H, Mg, Zn, Cu
Zinc (Zn) commonly forms a +2 ion (Zn²⁺) due to the loss of two valence electrons. Magnesium (Mg) also typically forms a +2 ion (Mg²⁺) for the same reason. Copper (Cu) can form two types of ions: a +1 ion (Cu⁺) and a +2 ion (Cu²⁺), depending on the chemical environment and the compounds it forms.
Zn + CuBr2 = Cu + ZnBr2
In this case, zinc will undergo oxidation and copper ions will experience reduction. The reduction half-reaction is Cu^2+ (aq) + 2e^- → Cu (s), and the oxidation half-reaction is Zn (s) → Zn^2+ (aq) + 2e^-. Overall, the reaction is Zn (s) + Cu^2+ (aq) → Zn^2+ (aq) + Cu (s).
Zn+CuSO4=Cu+ZnSO4 right?
Cu and Ga
Zinc (Zn) commonly forms a +2 ion (Zn²⁺) due to the loss of two valence electrons. Magnesium (Mg) also typically forms a +2 ion (Mg²⁺) for the same reason. Copper (Cu) can form two types of ions: a +1 ion (Cu⁺) and a +2 ion (Cu²⁺), depending on the chemical environment and the compounds it forms.
Copper (Cu) is more active than nickel (Ni) and less active than zinc (Zn).
(Na, K, Ca, Mg, P, Fe, Cu, Zn, Mn and B
N, P, K, Fe, Mg, Zn, S, Cu etc.
The molecular equation for Cu(NO3)2 and Zn is Cu(NO3)2 + Zn -> Zn(NO3)2 + Cu. The total ionic equation is Cu^2+ + 2NO3- + Zn -> Zn^2+ + 2NO3- + Cu. The net ionic equation is Cu^2+ + Zn -> Zn^2+ + Cu.
Zn(s) --> Zn2+(aq) + 2e : Oxidation Cu+(aq) + 1e --> Cu(s) : Reduction
Zn + CuBr2 = Cu + ZnBr2
In this case, zinc will undergo oxidation and copper ions will experience reduction. The reduction half-reaction is Cu^2+ (aq) + 2e^- → Cu (s), and the oxidation half-reaction is Zn (s) → Zn^2+ (aq) + 2e^-. Overall, the reaction is Zn (s) + Cu^2+ (aq) → Zn^2+ (aq) + Cu (s).
The balanced equation for the reaction between zinc (Zn) and copper (II) bromide (CuBr2) is: Zn + CuBr2 → ZnBr2 + Cu
Zn+CuSO4=Cu+ZnSO4 right?
Zn + CuSO4 --> ZnSO4 + Cu
Mg(s) + Zn(NO3)2 (aq) ---> Mg(NO3)2 (aq) + Zn(s)