33 Concentration Cells

# 33 Concentration - Electrolysis of Water How many moles of H2 will be produced in 1 hour by 1 amp of current A 0.019 mol Half reactions written as

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1 Half reactions written as reductions. O 2 (g) + 4H 3 O + + 4e - 6H 2 O 1.23 V 2 H 3 O + + 2e - H 2 (g) + 2H 2 O 0.00 V Cell reaction. 2H 2 O O 2 + 2H 2 E o = –1.23 V Electrolysis of Water How many moles of H 2 will be produced in 1 hour by 1 amp of current? A) 0.019 mol B) 0.19 mol C) 1.9 mol D) 0.034 mol E) 0.34 mol F = 96,485 C/mol 2 Cell reaction. 2H 2 O O 2 + 2H 2 E o = –1.23 Electrolysis of Water How many moles of H 2 will be produced in 1 hour by 1 amp of current? 1 C/s x 3600 s = 3600 C of electrons 3600 C / 96,485 C/mol = 0.0373mol electrons 2H+ H 2 requires 2 mol electrons/mol H 2 Produce 0.0187 mol H 2

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Today’s To Do Evaluate equilibrium constants from standard electrochemical cell potentials. Calculate cell potentials for different concentrations of reactants. Explain how and why a voltaic cell can be produced by using the same substances in both the anode and cathode compartments. 4
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## This note was uploaded on 04/23/2010 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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33 Concentration - Electrolysis of Water How many moles of H2 will be produced in 1 hour by 1 amp of current A 0.019 mol Half reactions written as

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