Electrochemistry Combining the Half-Reactions 5 C 2 O 4 2−  10 CO 2 + 10 e − 10 e − + 16 H + + 2 MnO 4 −  2 Mn 2+ + 8 H 2 O When we add these together,

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Presentation transcript:

Electrochemistry Combining the Half-Reactions 5 C 2 O 4 2−  10 CO e − 10 e − + 16 H MnO 4 −  2 Mn H 2 O When we add these together, we get: 10 e − + 16 H MnO 4 − + 5 C 2 O 4 2−  2 Mn H 2 O + 10 CO e −

Electrochemistry Combining the Half-Reactions 10 e − + 16 H MnO 4 − + 5 C 2 O 4 2−  2 Mn H 2 O + 10 CO e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H MnO 4 − + 5 C 2 O 4 2−  2 Mn H 2 O + 10 CO 2

Electrochemistry Balancing in Basic Solution If a reaction occurs in basic solution, one can balance it as if it occurred in acid. Once the equation is balanced, add OH − to each side to “neutralize” the H + in the equation and create water in its place. If this produces water on both sides, you might have to subtract water from each side.

Electrochemistry Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

Electrochemistry Voltaic Cells We can use that energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell.

Electrochemistry Voltaic Cells A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode.

Electrochemistry Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

Electrochemistry Voltaic Cells Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.  Cations move toward the cathode.  Anions move toward the anode.

Electrochemistry Voltaic Cells In the cell, then, electrons leave the anode and flow through the wire to the cathode. As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

Electrochemistry Voltaic Cells As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

Electrochemistry Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

Electrochemistry Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential, and is designated E cell.

Electrochemistry Cell Potential Cell potential is measured in volts (V). 1 V = 1 JCJC

Electrochemistry Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated.

Electrochemistry Standard Hydrogen Electrode Their values are referenced to a standard hydrogen electrode (SHE). By definition, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2 e −  H 2 (g, 1 atm)

Electrochemistry Standard Cell Potentials The cell potential at standard conditions can be found through this equation: E cell  = E red (cathode) − E red (anode)  Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

Electrochemistry Cell Potentials For the oxidation in this cell, For the reduction, E red = −0.76 V  E red = V 

Electrochemistry Cell Potentials E cell  = E red  (cathode) − E red  (anode) = V − (−0.76 V) = V

Electrochemistry Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials.

Electrochemistry Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell.

Electrochemistry Free Energy  G for a redox reaction can be found by using the equation  G = −nFE where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,485 C/mol = 96,485 J/V-mol

Electrochemistry Free Energy Under standard conditions,  G  = −nFE 

Electrochemistry Nernst Equation Remember that  G =  G  + RT ln Q This means −nFE = −nFE  + RT ln Q

Electrochemistry Nernst Equation Dividing both sides by −nF, we get the Nernst equation: E = E  − RT nF ln Q or, using base-10 logarithms, E = E  − RT nF ln Q

Electrochemistry Nernst Equation At room temperature (298 K), Thus the equation becomes E = E  − n ln Q RT F = V

Electrochemistry Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell,would be 0, but Q would not. E cell  Therefore, as long as the concentrations are different, E will not be 0.

Electrochemistry Applications of Oxidation-Reduction Reactions

Electrochemistry Batteries

Electrochemistry Alkaline Batteries

Electrochemistry Hydrogen Fuel Cells

Electrochemistry Corrosion and…

Electrochemistry …Corrosion Prevention