Answer
414.9k+ views
Hint: It is given that the displacement reaction is taking place so, to write the reaction involved, we can use the fact that a highly reactive metal can displace less reactive metal ion in a displacement reaction. To identify which metal is more reactive, reactivity series can be used.
Complete answer:
The reactivity series is:
Potassium (K)
Sodium (Na)
Calcium (Ca)
Magnesium (Mg)
Aluminium (Al)
Zinc (Zn)
Iron (Fe)
Tin (Sn)
Lead (Pb)
Copper (Cu)
Silver (Ag)
Gold (Au)
Platinum (Pt)
As we move down the series, the reactivity decreases.
When we compare silver (Ag) and copper (Cu):
Cu is more reactive than Ag as it lies above it.
The reaction taking place in refining can be given as:
Reactants:
Silver nitrate $\left[ {AgN{O_3}} \right]$ reacts with Copper metal [Cu].
Products:
Silver metal [Ag] (refined) and Copper nitrate $\left[ {CuN{O_3}} \right]$ (ion exchange)
Reaction:
$AgN{O_3}_{(aq)} + C{u_{(s)}} \to A{g_{(s)}} + CuN{O_3}_{(aq)}$
Here, (aq) means aqueous (dissolved in water)
Balancing the reaction:
$2AgN{O_3}_{(aq)} + C{u_{(s)}} \to 2A{g_{(s)}} + Cu{\left( {N{O_3}} \right)_2}_{(aq)}$
In balancing we equate the number of molecules of an element present on the reactant side equal to the product side.
Here, in the reaction, Copper (Cu) is displacing Silver from its nitrate solution $\left( {AgN{O_3}} \right)$ as it is more reactive compared to silver. Copper itself forms a compound by reacting with nitrate $\left( {N{O_3}} \right)$ and refined silver metal is obtained as another product.
Nitrate with silver forms $AgN{O_3}$ but with copper forms $Cu{\left( {N{O_3}} \right)_2}$ because valency of both silver and nitrate is 1 but that of copper is 2 and during the formation of compound the valencies are exchanges, so that 2 of copper’s valency goes to nitrate and it becomes ${\left( {N{O_3}} \right)_2}$
It can be stated that when the metallic copper is dissolved in silver nitrate solution during the recovery of silver, copper nitrate is formed, whereas silver is precipitated as metallic silver.
Therefore, the required reaction for given way of refining of silver is $2AgN{O_3}_{(aq)} + C{u_{(s)}} \to 2A{g_{(s)}} + Cu{\left( {N{O_3}} \right)_2}_{(aq)}$
Note: Displacement of metal ions by highly reactive metal can be learnt and observed with the help of activity series. The metal placed above can easily displace the metal placed below in the series as reactivity decreases moving downwards.
Complete answer:
The reactivity series is:
Potassium (K)
Sodium (Na)
Calcium (Ca)
Magnesium (Mg)
Aluminium (Al)
Zinc (Zn)
Iron (Fe)
Tin (Sn)
Lead (Pb)
Copper (Cu)
Silver (Ag)
Gold (Au)
Platinum (Pt)
As we move down the series, the reactivity decreases.
When we compare silver (Ag) and copper (Cu):
Cu is more reactive than Ag as it lies above it.
The reaction taking place in refining can be given as:
Reactants:
Silver nitrate $\left[ {AgN{O_3}} \right]$ reacts with Copper metal [Cu].
Products:
Silver metal [Ag] (refined) and Copper nitrate $\left[ {CuN{O_3}} \right]$ (ion exchange)
Reaction:
$AgN{O_3}_{(aq)} + C{u_{(s)}} \to A{g_{(s)}} + CuN{O_3}_{(aq)}$
Here, (aq) means aqueous (dissolved in water)
Balancing the reaction:
$2AgN{O_3}_{(aq)} + C{u_{(s)}} \to 2A{g_{(s)}} + Cu{\left( {N{O_3}} \right)_2}_{(aq)}$
In balancing we equate the number of molecules of an element present on the reactant side equal to the product side.
Here, in the reaction, Copper (Cu) is displacing Silver from its nitrate solution $\left( {AgN{O_3}} \right)$ as it is more reactive compared to silver. Copper itself forms a compound by reacting with nitrate $\left( {N{O_3}} \right)$ and refined silver metal is obtained as another product.
Nitrate with silver forms $AgN{O_3}$ but with copper forms $Cu{\left( {N{O_3}} \right)_2}$ because valency of both silver and nitrate is 1 but that of copper is 2 and during the formation of compound the valencies are exchanges, so that 2 of copper’s valency goes to nitrate and it becomes ${\left( {N{O_3}} \right)_2}$
It can be stated that when the metallic copper is dissolved in silver nitrate solution during the recovery of silver, copper nitrate is formed, whereas silver is precipitated as metallic silver.
Therefore, the required reaction for given way of refining of silver is $2AgN{O_3}_{(aq)} + C{u_{(s)}} \to 2A{g_{(s)}} + Cu{\left( {N{O_3}} \right)_2}_{(aq)}$
Note: Displacement of metal ions by highly reactive metal can be learnt and observed with the help of activity series. The metal placed above can easily displace the metal placed below in the series as reactivity decreases moving downwards.
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