E degrees (V)
A galvanic cell based on the following half reaction s Au3+ 3e- ----> Au 1.50
Mg2+ 2e- ------> Mg -2.37
The cell is set up at 25 C with [Mg2+] = 1.00 x 10^-5 M
The cell potential is observed to be 4.01 V. Calculate the [Au3+] that must be present.
Au3+ = ____________M
A galvanic cell based on the following half reaction s Au3+ 3e- ----> Au 1.50
Mg2+ 2e- ------> Mg -2.37
The cell is set up at 25 C with [Mg2+] = 1.00 x 10^-5 M
The cell potential is observed to be 4.01 V. Calculate the [Au3+] that must be present.
Au3+ = ____________M
-
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
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