The Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.
When NH3 is added to the first cell, Nh3 react with Cu(NO3) react to form complex.
Thus, Cu2+ ion concentration decrease in the first cell.
Anode
Cu ---- Cu(2+) + 2e-
Cathode
Cu(2+) + 2e- ------ Cu
Ecell can be calculated as
Ecell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2) cathode}
[Cu2+] cathode = 90ml × 0.01M = 9 × 10^(-4) moles
or,
0.129 = 0 - (0.059/2) log ( Cu(+2) / 9 × 10^(-4))
[Cu(2+) ] anode = 3.8 × 10^(-8) mol
<h3>Chemical reaction of Nh3 with Cu2+</h3>
(Cu2+) + 4 NH3 -----; Cu(NH3)4(2+)
Kf can be given as
Kf = [Cu(NH3)4(2+)]/ [Cu2+] [ NH3]^4
Concentration of NH3 = 19 ml × 0.5 M
= 0.005 m
Kf = 0.005/ (3.8 × 10^(-8) mol) × (0.005) ^4
= 2.09 × 10^14.
If 10ml NH3 id added in the solution, then the total concentration of NH3 can be 20ml and 0.5 M = 0.01mol
Now, we can calculate the [Cu2+] anode
[Cu2+] anode = [Cu(NH3)4(2+)]/ Kf × [ NH3]^4
By substituting all the values, we get
= 4.78 × 10^(-9) moles.
E cell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2)
0- (0.059/2) log{ 4.78 × 10^(-9) / 9 × 10^(-4))
E cell = 0.156 V.
Thus, we calculated that the Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.
learn more about Ecell:
brainly.com/question/861659
#SPJ4
