Electrochemistry


 * __Electrochemistry__**

Electrochemistry is the study of changes that cause electrons to flow, creating a current we call electricity. A reduction reaction and an oxidation reaction are the two halves that create this flow of current, and they are more commonly known as a redox reaction. These two cannot exist without the other; in other words, you can’t have a reduction without an oxidation, and vice versa. In a redox reaction, electrons are exchanged by the reactants, and the oxidation states of some of the reactants are changed over the course of the reaction. For a deeper look into the beginning of electrochemistry, click [|here]. **__Oxidation v. Reduction__**

- Oxidation o In an oxidation reaction, an atom is losing electrons. These electrons are giving the electrons to the reduced atom. o The oxidation number of an oxidized atom increases. - Reduction o In a reduction reaction, an atom is gaining electrons. These electrons come from the oxidized atom. o The oxidation number of an oxidized atom decreases. - Consider the following reaction:

Fe + 2HCL → FeCl2 + H2 The two half-reactions are

__Oxidation__ Fe → Fe2+ + 2 e- __Reduction__ 2 H+ + 2 e- → H2
 * Fe’s oxidation number increased from 0 to +2.
 * H+’s oxidation number decreased from +2 to 0.


 * In a reaction, the **oxidant** causes oxidation while being reduced. **
 * The **reductant** causes reduction while being oxidized.**


 * __Reduction Potentials__**

A reduction potential is the tendency of an atom to acquire electrons and be reduced. Every half-reaction has a potential, or voltage, associated with it. These potentials have been placed in a standard table, and the half-reactions in the table are all //reductions//, hence the name of the potential. It is denoted by Eº, and its units are V (voltage).

To find the oxidation potential, simply read the listed reaction in reverse and change the Eº sign. Also, NEVER MULTIPLY the potential for a half-reaction by a coefficient. For example:

The reduction reaction F2(g) + 2 e- → 2 F-(aq) is reversed to 2 F- → F2(g) + 2 e-. The Eº changes from +2.87 V to -2.87 V.
 * For a full table, click [|here]. *

To calculate the potential of the entire redox reaction, you simply add the potential for the oxidation half-reaction to the reduction half-reaction. //Remember, to switch the sign of Eº to get the oxidation potential.//

Take the previous reaction. Fe + 2HCL → FeCl2 + H2 The two half-reactions are Fe → Fe2+ + 2 e- Eº= +0.440 V 2 H+ + 2 e- → H2 Eº= 0.00 V

The standard potential, or Eºcell, for the redox reaction is Eºcell = Eºoxidation + Eºreduction = 0.440 V + 0.00 V = 0.440 V


 * __Voltage and Spontaneity__**

If a redox reaction has a positive value (+Eºcell), then it is spontaneous. In thermodynamics, we learned that a spontaneous reaction has a negative free-energy charge. Consider the following equation:

∆Gº = -nFEº This equation brings thermodynamics and electrochemistry together. ∆Gº = Standard Gibbs free energy charge (kJ/mol) n = Number of moles of electrons exchanged in the reaction (mol) F = Faraday’s Constant, 96,500 coulombs/mole Eº = Standard reaction potential (V, Eºcell)

//Note well://
 * If Eº is positive, ∆Gº will be negative and the reaction will be spontaneous.
 * If Eº is negative, ∆Gº will be positive and the reaction will be non- spontaneous.

For example: Fe + 2HCL → FeCl2 + H2 This reaction has an Eº of +0.44 V, which means it is a spontaneous reaction.


 * __Voltage and Equilibrium__**

Consider the following equation: This equation bridges the standard reaction potential, Eº, with the equilibrium constant, K. Eº = standard reaction potential R = gas constant, 8.31 (volt-coulombs)/(mol/K) T = absolute temperature (K) n = Number of moles of electrons exchanged in the reaction (mol) F = Faraday’s Constant, 96,500 coulombs/mole K = equilibrium constant

At 25ºC (298 K), this equation simplifies to Note well:
 * If Eº is positive, then K is greater than 1.
 * If Eº is negative, then K is less than 1.


 * __Galvanic Cells__**

A spontaneous redox reaction is used to generate a current in a galvanic, or voltaic cell.

These are the parts of a galvanic cell.

- The two parts of a galvanic cell are __anode__ and __cathode__, which are 2 separate chambers. o Oxidation occurs at the anode. o Reduction occurs at the cathode. Here’s a mnemonic device to remember which is which: AN OX – anode oxidation RED CAT – reduction cathode Or you can simply remember that the vowels – a & o – go together, and the consonants – r & c – go together. - A wire connects the two halves together and provides a passage for the electrons and the current to pass through (in opposite directions). - A salt bridge maintains electrical neutrality in the system by doing two things: 1) providing enough negative ions to equal the positive ions being created at the anode, and 2) providing positive ions being created at the cathode.

In a redox reaction, the electrons released by the oxidized atom are consumed by the reduced atom; in other words, electrons flow from the anode to the cathode. This electron exchange creates a current. Because current is the flow of positive charge, it is always in the opposite direction from the flow of electrons. Click [|here] to see an animation of a galvanic cell consisting of Zn(NO3)2 and Cu(NO3)2 solutions.

-- Nernst Equation -- This equation is used to calculate the cell voltage under nonstandard conditions. (Remember that the cell voltage under standard conditions – all concentrations are 1 M – is the same as the total voltage of the redox reaction.)

Ecell = cell potential under nonstandard conditions (V) Eºcell = cell potential under standard conditions (V) R = gas constant, 8.31 (volt/coulomb) (mol/K) T = absolute temperature (K) n = number of electrons exchanged in the reaction (mol) F = Faraday’s constant, 96,500 coulombs/mole Q = reaction quotient (same as equilibrium expressions, but with initial concentrations instead of equilibrium concentrations)

At 25ºC, the Nernst equation reduces to Note well: · As the concentration of products of a redox reaction decreases, the voltage decreases. · As the concentration of reactants of a redox reaction increases, the voltage increases.


 * __Electrolytic Cells__**

Electrolytic cells contain nonspontaneous redox reactions, and because of this, an outside source of voltage (e.g. a battery) is used to force the reaction to happen. Below is a picture of an electrolytic cell. To see an animation of an electrolytic cell, click [|here].

These cells are used for electroplating, which is a plating process that uses electrical current to reduce cations of a desired material from a solution and coat a conductive object with a thin layer of the material, such as metal. To calculate the number of grams created in a reaction, use the following four steps.


 * 1) If the current and time is given, use those to calculate the charge in coulombs, by using the following equation: [[image:untitled4.JPG]].
 * 2) Then use the charge (in coulombs) to find the number of electrons involved in the reaction:
 * 3) Using the half-reactions and the number of electrons, you can find the number of moles plated out. For example in this half-reaction,

2 H+ + 2 e- → H2 for every 2 moles of electrons consumed, there is 1 mole of hydrogen gas.
 * 1) Once you know the moles, you can use stoichiometry to find the number of grams.

Here’s an example. How many grams of solid silver were produced if a solution of 0.080 molar Ag+ solution was subjected to a current of 2.00 amperes for 10.0 minutes?

The half reaction is Ag+ + e- → Ag(s) (In case you wanted to know, the other half-reaction is 2 H2O → O2 + 4 H+ + 4 e-

C = sec x A =10 min x (60 sec/min)= 600 sec x 2.00 A = 1200 C 1200 C x (1 mol e- / 96500 C) x (1 mole Ag+ / 1 mole e- ) x (107.87 g / 1 mole Ag+)= 1.34 g.