CHEM 1412 Concept Review: Electrochemistry

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Oxidation: An increase in oxidation number (This happens when electrons are lost.)

Reduction: A decrease in oxidation number (This happens when electrons are gained.)

Oxidizing Agent or Oxidant: A substance which causes oxidation in something else (by being reduced itself)

Reducing Agent or Reductant: A substance which causes reduction in something else (by being oxidized itself)

 

 

Simplified Rules for Determining Oxidation Number

1.  Hydrogen always has an oxidation number of +1 when bonded to non-metals and -1 when bonded to metals.

2.  Fluorine has an oxidation number of -1 in ALL compounds.

3.  Oxygen nearly always has an oxidation number of -2. The only major exception is peroxides, such as H₂O₂, in which the oxidation number is -1.

4.  The oxidation number of any element in its elemental state is 0.

Examples:  O in O₂, S in S₈, Al by itself, etc.

5.  In any ionic compound, the oxidation number of any monatomic ion (single atom ion, as opposed to polyatomic ions) is the same as its charge.

6.  The sum of the oxidation numbers of all atoms within a compound must add up to the overall charge of the compound.

 

 

Steps for Balancing an Oxidation-Reduction Reaction

1. 1. Break the redox equation up into half-reactions.
      
      
   2. Balance all elements other than H’s and O’s.
      
      
   3. First, balance O’s with H₂O’s. Then, balance H’s with H⁺
      
      
   4. Balance charge by adding electrons (e^-) to the more positive (less negative) side of the equation.
      
      
   5. Add the two half-reactions back together, multiplying each half-reaction as needed in order to cancel out electrons on both sides of the equation. (Look for the lowest common multiple.)
      
      
   6. Extra Step for BASIC solutions:  Neutralize the H⁺ ions by adding an equal number of OH^- ions on both sides of the equation.  (Remember:  $H^++OH^-\longrightarrow H_2O$)

 

Electrochemical Cells and Related Equations

Voltaic Cell (also called a Galvanic Cell): A cell that uses a redox reaction to produce electricity

Electrolytic Cell:  A cell that uses electricity to drive forward a redox reaction that would normally be non-spontaneous (also called an electrolysis reaction)

In cells, oxidation occurs at the anode, and reduction occurs at the cathode (AN OX and a RED CAT.)

Electromotive Force (emf):  The force produced by a voltaic cell which moves electrons from the anode to the cathode.  This is also called cell potential or E_cell.

$E^\circ_{cell}=E^\circ_{red_{cathode}}-E^\circ_{red_{anode}}$          OR          $E^\circ_{cell}=E^\circ_{red_{cathode}}+E^\circ_{ox_{anode}}$

 

Note:  $1\:Volt=1\:\frac{Joule}{Coulomb}\:or\:1\:\frac{J}{C}$

 

Standard Conditions for a Cell: All concentrations are 1 M and all pressures of any gasses would be 1 atm.  The temperature for standard conditions is 25^oC.  When a cell is at standard conditions, we use the symbol E^o_cell.

 

 

Gibbs free energy relates to cell potential according to the following equation:

$\Delta G=-nFE$
Where “∆G” is Gibbs free energy, “n” is the number of moles of electrons transferred in the reaction, “F” is Faraday’s constant (96,485 C/mol), and “E” is the cell potential.

The maximum amount of work that a cell is capable of is equal to the Gibbs Free Energy:

$w_{\max}=-nFE$
The work is going to be negative in this case because it is done by the system (cell) to the surroundings.

 

Manipulation of the Gibbs energy equation for non-standard conditions, $\Delta G=\Delta G^\circ+RT\ln Q$, with the above equation relating ∆G and E leads to the Nernst Equation, which allows us to solve for the cell potential at non-standard conditions:

$E=E^\circ-\frac{RT}{nF}\ln Q$

Where E is the cell potential at non-standard conditions, E^o is the cell potential at standard conditions,   R is the ideal gas law constant of 8.314 J/mol*K, T is the temperature in Kelvins, n is the number of moles of electrons transferred, F is Faraday’s Constant, and Q is the reaction quotient.

At 25^oC, this equation can be simplified to the following:

$E=E^\circ-\frac{0.0592\:V\cdot mol}{n}\log Q$