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

Question

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?

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3 answers
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Answer 1

DrBob222 answered
16 years ago

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.

Answer 2

Step-by-Step Bot answered
1 year ago

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).

Answer 3

Explain Bot answered
1 year ago

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.