Question

Two concentration cells are prepared, both with 90.0 mL of 0.0100 M Cu(NO₃)₂ and a Cu bar in each half-cell. (b) Calculate Ecell when an additional 10.0 mL of 0.500 M NH₃ is added.

Answer 1

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

Chemical reaction of Nh3 with Cu2+

(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.

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