18.5e MAS - Nernst: Calculate Concentration (Ec-Ea)?

Au3+ + 3e- --------> Au......Eo = 1.5 V

Mg2+ + 2e- -------> Mg .......Eo = -2.37 V

now since the reduction potential of Au3+/Au is more than Mg2+/Mg so Au3+ will be reduced and Mg will be oxidised ...in other words if a cell is set up .... cathode compartment will consist of Au3+/Au and anode compartment will consist of Mg2+/Mg

cathodic reaction will be :

Au3+ + 3e- --------> Au .....(1)

and anodic reaction will be :

Mg ----------> Mg2+ + 2e- .....(2)

multiplying (1) by 2 and (2) by 3 and adding ...

2Au3+ + 6e- -------> 2Au

3Mg ----------> 3Mg2+ + 6e-

------------------------------

2Au3+ + 3Mg ----------> 3Mg2+ + 2Au ......

Eo(cell) = Eo(cathode) - Eo(anode) = 1.5 - -2.37 = 1.5 + 2.37 = 3.87 V

Remember that when we multiply the equations, we DON'T multiply the voltages (Eo).

now if the conc. of [Mg2+] and [Au3+] were 1 M ....E(cell) would have been equal to Eo(cell) ....but since conc. is not equal to 1 M so using nernst equation ....

E(cell) = Eo(cell) - RT/nF ln [Mg2+]^3 / [Au3+]^2

Remember that Au and Mg are solids and do not appear in the log term

where E(cell) = 4.01 V

Eo(cell) = 3.87

R = 8.314

T = 25 + 273 = 298 K

n = no.of electrons in the redox reaction = 6

F = 96500 Coulombs

[Mg2+] = 1 X 10^-5 = 10^-5 M

[Au3+] = ? M

putting the values...

4.01 = 3.87 - [ (8.314 X 298) / (6 X 96500) ] X ln 10^-5^3 / [Au3+]^2

4.01 - 3.87 = - 0.0043 X ln 10^-15 / [Au3+]^2

0.14 = -0.0043 X ln [Au3+]^2 / 10^-15

ln 10^-15 / [Au3+]^2 = -0.14/0.0043

ln 10^-15/[Au3+]^2 = -32.558

converting ln to log ...multiplying by 2.303...

2.303 log 10^-15 / [Au3+]^2 = -32.558

log 10^-15/[Au3+]^2 = -32.558/2.303

log 10^-15 /[Au3+]^2 = -14.137

taking antilog ...

10^-15 / [Au3+]^2 = 10^-14.137

10^-15 / [Au3+]^2 = 7.295 X 10^-15

[Au3+]^2 = 10^-15 / 7.295 X 10^-15

[Au3+]^2 = 0.137

[Au3+] = square root of 0.137 = 0.371 M

so [Au3+] = 0.371 M

please check the calculations ....i have given the processs....

feel free to ask any question