In: Chemistry
A concentration cell based on the following half reaction at 305 K Mg2+ + 2 e- → Mg SRP = -2.37 V has initial concentrations of 1.27 M Mg2+, 0.255 M Mg2+, and a potential of 0.02110 V at these conditions. After 10 hours, the new potential of the cell is found to be 0.009984 V. What is the concentration of Mg2+ at the cathode at this new potential?
Nernst equation for a single half Mg/Mg2+ copper half
cell is just
E=E0+RT/2F(ln[Mg2+])
If you would write such equations for each half cell (they have
different concentrations) you would find that
E=E0+RT/2F(ln[Mg2+]1)−E0−RT/2F(ln[Mg2+]2)
=RT/2F(ln[Mg2+]1[Mg2+]2)
oxidation @ anode : Mg --> Mg2+ + 2 e- Eo = +2.37 V
reduction @ cathode: Mg2+ + 2 e- → Mg Eo = - 2.37 V
your concentration cell:
Mg2+ + Mg - - --> Mg2+ & Mg E o = zero
The above Nernst equation is
E = RT/2F(ln[Mg2+]1[Mg2+]2)
E = (8.314*305/n*96500)*2.303log Q
E = Eo - [(0.06052 / n) (log Q)]
E = 0 - [(0.06052 / 2 mol e-) (log [products] / [reactants])]
having a higher concentration as the reaction will drive a
concentration cell to produce voltage:
oxidation @ anode : Mg --> 0.255M Mg2+ + 2 e-
reduction @ cathode: 1.27M Mg2+ + 2 e- → Mg
E = - [(0.03026) (log [0.255] / [1.27])]
E = - [(0.03026) (log 0.201])]
E = - [(0.03026) (- 0.6968)]
E = - [- 0.021]
E = + 0.021 volts
======================================...
What will be the new concentration of Mg2+ at the cathode when new
potential of this cell has changed by 0.009984 V
oxidation @ anode : Mg --> 0.255M Mg2+ + 2 e-
reduction @ cathode: xM Mg2+ + 2 e- → Mg
the reaction will reverse, without switching the elcetrodes of the
volt meter, it will read a minus
E = - [(0.03026) (log [0.255] / [x])]
0.009984 = - [(0.03026) (log 0.255/x)]
0.3299405 = - [log 0.255/x]
0.4677992 = 0.255/x
x = 1.83 M
concentration of Mg2+ at the cathode at this new potential is 1.83
M.