Determining Standard Reduction Potentials, Equilibrium Constants and Investigating a Lead-Acid Electrolytic Cell

Topics: Electrochemistry, Redox, Cathode Pages: 4 (1344 words) Published: February 15, 2010
Determining Standard Reduction Potentials, Equilibrium Constants and Investigating a Lead-Acid Electrolytic Cell Purpose
Experimental Methods
All procedures were followed according to the lab manual (experiment 10 – Electrochemistry Laboratory). Data and Observations
Part A:
Grams of FeSO4 used: 0.759
Voltage during voltmeter and battery check: 9.23V
Table 1. Electrochemical Cells Data
Part B:
Table 2. Lead-Acid Battery
Results and Calculations
In order to make 5ml of 1M FeSO4(aq) (molar mass = 151.92), we calculated the needed grams according to c= n/V; 1 = n/(0.005), n = 0.005. Since n = m/M, 0.05X151.92 = 0.7596g. Part A:
Cu2+ (aq) + Zn (s)  Cu (s) + Zn2+ (aq) E°cell = 1.06 V Cu2+ (aq) + 2e-  Cu(s) E°red = 0.34 V (known standard reduction potential of copper) 0.34 V – 1.06 V
E°red (Zn) = -0.72 V
E° = RT/nF · lnK; n = 2 as 2 electrons were transferred from Zn to Cu 1.06 = (8.314 · 298)/(2 · 96485) · lnK
K = 7.16 · 1035
Zn2+ (aq) + Mg (s)  Zn (s) + Mg2+ (aq), E°cell = 0.66 V K = 2.11 · 1022
Zn (s) + Fe2+ (aq)  Zn2+ (aq) + Fe (s) E°cell = 0.96 V K = 2.97 · 1032
Cu2+ (aq) + Mg (s)  Cu (s) + Mg2+ (aq) E°cell = 1.72 V K = 1.51 · 1058
E°red(Mg)= 0.34 – 1.72 = -1.38V
K = 1.72 · 1016
E°red(Fe) = 0.34 – 0.48 = -0.14
Mg (s) + Fe2+ (aq)  Fe (s) + Mg2+ (aq) E°cell = 1.28 V K = 1.98 · 1043
Part B:
Cathode: PbO2 (s) + HSO4- (aq) + 3H+ (aq) + 2e-  PbSO4 (s) + 2H2O (l) Anode: Pb(s) + HSO4- (aq)  PbSO4 (s) + H+ (aq) + 2e-
Overall: PbO2 (s) + Pb(s) + 2HSO4- (aq) + 2H+ (aq)  2PbSO4 (s) + 2H2O (l) When an external power source applies energy to the cell, the reverse reaction occured, thus regenerating the reactants. 2PbSO4 (s) + 2H2O (l)  PbO2 (s) + Pb(s) + 2HSO4- (aq) + 2H+ (aq) Discussion and Conclusion

In part A, the large equilibrium constants that were calculated (K>106) tell us that for each set of reactions, the equilibrium lies far to the right (product side)...
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