[13%] Copper(II) forms a complex ion with ammonia, Cu(NH3)4^2+, with Kf = 1.7 × 10^13. An electroche — Electrochemistry Chemistry Question
Problem Context
[13%] Copper(II) forms a complex ion with ammonia, Cu(NH3)4^2+, with Kf = 1.7 × 10^13. An electrochemical cell is set up as shown below at 298 K. Half-cell A contains 100 mL of 1.00 M Cu(NO3)2, while half-cell B contains 100 mL of a solution that contains a small amount of copper(II) and is 0.100 M in NH3. A solution of nitric acid is slowly added to half-cell B and the potential measured by the voltmeter is recorded as a function of the added volume of HNO3.
Which half-cell is the cathode and which is the anode? Justify your answer.
Model Answer
Half-cell B has a much lower concentration of Cu2+(aq). Not only is it stated that there is only a small amount of Cu(II), at [NH3] = 0.1 M, almost all copper will be in the form of Cu(NH3)4^2+. Since current will flow spontaneously in a concentration cell in a direction that tends to equalize the concentrations in the two half-cells, reduction of Cu(II) will happen preferentially in half-cell A, which is thus the cathode. Half-cell B is the anode.
Qualitatively explain the shape of the graph.
Model Answer
As HNO3 is added, it protonates NH3 and decreases its concentration. Less NH3 means a higher concentration of uncomplexed Cu2+ and hence a smaller voltage. When all the NH3 is protonated, then all the Cu(II) is in the form of Cu2+(aq), which does not change further, so the voltage is stable after the endpoint.
What is the total concentration of copper(II) in the solution in half-cell B?
Model Answer
The easiest way to solve this is to use the potential past the endpoint, where the copper is in the form of Cu2+(aq). From the graph, E = 0.14 V:
E = E° – (RT/nF)ln([Cu2+]B/[Cu2+]A)
With n = 2, [Cu2+]A = 1.00:
0.14 = 0 – (0.0128)ln([Cu2+]B)
[Cu2+] = 1.8 × 10^-5 M
What is the concentration of nitric acid in the buret?
Model Answer
The endpoint is at 3.15 mL, at which point the NH3 has been neutralized:
0.100 L solution × 0.100 mol L^-1 NH3 = C × 0.00315 L HNO3 solution
C = 3.2 M
Suppose that the experiment is set up again with silver metal in place of copper metal and silver(I) ion in place of copper(II) ion, but with all concentrations and all other reagents identical. What would the graph of E vs. mL added HNO3 look like in this experiment? Sketch your result on the grid below (the graph shown above is redrawn for your convenience), and explain your answer. Silver(I) forms a complex ion with ammonia, Ag(NH3)2+, with Kf = 1.7 × 10^7.
Model Answer
In the original titration, before the endpoint, the Cu(II) is almost all complexed, so:
Kf = 1.7 × 10^13 = [Cu(NH3)4^2+] / ([Cu2+][NH3]^4) = [1.8 × 10^-5] / ([Cu2+][NH3]^4)
[Cu2+] = 1.1 × 10^-18[NH3]^-4
E = (–0.0128)ln((1.1 × 10^-18)[NH3]^-4)
E = 0.53 + 0.51•ln[NH3]
Repeating the calculation for Ag(I), which differs in the magnitude and expression for Kf and has n = 1 rather than n = 2:
Kf = 1.7 × 10^7 = [Ag(NH3)2^+] / ([Ag+][NH3]^2) = [1.8 × 10^-5] / ([Ag+][NH3]^2)
[Ag+] = 1.1 × 10^-12[NH3]^-2
E = (–0.0257)ln((1.1 × 10^-12)[NH3]^-2)
E = 0.71 + 0.51•ln[NH3]
So before the endpoint, the Ag experiment will give potentials that are 0.18 V higher than the corresponding values for the Cu experiment. After the endpoint, the only difference between the Cu and Ag experiments is that n = 1 for Ag (vs. n = 2 for Cu). This results in a doubling of the cell potential for Ag. So: