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ELECTROCHEMISTRY - PowerPoint PPT Presentation

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ELECTROCHEMISTRY. During electrolysis positive ions (cations) move to negatively charged electrode (catode) and negative ions (anions) to positively charged electrode (anode) For the case of NaCl we have: cathodic reduction: Na + + electron = Na and anodic oxidation:

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During electrolysis positive ions (cations) move to negatively charged

electrode (catode) and negative ions (anions) to positively charged

electrode (anode)

For the case of NaCl we have:

cathodic reduction:

Na + + electron = Na


anodic oxidation:

Cl - - electron = Cl


Faraday laws:

  • The mass of product formed in an electrolysis is directly
  • proportional to the electric charge moved during the process.
  • The masses of different compounds formed by the same electric
  • charge are chemically equivalent.
  • The amount of electric charge needed for formation of 1 gramion
  • of a substance:
  • N . e (N is Avogadro number, e electron charge)
  • N . e = F (Faraday constant, 1Faraday = 96 494 Coulomb)
  • Work produced by electric current:
  • w = F . D E

Electrochemical cell is composed of two half-cells, realized e.g. as

metal electrode immersed in the solution of its salt.

The half-cells are conductively connected, e.g. by salt bridge.

Each half-cell contains oxidized and reduced component, which

create a redox couple




p .... osmotic pressure

P .... solvatation pressure

P > p negative electrode charge

P < p positive electrode charge


Hydrogen electrode

Precious metals such as platinum or palladium absorb vigorously

hydrogen. A solid solution is formed, analogical to the metal alloys.

Here we have hydrogen present in its atomic form, not as a two atom

molecule. Thus, in this state hydrogen has properties of a metal.

If we saturate a platinum electrode coated with platinum black by a

stream of hydrogen and immerse this electrode to the solution,

protons will be released into the solution due to the solvatation

pressure until they balance the proton osmotic pressure. This leads

to the generation of a potential, dependent on hydrogen partial


Standard hydrogen electrode

is realized under conditions of [ H+] = 1 (i.e. pH = 0) and hydrogen

pressure 1 atm. By convention its potential = 0


By comparison of the potential of a half-cell, realized as a metal

electrode immersed in 1 N solution of its salt, with standard hydrogen

electrode we obtain electrochemical series.

Some examples:

Electrode Potential (Volt)

Li/Li+ - 3, 02

K/K+ - 2, 92

Na/Na+ - 2, 71

Zn/Zn2+ - 0, 76

Fe/Fe2+ - 0, 43

Fe/Fe3+ - 0, 04

H/H+ 0, 00

Cu/Cu2+ + 0, 34

Cu/Cu+ + 0, 51

Ag/Ag+ + 0,80

Au/Au+ + 1, 50


Metals, placed above hydrogen in this table, have a tendency to form

positive cations and with distance from hydrogen, their

electropositivity increases.

More electropositive metal displaces less electropositive metal from

the solution.

Potential of a metal electrode dissolving metal cations into solution

is given by

Nernst equation:

E = - RT/nF . ln c

where R ...universal gas constant

n ... number of electrons representing the difference between

the metal and its ion

c ... concentration of the ions in solution


We can express the amount of energy released in electrochemical

process as:

DG = - nFE

Under standard conditions (concentration 1 M, pressure 1 atm) we


DG0 = - nFE0

where E0 is the standard potential of the cell

Standard potential of the cell can be calculated as a sum of standard

potentials of electrodes:

E0 = E0(anode) + E0(catode)


We can express the amount of energy released under standard

  • conditions in a general form:
  • DG0 = - RT lnK
  • Another expression can be used for the electrochemical
  • process:
  • DG0 = - nFE0
  • If we consider a real process out of standard conditions we get:
  • DG = DG0 + RT lnQ
  • where Q corresponds to the actual ratio of products and reactants
  • For electrode potential we get:
  • -nFE = -nFE0 + RT lnQ and hence
  • E = E0 - RT/nF ln Q
  • This is an important expression of Nernst equation

Concentration cell

E = E0 – RT/2F ln (c2/c1)


We can use Nernst equation for the calculation of a potential

generated in redox reactions in the living cell. The reaction:

reductant + oxidant = oxidized reductant + reduced oxidant

can be simplified:

electron donor = electron acceptor + electron

then DE0 = E0(acceptor) - E0(donor)

as DG0 = - nF DE0

we get:

E = E0 + RT/nF ln( [acceptor]/[donor] )


To calculate membrane potential we first consider the amount of

  • energy needed for the transport of a substance across the membrane.
  • For the transport of 1 mol of a substance from the region of concentration c1 to the region of concentration c2 we get:
  • DG = RT ln (c2/ c1)
  • When c2 is lower than c1, DG is negative and the transport proceeds.
  • Under equilibrium DG = 0, concentrations are equal and transport is stopped.
  • If we have an ion with a charge Z, the change of Gibbs function during its transport will contain two components – a concentration part and a part describing charge movement:
  • DG = RT ln (c2/ c1) + ZF Dy
  • where Dy is the membrane potential in Volts