Welcome to the discussion of the examination
" Physical Chemistry part II " (Kinetics,
Electrochemistry).
The first exam question deals with the redox
potential of the metabolism of electrodes
. A silver chloride electrode and an oxygen
electrode are connected together at 25 ° C
. Determine the redox potential of the silver
chloride electrode , and the redox potential
of the oxygen electrode.
A silver chloride electrode is a second electrode
Art on her are in equilibrium : silver, solid
silver chloride and a chloride solution.
The Nernst equation for this electrode is
formulated here . We set the number of stations
one : since there are all involved species
in the standard state corresponds to the potential
of the standard potential : 0.222 V.
For the oxygen electrode, the Nernst equation
is as follows: We set the corresponding numerical
values: 4 exchanged electrons [ partial pressure
oxygen ] = 00:21 bar; [ mole fraction of water
] = 1 and [ OH concentration of (-)- ion ] = 10
^ (-7) mol / L.
We calculate a potential of 0.805 V.
This value is more positive than the potential
of the silver chloride electrode , ie, the
oxygen electrode is the cathode and the positive
pole . The circuit is closed and a current
of strength 0.156 A flows . For the duration
of such current must flow so that 1 mg of
silver is it implemented?
To solve this problem we need the Faraday
equation.
This is: n = I xt / ( nu x F) The amount of
substance n we will replace with mass and
molar mass : 0.001 g of silver divided by
107.8682 g / mol (molar mass of silver ) current
I is equal to 0.156 amperes nu equal to 1
( one electron is required to reduce silver
(+) ions. the Faraday constant F is 96,485
C / mol. we solve the equation with respect
to time and obtain 5.73 seconds.
