Two concentration cells are prepared, both with 90.0 mL of 0.0100 M Cu(NO3)2 and

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.
(a) In the first concentration cell, 10.0 mL of 0.500 M NH3 is added to one half-cell; the complex ion Cu(NH3)42 forms, and Ecell is 0.129 V. Calculate Kf for the formation of the complex ion.

b) Calculate Ecell when an additional 10.0 mL of 0.500 M NH3 is added.
(c) In the second concentration cell, 10.0 mL of 0.500 M NaOH is added to one half-cell; the precipitate Cu(OH)2 forms (Ksp 2.2 10 20). Calculate E°cell.

(d) What would the molarity of NaOH have to be for the addition

of 10.0 mL to result in an E°cell of 0.340 V? [Please do a b c d Thank you!]

Explanation / Answer

When the concentration of Cu(NO3)2 is same on both the cells (0.001M), net voltage is zero. However when NH3 is added to the first cell, some of the Cu(NO3)2 reacts with NH3 to form complex. Thus Cu^2+ ion concentration is the first cell is decreased. At this point E(cell) is 0.129 V.

The first cell where Cu^2+ concentration is less will try to oxidise Cu to Cu^2+ and this cell will act as anode. The second cell will act as cathode. The half cell reacts are as follows:

Anode : Cu ==> Cu^2+ + 2e

cathode : Cu^2+ + 2e ==> Cu

E(cell) = E^0(cell) - 0.059/2 log {[Cu^2+]anode/[Cu^2+]cathode}

[Cu^2+]cathode = 90mL 0.01M = 9*10^-4 moles

or, 0.129 = 0 - 0.059/2 log {[Cu^2+]anode/9*10^-4 moles}

or, [Cu^2+]anode = 3.8*10^-8 moles

The reaction of NH3 with Cu^2+ is as follows:

Cu^2+ + 4NH3 ==> Cu(NH3)4^2+

Kf = [Cu(NH3)4^2+]/[Cu^2+][NH3]^4

[NH3] = 10mL 0.5 M = 0.005 moles

Kf = 0.005 /(3.8*10^-8) * (0.005) ^4 = 2.09*10^14

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If additional 10mL 0.5 NH3 is added, then total NH3 concentration = 20mL and 0.5 M =0.01 moles

Using this value, calculate the [Cu^2+]anode.

[Cu^2+]anode = [Cu(NH3)4^2+]/ Kf* [NH3]^4

= 0.01 / (2.09*10^14) * (0.01)^4 = 4.78 *10^-9 moles

E(cell) = E^0 - 0.059/2 log{[Cu^2+]anode/[Cu^2+]cathode}

= 0 - 0.059/2 log{4.78 *10^-9 moles/9*10^-4} = 0.156 V

The E(cell ) will be 0.156 V

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moles of NaOH added = 10mL & 0.5 M = 0.005 moles

The reaction occuring in the cell is : Cu^2+ + 2OH- ==> Cu(OH)2

Ksp = [Cu^2+][OH-]^2

or, 2.2*10^-20 = [Cu^2+](0.005)^2

or, [Cu^2+] = 8.8 *10^-16 M

Total volume of the solution is 90mL + 10mL = 100mL

moles of [Cu^2+] = 100mL & 8.8 *10^-16 M = 8.8*10^-17 moles

E(cell) = E^0 - 0.059/2 log (8.8*10^-17)/(9*10^-4) = 0.384 V

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If E(cell) = 0.34 V, then calculate [Cu^2+]anode.

0.34 = 0 - 0.059/2 log {[Cu^2+]anode/[Cu^2+]cathode}

or, 0.34 = -0.059/2 log {[Cu^2+]anode/(9*10^-4)}

or, [Cu^2+]anode = 2.68 *10^-15 moles

Molarity of [Cu^2+]anode = 2.68 *10^-15 moles/0.1L = 2.68 *10^-14 M

Ksp = [Cu^2+]anode [OH-]^2

or, 2.2*10^-20 = 2.68 *10^-14 M * [OH-]^2

or, [OH-] = 9.06*10^-4 M

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