
The given equation represents the reaction of copper sulphate with an element X.
\[CuS{O_4} + X \to Cu + Y\]
Which element is represented by X, among Fe and Ag? Justify your answer. Write the molecular formula of the compound represented by Y.
Answer
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Hint: We know from the electrochemical series that the standard reduction potential of iron (Fe) is less than that of copper’s (Cu). The standard reduction potential of silver (Ag) is higher than copper (Cu).
Complete step by step answer:
We know that if the electrons are accepted by the metal electrode during the process of electrochemical reaction, this is called the reduction potential of that electrode. Thus, it is known that the reduction potential of a metal electrode as determined with respect to the standard hydrogen electrode is called standard reduction potential \[({E^ \circ })\].
According to the electrochemical series, we can write the standard reduction potential \[({E^ \circ })\] of Ag, Cu, and Fe as follows.
Generally, the oxidation state of silver is +1 in an aqueous state.
\[A{g^ + }_{(aq)} + {e^ - } \to A{g_{(s)}}\] \[{E^ \circ } = 0.80V\]
Generally, the oxidation state of copper is +2 in an aqueous state.
\[C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}\] \[{E^ \circ } = 0.34V\]
Generally, the oxidation state of iron is +2 in an aqueous state.
\[F{e^{2 + }}_{(aq)} + 2{e^ - } \to F{e_{(s)}}\] \[{E^ \circ } = - 0.45V\]
We know that the smaller the standard reduction potential, the stronger is the reducing agent.
Hence, from the above chemical reactions and its standard reduction potential value, we know that Fe is a stronger reducing agent than Ag.
Therefore, Fe can replace Cu from copper sulphate as shown in the following reaction:
\[CuS{O_4} + Fe(X) \to Cu + FeS{O_4}(Y)\]
Therefore, we can conclude that X is Fe and the molecular formula for Y is \[FeS{O_4}\].
Note:
The elements present below in the electrochemical series are stronger reducing agents and the elements present above in the electrochemical series are the stronger oxidizing agents.
Complete step by step answer:
We know that if the electrons are accepted by the metal electrode during the process of electrochemical reaction, this is called the reduction potential of that electrode. Thus, it is known that the reduction potential of a metal electrode as determined with respect to the standard hydrogen electrode is called standard reduction potential \[({E^ \circ })\].
According to the electrochemical series, we can write the standard reduction potential \[({E^ \circ })\] of Ag, Cu, and Fe as follows.
Generally, the oxidation state of silver is +1 in an aqueous state.
\[A{g^ + }_{(aq)} + {e^ - } \to A{g_{(s)}}\] \[{E^ \circ } = 0.80V\]
Generally, the oxidation state of copper is +2 in an aqueous state.
\[C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}\] \[{E^ \circ } = 0.34V\]
Generally, the oxidation state of iron is +2 in an aqueous state.
\[F{e^{2 + }}_{(aq)} + 2{e^ - } \to F{e_{(s)}}\] \[{E^ \circ } = - 0.45V\]
We know that the smaller the standard reduction potential, the stronger is the reducing agent.
Hence, from the above chemical reactions and its standard reduction potential value, we know that Fe is a stronger reducing agent than Ag.
Therefore, Fe can replace Cu from copper sulphate as shown in the following reaction:
\[CuS{O_4} + Fe(X) \to Cu + FeS{O_4}(Y)\]
Therefore, we can conclude that X is Fe and the molecular formula for Y is \[FeS{O_4}\].
Note:
The elements present below in the electrochemical series are stronger reducing agents and the elements present above in the electrochemical series are the stronger oxidizing agents.
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