/R Zn Changed to black color , precipitation Changed to dark gray, precipitation X N/R Mg Changed to black , Bubble, Mg crashed and perfectly disappeared Rusted, precipitation Changed little bit, dark gray X Well # Net Reaction Theoretical E°cell (V) Spontaneous? 1 Cu (s) + Pb2+ (aq) -> Cu2+ (aq) + Pb (s) -0.13 – (+0.34) = -0.47 No 2 Cu (s) + Zn2+ (aq) -> Cu2+ (aq) + Zn (s) -0.76 – (+0.34) = -1.1 No 3 Cu (s) + Mg2+ (aq) -> Cu2+ (aq) + Mg (s) -2.37 – (+0.34) = -2.71 No 4 Pb (s) + Cu2+ (aq) ->Pb2+ (aq) + Cu (s) +0.34 + (+0.13) = +0.47 Yes 5 Pb (s) + Zn2+ (aq) ->Pb2+ (aq) + Zn (s) -0.76 + (+0.13) = -0.63 No 6 Pb (s) + Mg2+ (aq) ->Pb2+ (aq) + Mg (s) -2.37 + (+0.13) = -2.24 No 7 Zn (s) + Cu2+ (aq) -> Zn2+ (aq) + Cu (s) +0.34 + (+0.76) = 1.1 Yes 8 Zn (s) + Pb2+ (aq) -> Zn2+ (aq) + Pb (s) +0.13 + (+0.76) = 0.89 Yes 9 Zn (s) + Mg2+ (aq) -> Zn2+ (aq) + Mg (s) -2.37 + (+0.76) = -1.61 No 10 Mg (s) + Cu2+ (aq) -> Mg2+ (aq) + Cu (s) +0.34 + (+2.37) = 2.71 Yes 11 Mg (s) + Pb2+ (aq) -> Mg2+ (aq) + Pb (s) -0.13 + (+2.37) = 2.24 Yes 12 Mg (s) + Zn2+ (aq) ->Mg2+ (aq) + Zn (s) -0.76 + (+2.37) = 1.61 Yes Part II: Reduction Potentials 1.       Cu. E°cell = +0.34 + +.76 = 1.1V Pb. E°cell = -.13 + .76 = .63V Fe. E°cell = -.44 + +.76 = .32V Ag. E°cell = .80 + +.76 = 1.56V Voltaic cell diagram Voltage Zn (s) | Zn(NO3)2 (aq, 1.0 M) || Cu(NO3)2 (aq, 1.0 M) | Cu (s) 1.015 V Zn (s) | Zn(NO3)2 (aq, 1.0 M) || Pb(NO3)2 (aq, 1.0 M) | Pb (s) 0.604 V Zn (s) | Zn(NO3)2 (aq, 1.0 M) || FeSO4 (aq, 1.0 M) | Fe (s) 0.314 V Zn (s) | Zn(NO3)2 (aq, 1.0 M) || AgNO3 (aq, 1.0 M) | Ag (s) 1.516 V Half-Reaction (Reduction) Theoretical Eo (V) Experimental Eo (V) % Error Cu2+ (aq) + 2e- -> Cu (s) 1.1V 1.015 7.7% Pb2+ (aq) + 2e- -> | Pb (s) .63V .604 4.13% Fe2+ (aq) + 2e- ->Fe (s) .32V .314 1.88% Ag+ (aq) + 1e- -> Ag (s) 1.56V 1.516 2.82% Discussion: In this experiment, part 1 shows that the visible change of Cu,Pb,Zn,Mg with each solution Cu(NO3)2, Pb(NO3)2, Zn(NO3)2, MgCl2. MgCl2 doesn’t react with any metals because most strong oxidizing metal. As result table 1 shows, if one solution has stronger oxidizing metal, it doesn’t react with any other metals. And table 2 shows that the voltaic cell diagram and voltage. Zn (s) | Zn(NO3)2 (aq, 1.0 M) || AgNO3 (aq, 1.0 M) | Ag (s) has strongest voltage, and Zn (s) | Zn(NO3)2 (aq, 1.0 M) || FeSO4 (aq, 1.0 M) | Fe (s) has weakest voltage. That’s because Ag is the strongest oxidizing agent, so that means Ag has the strongest voltage.