2 Types of Electrochemical Cells 1)Voltaic Cells  Spontaneous reaction  Reaction itself creates electric current  Main concept for batteries 2)Electrolytic.

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Presentation transcript:

2 Types of Electrochemical Cells 1)Voltaic Cells  Spontaneous reaction  Reaction itself creates electric current  Main concept for batteries 2)Electrolytic Cells

Electrolytic CellsElectrolytic Cells  Reaction is NONSPONTANEOUS  Electric current drives redox reaction  External source for electrical current—serves as “electron pump”  Electrolysis  Process of using an electric current to drive chemical reaction Ex. Electroplating

Electrolytic Cells (cont.)Electrolytic Cells (cont.)  Electrodes submerged into a salt solution, NO salt bridge  Anode  Picks up electrons from solution, negative ions give up electrons  Oxidation still occurs here  Cathode  Receives electrons from anode/wire  Positive ions accept electrons  Reduction still occurs here

Electrolyic Cells (cont.)Electrolyic Cells (cont.)  Polarities of electrodes (cathode/anode) change  Electrons pulled away from ANODE so it takes POSITIVE charge  External electric source gives excess electrons to CATHODE so it takes NEGATIVE charge  Oxidation and reduction still occur in same place  External electric source controlling electrons

Electrolytic Cells and Redox Reactions in Aqueous Solutions  Possible to have multiple redox reactions occurring  Reduction/Oxidation processes could happen with water and not with the chemical reaction of interest

How do we know which redox reaction is favored?  Nonspontaneous  Negative cell potential, requires a lot of energy from electric source  Spontaneous  Positive cell potential, less energy needed from electric source **Favored redox reaction is the MOST SPONTANEOUS or LEAST NEGATIVE of the possible reactions. **Look at half-reactions

Example 1:Example 1:  Electrolysis of CaI 2 Reduction: Ca e -  Ca (s) Ered = -2.87V 2H 2 O + 2e -  H 2 + 2OH - Ered = -0.83V Oxidation: 2I -  I 2 + 2e - Ered = +0.53V 2H 2 O  O 2 + 4H + + 4e - Ered = V

Example 2:Example 2:  A direct current is applied to an aqueous copper(II) bromide solution.  2H 2 O + 2e -  H 2 + 2OH - Ered = -0.83V  O 2 + 4H + + 4e -  2H 2 O Ered = +1.23V  Cu e -  Cu Ered = +0.34V  Br 2 + 2e -  2Br - Ered = +1.07V  a) Identify reduction half-reaction  b) Identify oxidation half-reaction

Electric Current (I)Electric Current (I)  Amount of charge transferred through an area per second  I = q/T  I = current (A, amperes)  q = charge (C, coulombs)  T = time (seconds, s)

Faraday’s LawFaraday’s Law  Relates amount of electric current in an electrochemical cell to the mass of a chemical substance  “amount of chemical consumed/made at electrode in electrolytic cell DIRECTLY proportional to amount of electric current through cell” (Spencer, p. 517)

Faraday’s constant (F)Faraday’s constant (F)  States the amount of charge (C) transferred by one mole of electrons  F = 96,500 C/mol of electrons  Stoichiometric relationship between mole of electrons and mole of chemical substance

Example 3:Example 3:  How much charge goes through a wire containing 0.987A of current for 12.3 minutes?

Example 4:Example 4:  How many moles of electrons went through the area in Example 3?

Example 5:Example 5:  Calculate the volume of H 2 gas at 25°C and 1atm that will be produced at the cathode when water undergoes electrolysis for 2.0 hours at a 10.0A electric current.

Homework  p. 795 #77, 79, 81