ma Quiz 7 Key

Ma Quiz 7 Key - Chemistry 106 Recitation 310 Name 1 Given the following reactions Pb2(aq 2e hPb(s E=-0.126 V cathode Cd2(aq 2e hCd(s)E=-0.403 V

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Chemistry 106 Recitation 3- 10 Name: Section: 1. Given the following reactions : Pb 2+ (aq) + 2e- h Pb (s) E ° = -0.126 V cathode Cd 2+ (aq) + 2e - h Cd (s)      E ° = -0.403 V anode a. Write the balanced chemical reaction for the cell Pb 2+ (aq) + Cd (s)   Pb (s) + Cd 2+ (aq) b. Sketch the voltaic cell for the reaction given in question 2. On your diagram indicate the flow of electrons, the direction of the flow of cations and anions, label the electrodes and solutions (indicate anions and cations), and also label which electrode serves as the cathode and the anode. Show the direction of electron flow with arrows and all of the chemical species in the reaction.
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Chemistry 106 Recitation 3- 10 This is correct only for ours we have Cd solid on the left (anode position) with Cd 2+ coming off in solution . The Pb 2+ (aq) is is solution in the cathode beaker and as electrons flow to it the lead ion becomes lead solid. c. Calculate the E
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This note was uploaded on 04/07/2011 for the course CHEM 106 taught by Professor Herron,s during the Winter '08 term at BYU.

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Ma Quiz 7 Key - Chemistry 106 Recitation 310 Name 1 Given the following reactions Pb2(aq 2e hPb(s E=-0.126 V cathode Cd2(aq 2e hCd(s)E=-0.403 V

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