In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the chemical reaction involved.
Answer
Verified
468.6k+ 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.
Recently Updated Pages
If the perimeter of the equilateral triangle is 18-class-10-maths-CBSE
How do you make the plural form of most of the words class 10 english CBSE
Quotes and Slogans on Consumer Rights Can Anybody Give Me
What is the orbit of a satellite Find out the basis class 10 physics CBSE
the period from 1919 to 1947 forms an important phase class 10 social science CBSE
If the average marks of three batches of 55 60 and class 10 maths CBSE
Trending doubts
Assertion The planet Neptune appears blue in colour class 10 social science CBSE
The term disaster is derived from language AGreek BArabic class 10 social science CBSE
Imagine that you have the opportunity to interview class 10 english CBSE
Find the area of the minor segment of a circle of radius class 10 maths CBSE
Differentiate between natural and artificial ecosy class 10 biology CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE