1) List a series of tests that would determine the order of activity (least active to most active) for Zn, Mn, and Ag.
2) What would be the reaction for the addition of excess OH- ions to Mg(NO3)2 and Cu(NO3)2?
This is a complex ion forming reaction.
Would the complete ionic equation be 2OH-(aq) + Cu(OH)2(s)= Cu^2+(aq) + 4OH-(aq) for the Cu(NO3)2 and 2OH- + Mg(OH)2 (s)= Mg^2+(aq) + 4OH-(aq)for Mg(NO3)2? What would the net ionic be then for both equations? Would the Mg(NO3)2 be able to form a precipitate in this reaction with the OH-ions?
3) Likewise, What are the reactions for the addition of excess NH3(aq) to Mg(NO3)2 and Cu(NO3)2? Net ionic and complete ionic formulas?
1.Zn metal in MnSO4
Zn metalin AgNO3
Mn metal in Zn(NO3)2
Mn metal in AgNO3
Ag metal in Zn(NO3)2
Ag metal in MnSO4.
2. Is the note about complex ion formation your observation or is that a part of the question. The most likely products are Cu(OH)2 and Mg(OH)2. Mg(OH)2 is not likely to form a complex ion and Cu(OH)4^2- is known but it isn't a common reaction.
3. The most likely reaction is the formation of the complex ion Cu(NO3)4^2+. It's difficult to write the formula on this forum. The 3 is a subscript, the 4 is a subscript and the 2+ refers t0 the charge on the ion.
1) To determine the order of activity from least active to most active for Zn, Mn, and Ag, you can conduct a series of tests such as:
- Reacting each metal with dilute hydrochloric acid (HCl). The metal that reacts the slowest or does not react at all will be the least active.
- Performing a displacement reaction by placing each metal in a solution of a different metal salt (e.g., Zn in AgNO3, Mn in ZnSO4, Ag in MnCl2). The metal that displaces another metal from its salt solution will be the more active metal.
By comparing the results of these tests, you can determine the order of activity for Zn, Mn, and Ag.
2) The reaction for the addition of excess OH- ions to Mg(NO3)2 and Cu(NO3)2 would indeed involve the formation of complex ions. The complete ionic equation for the reaction between OH- and Cu(NO3)2 would be:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
The complete ionic equation for the reaction between OH- and Mg(NO3)2 would be:
Mg2+(aq) + 2OH-(aq) → Mg(OH)2(s)
The net ionic equation for both reactions would remove the spectator ions (the nitrate ions):
For Cu(NO3)2:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
For Mg(NO3)2:
Mg2+(aq) + 2OH-(aq) → Mg(OH)2(s)
Mg(OH)2 is slightly soluble in water, so it would form a precipitate in this reaction.
3) The reactions for the addition of excess NH3(aq) to Mg(NO3)2 and Cu(NO3)2 would involve the formation of complex ions. The net ionic and complete ionic formulas would be as follows:
For Cu(NO3)2:
Complete ionic equation:
Cu2+(aq) + 6NH3(aq) + 4NO3-(aq) → [Cu(NH3)6]2+(aq) + 4NO3-(aq)
Net ionic equation:
Cu2+(aq) + 6NH3(aq) → [Cu(NH3)6]2+(aq)
For Mg(NO3)2:
Complete ionic equation:
Mg2+(aq) + 6NH3(aq) + 4NO3-(aq) → [Mg(NH3)6]2+(aq) + 4NO3-(aq)
Net ionic equation:
Mg2+(aq) + 6NH3(aq) → [Mg(NH3)6]2+(aq)
1) To determine the order of activity for Zn, Mn, and Ag, you would need to perform a series of tests. Here is a suggested series of tests that would help determine the order from least active to most active:
a) React each metal with water: Zn + H2O, Mn + H2O, Ag + H2O.
- If there is no reaction, move to the next test.
- If there is a reaction, observe the intensity and speed of the reaction.
b) React each metal with an acid (e.g., HCl): Zn + HCl, Mn + HCl, Ag + HCl.
- Again, observe the intensity and speed of the reaction.
c) React each metal with a stronger oxidizing agent (e.g., H2SO4): Zn + H2SO4, Mn + H2SO4, Ag + H2SO4.
- Note any noticeable changes or reactions.
Based on the observations from these tests, you can determine the order of activity for Zn, Mn, and Ag. The metal that reacts the least or shows the slowest reaction would be the least active, while the metal that reacts the most or shows the fastest reaction would be the most active.
2) When excess OH- ions are added to Mg(NO3)2 and Cu(NO3)2, complex ion formation occurs. The complete ionic equations you provided are correct:
a) For Cu(NO3)2, the complete ionic equation is:
2OH-(aq) + Cu(OH)2(s) = Cu^2+(aq) + 2OH-(aq)
To get the net ionic equation, we need to eliminate the spectator ions (OH-):
Cu(OH)2(s) = Cu^2+(aq) + 2OH-(aq)
2OH- ions cancel out on both sides:
Cu(OH)2(s) = Cu^2+(aq)
b) For Mg(NO3)2, the complete ionic equation is:
2OH-(aq) + Mg(OH)2(s) = Mg^2+(aq) + 2OH-(aq)
To get the net ionic equation, we again eliminate the spectator ions (OH-):
Mg(OH)2(s) = Mg^2+(aq)
In both cases, the net ionic equations show the formation of complex ions, where the metal cation (Cu^2+ or Mg^2+) is complexed by the OH- ions.
Regarding the precipitation of Mg(NO3)2 with OH- ions, Mg(OH)2 is only slightly soluble in water. However, adding excess OH- ions can cause the solubility product to be exceeded, resulting in the formation of a precipitate of Mg(OH)2.
3) When excess NH3(aq) is added to Mg(NO3)2 and Cu(NO3)2, complex ion formation also occurs. The reactions would be as follows:
a) For Cu(NO3)2, the complete ionic equation is:
6NH3(aq) + Cu^2+(aq) + 4NO3-(aq) = [Cu(NH3)6]2+(aq) + 4NO3-(aq)
The net ionic equation is:
6NH3(aq) + Cu^2+(aq) = [Cu(NH3)6]2+(aq)
b) For Mg(NO3)2, the complete ionic equation is:
6NH3(aq) + Mg^2+(aq) + 4NO3-(aq) = [Mg(NH3)6]2+(aq) + 4NO3-(aq)
The net ionic equation is:
6NH3(aq) + Mg^2+(aq) = [Mg(NH3)6]2+(aq)
In both cases, the ammonia (NH3) forms coordination compounds with the metal cations (Cu^2+ and Mg^2+), resulting in the formation of complex ions.