Qualitively explain the effect of changing concentration of Cu^2+ using your experimental observations and the Keq values below (no calculations required)

Cu^2+(aq) + 4 NH3(aq) --> [Cu(NH3)4]^2+(aq) , Keq = 1.2x10^12
CuS(s) --> Cu^2+(aq) + S^2-(aq) , Keq = 8.5 x 10^-45

Cu/Zn = 0.525 V
after 1 drop NH3: 0.505 V
after 2 drops NH3: 0.435 V

Cu/Zn = 0.510 V
after 1 drop Na2S: 0.452 V
after 2 drops Na2S: 0.393 V

So I know Voltage obviously decreases with added concentration, but how does the Keq account for that?

1. The Keq is a huge number which means that the complex (the product) is favored (big time) over the reactants. So adding ammonia obviously decreases Cu^+2 (big time since Keq is so large).

2. Look at Keq for #2. The way it is written is very small BUT if you reverse it, the number is gain, a very large number. You add Na2S, that ppts CuS (big time) and that reduces the concn of Cu^+2 quickly because Keq is so large.

Based on the provided experimental observations and Keq values, we can make the following qualitative observations:

1. Cu^2+ + 4 NH3 ⇌ [Cu(NH3)4]^2+:
- By adding NH3, the voltage decreases from 0.525 V to 0.435 V.
- This indicates that an increase in NH3 concentration causes a shift towards the products, [Cu(NH3)4]^2+.
- The positive value of Keq (1.2x10^12) suggests that the reaction is highly favored towards the products. Therefore, an increase in Cu^2+ concentration will lead to an increase in the concentration of [Cu(NH3)4]^2+.

2. CuS ⇌ Cu^2+ + S^2-:
- By adding Na2S, the voltage decreases from 0.510 V to 0.393 V.
- This indicates that an increase in Na2S concentration causes a shift towards the reactants, CuS and S^2-.
- The extremely small value of Keq (8.5 x 10^-45) suggests that the reaction strongly favors the reactants. Therefore, an increase in Cu^2+ concentration will lead to an increase in the concentration of CuS.

In summary, the voltage changes indicate that increasing the concentration of a component in a reaction can shift the equilibrium towards the products (if Keq is large and positive) or towards the reactants (if Keq is small).

From your experimental observations, you have noticed that the voltage decreases as the concentration of Cu^2+ changes. The equilibrium constant (Keq) provides insights into how the concentrations of reactants and products affect the position of the equilibrium. In this case, you have two different equilibrium reactions:

1. Cu^2+(aq) + 4 NH3(aq) → [Cu(NH3)4]^2+(aq) (Keq = 1.2x10^12)
2. CuS(s) → Cu^2+(aq) + S^2-(aq) (Keq = 8.5 x 10^-45)

Let's focus on the first equation. The equilibrium constant (Keq) for the reaction indicates the relationship between the concentrations of the reactants and products at equilibrium. A large Keq value (1.2x10^12 in this case) suggests that at equilibrium, the concentration of the products is much greater compared to the concentration of the reactants.

When you add NH3(aq) to the Cu^2+(aq) solution, it forms the complex [Cu(NH3)4]^2+(aq). This reaction consumes Cu^2+ and produces the complex ion. As a result, the concentration of Cu^2+ decreases, leading to a decrease in voltage. The Keq value indicates that this reaction favors the formation of the complex ion, so the decrease in Cu^2+ concentration is expected as the reaction shifts towards equilibrium.

Now, let's consider the second equation. The equilibrium constant (Keq) for this reaction is very small (8.5 x 10^-45). This indicates that at equilibrium, the concentration of the reactant (CuS) is much greater compared to the concentrations of the products (Cu^2+ and S^2-).

When you add Na2S to the Cu^2+ solution, it reacts to form CuS. This reaction consumes Cu^2+ and produces CuS as a solid. As a result, the concentration of Cu^2+ decreases, which leads to a decrease in voltage. The tiny Keq value suggests that the reaction strongly favors the formation of CuS and, to a negligible extent, the formation of Cu^2+ and S^2- ions.

In summary, the experimental observations of decreasing voltage align with the changes in concentration due to the equilibrium constants. In both cases, the reactions consume Cu^2+ and decrease its concentration, affecting the voltage readings. The large Keq for the first reaction reinforces the decrease in Cu^2+ concentration, while the extremely small Keq for the second reaction emphasizes the strong formation of CuS.