Assertion: $Zn$ is able to displace $Ag$ from $AgN{O_3}$ Solution.
Reason: The standard oxidation electrode potential of $Zn$ is more positive than that of $Ag$ .
1. Both the assertion and reason are true and the reason is the correct explanation of the assertion.
2. Both the assertion and the reason are true but the reason is not the correct explanation for the assertion
3. Assertion is true but the reason is false
4. Assertion is false but the reason is true.
5. Both assertion and reason are false.
Answer
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Hint: A metal can only be displaced from its compound by another metal, if the metal which is being added is more electropositive than the metal which is to be displaced.
If a metal is more electropositive then it will have a more positive oxidation electrode potential.
Complete step by step solution:
Let us understand the terms in the given question. Zinc ( $Zn$ ) and Silver ( $Ag$ ) are both metals. Metals undergo loss of electrons to attain octet and hence they are electropositive.
The type of reaction taking place in the question is called a displacement reaction.
$Zn + AgN{O_3}\xrightarrow{{}}ZnN{O_3} + Ag$
This reaction is called a displacement reaction, because the more electropositive metal, (which in this case is Zinc), displaces the less electropositive metal, (which in this case is Silver) from its compound, (which in this case is Silver Nitrate) and the product formed is a compound with the more electropositive metal, (which is this case is Zinc Nitrate) and the less electropositive metal (which in this case is Silver), is displaced.
Oxidation electrode potential is the ease with which a metal will accept electron. It is measured in $Volts$. Oxidation potential is more positive for more electropositive metals. An Electropositive metal and an Electronegative non-metal form an ionic bond. If the solution has a more electropositive metal, the electronegative non-metal will have a higher affinity for that metal.
We can say this is another term that, If a metal has higher positive oxidation electrode potential than the metal in the ionic compound, the compound will break and the metal with lower value of oxidation potential will be displaced by the metal with higher oxidation potential.
Therefore , Both the assertion and reason are true and the reason is the correct explanation of the assertion.
Hence, Option 1 is the correct answer.
Note:
Reduction electrode potential is the exact opposite of Oxidation electrode potential in terms of magnitude but with opposite sign. For example: The Standard Oxidation potential of Zinc is $0.763V$and its Standard reduction potential will be $ - 0.763V$. In such a case, the metal with more negative value of reduction potential will displace a metal with less negative value of reduction potential.
If a metal is more electropositive then it will have a more positive oxidation electrode potential.
Complete step by step solution:
Let us understand the terms in the given question. Zinc ( $Zn$ ) and Silver ( $Ag$ ) are both metals. Metals undergo loss of electrons to attain octet and hence they are electropositive.
The type of reaction taking place in the question is called a displacement reaction.
$Zn + AgN{O_3}\xrightarrow{{}}ZnN{O_3} + Ag$
This reaction is called a displacement reaction, because the more electropositive metal, (which in this case is Zinc), displaces the less electropositive metal, (which in this case is Silver) from its compound, (which in this case is Silver Nitrate) and the product formed is a compound with the more electropositive metal, (which is this case is Zinc Nitrate) and the less electropositive metal (which in this case is Silver), is displaced.
Oxidation electrode potential is the ease with which a metal will accept electron. It is measured in $Volts$. Oxidation potential is more positive for more electropositive metals. An Electropositive metal and an Electronegative non-metal form an ionic bond. If the solution has a more electropositive metal, the electronegative non-metal will have a higher affinity for that metal.
We can say this is another term that, If a metal has higher positive oxidation electrode potential than the metal in the ionic compound, the compound will break and the metal with lower value of oxidation potential will be displaced by the metal with higher oxidation potential.
Therefore , Both the assertion and reason are true and the reason is the correct explanation of the assertion.
Hence, Option 1 is the correct answer.
Note:
Reduction electrode potential is the exact opposite of Oxidation electrode potential in terms of magnitude but with opposite sign. For example: The Standard Oxidation potential of Zinc is $0.763V$and its Standard reduction potential will be $ - 0.763V$. In such a case, the metal with more negative value of reduction potential will displace a metal with less negative value of reduction potential.
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