Consider a cell composed of the following half cells:
- Mg(s) | Mg2+ (aq)
- Ag(s) | Ag+ (aq)
The emf of the cell is 2.96 V at [Mg2+] = 0.130 M and [Ag+] = 1.0 * 10-4 M. Calculate the standard emg of the cell ( R = 8.3 JK-1 mol-1 , F = 96500 C )
I am asking this question 2nd time soo plzz give me the answer this time.
Dear Student,
The standard EMF of the cell has to be calculated
Mg (s) / Mg2+ (aq) Ag (s) / Ag+ (aq)
The reactions would be:
Mg (s) ---> Mg2+ (aq) + 2e- ------ (-2.37)
Ag+ (aq) + e- ----> Ag (s) -----(+0.8)
Mg (s) + 2Ag+ (aq) Mg2+ (aq) + 2Ag (s)
E ocell = 2.96 V
According to Nernst Equation:
Ecell = E ocell - log kc
Because, 2 electrons are transferring, therefore, n = 2
Placing the values we get,
E cell = 2.96 - log
= 2.96 - log
= 2.96 - log
= 2.96 - log (13000000) = 2.96 - (-7.11)
= 2.96 + 0.210 = 3.17 V
[Another Approach:
Eocell = Eoright - Eoleft
Eo Mg2+/ Mg = -2.36 V and Eo Ag+/Ag = +0.80
So, Eocell = (-2.36 - 0.80) V = +3.17 V]
Regards
The standard EMF of the cell has to be calculated
Mg (s) / Mg2+ (aq) Ag (s) / Ag+ (aq)
The reactions would be:
Mg (s) ---> Mg2+ (aq) + 2e- ------ (-2.37)
Ag+ (aq) + e- ----> Ag (s) -----(+0.8)
Mg (s) + 2Ag+ (aq) Mg2+ (aq) + 2Ag (s)
E ocell = 2.96 V
According to Nernst Equation:
Ecell = E ocell - log kc
Because, 2 electrons are transferring, therefore, n = 2
Placing the values we get,
E cell = 2.96 - log
= 2.96 - log
= 2.96 - log
= 2.96 - log (13000000) = 2.96 - (-7.11)
= 2.96 + 0.210 = 3.17 V
[Another Approach:
Eocell = Eoright - Eoleft
Eo Mg2+/ Mg = -2.36 V and Eo Ag+/Ag = +0.80
So, Eocell = (-2.36 - 0.80) V = +3.17 V]
Regards