Of the metals Zn, Mg and Fe, which removes copper (II) ion from the solution mostcompletely? The following equilibrium constants are measured at room temperature.Zn (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1037Mg (s) + Cu2+ (aq) → Cu(s) + Mg2+(aq) Kc = 2 x 1090Fe (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1026Give reason for your answer
Question
Of the metals Zn, Mg and Fe, which removes copper (II) ion from the solution mostcompletely? The following equilibrium constants are measured at room temperature.Zn (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1037Mg (s) + Cu2+ (aq) → Cu(s) + Mg2+(aq) Kc = 2 x 1090Fe (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1026Give reason for your answer
Solution
To determine which metal removes copper (II) ion from the solution most completely, we need to compare the equilibrium constants (Kc) for the reactions involving Zn, Mg, and Fe.
The equilibrium constant (Kc) represents the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. A higher value of Kc indicates a greater extent of reaction and a more complete removal of copper (II) ion from the solution.
Comparing the given equilibrium constants:
- For the reaction involving Zn: Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq), Kc = 2 x 10^37
- For the reaction involving Mg: Mg (s) + Cu2+ (aq) → Cu (s) + Mg2+ (aq), Kc = 2 x 10^90
- For the reaction involving Fe: Fe (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq), Kc = 2 x 10^26
From the given equilibrium constants, we can see that the reaction involving Mg has the highest value of Kc (2 x 10^90). This indicates that the reaction between Mg and Cu2+ is more favorable and proceeds to a greater extent compared to the reactions involving Zn and Fe.
Therefore, Mg removes copper (II) ion from the solution most completely among the metals Zn, Mg, and Fe.
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