Free+Energy,+the+Equilibrium+Constant,+and+Electrode+Potentials

The Change in Free Energy and the Equilibrium Constant

Within any reaction, the change in free energy, ∆G, can predict whether it will be spontaneous or not. Under standard conditions (1 atm and 298 Kelvin) it is known as ∆G0. So, how do you calculate the change in free energy under non-standard conditions? The equation used is: ∆G = ∆G 0 +RT ln Q - ∆G is equal to zero, as there is no change in free energy. - Q becomes K, the equilibrium constant In equilibrium conditions, the equation can now be rearranged as:
 * ∆G0 is the standard free energy (J/mol or kJ/mol)
 * R is the gas constant 8.314 J/mol-K
 * T is the temperature in Kelvin
 * Q is the reaction quotient
 * When a reaction is at equilibrium:

∆G0= -RT ln K = -2.303 RT log K   ** When you plug in numbers into the formula, make sure you use the same units! ** //Have you ever wondered how batteries work?// courtesy of: http://www.electronicplus.com/images/products/MN1500B4.jpg
 * Free Energy and Electrode Potential **

A battery is an example of a voltaic cell.

**Fun Fact: Why is the cell called a voltaic cell? **   **An 18th century physicist, Alessandro Volta, showed that when placing two metals into the bloodstream of an animal specimen, an electric current would flow. Voltaic cells were named after him because of his great contributions to electrochemistry. ** The voltaic or galvanic cell generates an electrical potential through the differences in the ability of elements to donate or accept electrons that determines the reduction potential. The reactions in a voltaic cell can be broken up into two half-reactions, one oxidation and one reduction. Reduction occurs at the elemental electrode which gains electrons and is known as the cathode. Oxidation occurs at the elemental electrode that loses electrons and is known as the anode. Reduction potentials are assigned to elements relative to the Standard Hydrogen Electrode which has a reduction potential of 0 V. The circuit is completed by a salt bridge between the anode and cathode compartments. The electrode potential of a cell (E 0 ) in volts (V) can be calculated as:

E 0 = E cathode – E anode, at standard conditions (1M solution at 298 K and 1 atm) E cathode = Reduction potential of cathode E anode = Reduction potential of anode

If E0 is positive, electrons will flow from the anode to the cathode through an external wire, and now we have electricity. This leads to a change in free energy. The change in free energy and electrode potential is given by

ΔG0 = -nFE°

ΔG0 = standard free energy n = moles of electrons transferred F = Faraday’s constant, 96485 J/V-mol (Light bulb icon-who was Faraday?)

If the electrode potential is positive, ΔG will be negative, the reactions in the oxidation/reduction half-cells will occur spontaneously, and voltage will be generated. Conversely, when the electrode potential is negative, the free energy will be a positive number, the reaction will not be spontaneous and no voltage will be generated.

 Practice Problem:

Given the following half reaction reduction potentials: **Cu2+(aq) + 2e- -> Cu(s) (0.34 volts) ** Zn(s) -> Zn2+(aq) + 2e- **<span style="color: rgb(128, 0, 128); font-size: 130%;">(-0.76 volts) ** <span style="color: rgb(128, 0, 128); font-family: 'Arial Black',Gadget,sans-serif; font-size: 130%;"> <span style="color: rgb(128, 0, 128); font-size: 120%;"> <span style="font-family: 'Arial Black',Gadget,sans-serif; color: rgb(128, 0, 128);">Figure out the direction of the electron flow in the following cell: <span style="font-family: 'Arial Black',Gadget,sans-serif; color: rgb(128, 0, 128);">* courtesy of: http://upload.wikimedia.org/wikipedia/commons/3/3b/Electrochemical_element_with_salt_bridge<span style="font-family: 'Arial Black',Gadget,sans-serif; color: rgb(128, 0, 128);">

The Nernst Equation When dealing with reactions that are not in standard conditions, e.g. using different molarities of solutions in the anode and cathode compartments, the Nernst Equation can be used: Ecell = E0 – (RT/nF) ln Q = E0 – (0.0592/n) log Q at 298 K <span style="color: rgb(239, 72, 72);">

Finally, the cell potentials will vary depending on the Q value of the products divided by the reactants.

**References:** BROWN, THEODORE L., LEMAY, JR., H. EUGENE, BURSTEN, BRUCE E., and MURPHY, CATHERINE J. __Chemistry, The Central Science__. 10th ed. New Jersey: Pearson Education, Inc. 2006. - www.scienceclarified.com/Ca-Ch/**Cell**-Electrochemical.html - www.engineeringtoolbox.com/**electrode**-**potential**-d_482.html http://www.substech.com/dokuwiki/doku.php?id=electrode_potentials