What coefficient of $ AgN{O_3} $ is required to balance the following reaction:
$ Cu{\text{ + AgN}}{{\text{O}}_3}{\text{ }}\xrightarrow{{}}{\text{ Cu}}{\left( {N{O_3}} \right)_2}{\text{ + Ag}} $
$ A. $ 0
$ B. $ 1
$ C. $ 2
$ D. $ 3
Answer
524.1k+ views
Hint: A balanced chemical reaction is one in which the number of atoms present in the reactants are equal to the number of atoms present in the products. Therefore the atomicity of each individual atom must be the same before the reaction and after the reaction.
Complete answer:
We can balance a chemical reaction by using a hit and trial method. Using the hit and trial method, we can balance the reaction verbally. But we will balance the reaction by separating the anodic and cathodic reactions separately. The above reaction is an example of a redox reaction. We will write two half-cell reactions which are anodic and cathodic reactions.
At anode;
$ Cu{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 2}}{\text{ + 2e}} $ ________ $ \left( 1 \right) $
At cathode;
$ A{g^ + }{\text{ + 1e }}\xrightarrow{{}}{\text{ Ag }} $ _________ $ \left( 2 \right) $
We will balance these half-cell reactions which will balance complete reactions. Balancing the electrons in the reaction:
Multiply equation $ \left( 2 \right) $ by $ 2 $ and then adding it with equation $ \left( 1 \right) $ , we get
$ Cu{\text{ + 2A}}{{\text{g}}^ + }{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 2}}{\text{ + 2Ag}} $
This is the balanced half-cell reaction. Now substituting the ions to their original molecule we get,
$ Cu{\text{ + 2AgN}}{{\text{O}}_3}{\text{ }}\xrightarrow{{}}{\text{ Cu}}{\left( {N{O_3}} \right)_2}{\text{ + 2Ag}} $
Therefore by taking two moles of $ AgN{O_3} $ the reaction becomes a balanced chemical reaction. Such reactions are balanced by balancing the number of electrons shared at anode and cathode respectively.
Note:
The atomicity of atoms must be the same at both sides of the reaction. By using hit and trial methods, the probability of error is large. So we used to write half-reaction separately and balanced them and added them to get complete reactions. This also follows the law of conservation of mass. For the redox reaction we have separate anode and cathode reactions. Balance the anode and cathode reaction by balancing the number of shared reactions separately.
Complete answer:
We can balance a chemical reaction by using a hit and trial method. Using the hit and trial method, we can balance the reaction verbally. But we will balance the reaction by separating the anodic and cathodic reactions separately. The above reaction is an example of a redox reaction. We will write two half-cell reactions which are anodic and cathodic reactions.
At anode;
$ Cu{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 2}}{\text{ + 2e}} $ ________ $ \left( 1 \right) $
At cathode;
$ A{g^ + }{\text{ + 1e }}\xrightarrow{{}}{\text{ Ag }} $ _________ $ \left( 2 \right) $
We will balance these half-cell reactions which will balance complete reactions. Balancing the electrons in the reaction:
Multiply equation $ \left( 2 \right) $ by $ 2 $ and then adding it with equation $ \left( 1 \right) $ , we get
$ Cu{\text{ + 2A}}{{\text{g}}^ + }{\text{ }}\xrightarrow{{}}{\text{ C}}{{\text{u}}^{ + 2}}{\text{ + 2Ag}} $
This is the balanced half-cell reaction. Now substituting the ions to their original molecule we get,
$ Cu{\text{ + 2AgN}}{{\text{O}}_3}{\text{ }}\xrightarrow{{}}{\text{ Cu}}{\left( {N{O_3}} \right)_2}{\text{ + 2Ag}} $
Therefore by taking two moles of $ AgN{O_3} $ the reaction becomes a balanced chemical reaction. Such reactions are balanced by balancing the number of electrons shared at anode and cathode respectively.
Note:
The atomicity of atoms must be the same at both sides of the reaction. By using hit and trial methods, the probability of error is large. So we used to write half-reaction separately and balanced them and added them to get complete reactions. This also follows the law of conservation of mass. For the redox reaction we have separate anode and cathode reactions. Balance the anode and cathode reaction by balancing the number of shared reactions separately.
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