CHEM 125 LAB 5.docx

# The following shows the calculations using the

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for the three half-cells of Cu, Fe, and Mg. The following shows the calculations using the formula Ecell = Ecathode - Eanode: (both of these values from the table of reduction potentials). All of the electrodes that were touching the red were the cathodes, and the ones touching the black were the anodes. Cu/Zn Ecell = 0.34 -(-0.76) = 1.1V Fe/Zn Ecell = 0.77-(-0.76) = 1.53V Mg/Zn Ecell = (-0.76)-(-2.37) = 1.61V Mg/Cu Ecell = (-2.37)-(0.34) = 2.71V Mg/Fe Ecell = (-0.44)-(-2.37) = 1.93V Cu/Fe Ecell = (0.34)-(-0.44) = 0.78V These potentials are very far off from our measured voltages with the multimeter, which means there must be some inaccuracy with equipment or usage. These reactions were done to identify the direction of electron flow. Zn (s) + CuSO 4(aq) ➔ Zn(NO 3 ) 2(aq) + Cu (s) Anode = Zn Cathode = Cu 2+ Electrons flow from Zn to Cu 2+ Mg (s) + Zn(NO 3 ) 2(aq) ➔ MgSO 4(aq) + Zn (s) Anode = Mg Cathode = Zn 2+ Electrons flow from Mg to Zn 2+ Mg (s) + CuSO 4(aq) ➔ MgSO 4(aq) + Cu (s) Anode = Mg Cathode = Cu 2+ Electrons flow from Mg to Cu 2+ Zn (s) + FeSO 4(aq) ➔ Zn(NO 3 ) 2(aq) + Fe (s) Anode = Zn Cathode = Fe 2+ Electrons flow from Zn to Fe 2+ Mg (s) + FeSO 4(aq) ➔ MgSO 4(aq) + Fe (s) Anode = Mg Cathode = Fe 2+ Electrons flow from Mg to Fe 2+ Part 2: Dilutions of Copper Sulfate 1 CuSO4 (1 M) w/Cu electrode + CuSO4 (0.001 M) w/Cu electrode = 0.041 V 2 Zn(NO3)2 (0.1M) w/ Zn electrode + CuSO4 (0.01 M) w/Cu electrode = 0.673 V 3 Zn(NO3)2 (0.1M) w/ Zn electrode + CuSO4 (0.001 M) w/Cu electrode = 0.605 V 4 Zn(NO3)2 (0.1M) w/ Zn electrode + CuSO4 (0.0001 M) w/Cu electrode = 0.578 V 5 Zn(NO3)2 (0.1M) w/ Zn electrode + CuSO4 (0.1 M) w/Cu electrode = 0.650 V 6 Zn(NO3)2 (0.1M) w/ Zn electrode + CuSO4 (unknown) w/Cu electrode = 0.580 V

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