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?

## Expert Answer

Nernst equation for a single half Mg/Mg^{2+} 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

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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.