
Statement 1: A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless \[Z{n^{2 + }}\] solution.
Statement 2: Copper is a much more active metal than zinc.
A.Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1.
B.Both statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1.
C.Statement 1 is correct but statement 2 is not correct.
D.Statement 1 is not correct but statement 2 is correct.
E.Both the statement 1 and statement 2 are not correct
Answer
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Hint:We can solve this question using the concept of electrochemical series. Electrochemical series also referred to as activity is a list of elements arranged in order of their increasing electrode potential values. The series has been made by measuring the potential of various electrodes concerning the standard hydrogen electrode (SHE). It is a series of electrodes arranged in order of their increasing standard oxidation potentials or decreasing order of their standard reduction potentials. It helps us to select good oxidizing and reducing agents. In the electrochemical series, the element having a high negative value of standard reduction potential readily loses electrons and is converted into a cation. An element that is higher in the series is more reactive than an element that is lower than it which means that it can displace it from its solution.
Complete step by step answer:
Firstly, we will analyze the given statements.
Statement 1 is: A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless \[Z{n^{2 + }}\] solution.
From the electrochemical series, we can see that zinc is present above copper in the electrochemical series. Hence, we can infer that zinc is more reactive than copper. Hence zinc will displace copper from its solution. Therefore statement 1 is correct. Zinc will displace copper from copper nitrate solution and will form colorless zinc nitrate. The balanced chemical equation for the reaction is:
\[Zn(s) + Cu{(N{O_3})_2}(aq) \to Zn{(N{O_3})_2}(aq) + Cu(s)\]
Statement 2: Copper is a much more active metal than zinc.
This statement is wrong as zinc is present above copper in the electrochemical series and is more reactive than copper.
Hence option (C) is the correct answer.
Note:
The electrode potential of SHE is zero.
The element with greater reduction potential gets reduced easily.
Elements with low reduction potential get oxidized easily.
Complete step by step answer:
Firstly, we will analyze the given statements.
Statement 1 is: A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless \[Z{n^{2 + }}\] solution.
From the electrochemical series, we can see that zinc is present above copper in the electrochemical series. Hence, we can infer that zinc is more reactive than copper. Hence zinc will displace copper from its solution. Therefore statement 1 is correct. Zinc will displace copper from copper nitrate solution and will form colorless zinc nitrate. The balanced chemical equation for the reaction is:
\[Zn(s) + Cu{(N{O_3})_2}(aq) \to Zn{(N{O_3})_2}(aq) + Cu(s)\]
Statement 2: Copper is a much more active metal than zinc.
This statement is wrong as zinc is present above copper in the electrochemical series and is more reactive than copper.
Hence option (C) is the correct answer.
Note:
The electrode potential of SHE is zero.
The element with greater reduction potential gets reduced easily.
Elements with low reduction potential get oxidized easily.
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