
Name a metal the does not react with dil.\[{\text{HCl}}\] or \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\].
A) Copper
B) Silver
C) Gold
D) All of the above
Answer
520.8k+ views
Hint: From the electrochemical series the standard reduction potential of metals greater than that of hydrogen cannot replace hydrogen from the non-metal anion. Hence, we have to find which metal had greater standard reduction potential than hydrogen from the following option.
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 \[\left( {{E^ \circ }} \right)\].
The hydrogen electrode is kept as a reference because its standard reduction potential is zero.
According to the electrochemical series, we can write the standard reduction potential \[\left( {{E^ \circ }} \right)\] of silver, copper and gold as follow:
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 gold is \[ + 3\] in an aqueous state.
\[Au_{(aq)}^{ + 3} + 3{e^ - } \to A{u_{(s)}}\]\[{E^ \circ } = 1.50V\]
It is known that the greater the standard reduction potential stronger is the oxidizing agent.
But to replace the hydrogen from the dilute acid we need a strong reducing agent. All the metal given in the options will act as a strong reducing agent. So, none of the metals can react with dil. \[{\text{HCl}}\] or \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\].
Therefore, we can conclude that the correct option for the following question is option D.
Note:
Below in the electrochemical series are stronger reducing agents and 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 \[\left( {{E^ \circ }} \right)\].
The hydrogen electrode is kept as a reference because its standard reduction potential is zero.
According to the electrochemical series, we can write the standard reduction potential \[\left( {{E^ \circ }} \right)\] of silver, copper and gold as follow:
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 gold is \[ + 3\] in an aqueous state.
\[Au_{(aq)}^{ + 3} + 3{e^ - } \to A{u_{(s)}}\]\[{E^ \circ } = 1.50V\]
It is known that the greater the standard reduction potential stronger is the oxidizing agent.
But to replace the hydrogen from the dilute acid we need a strong reducing agent. All the metal given in the options will act as a strong reducing agent. So, none of the metals can react with dil. \[{\text{HCl}}\] or \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\].
Therefore, we can conclude that the correct option for the following question is option D.
Note:
Below in the electrochemical series are stronger reducing agents and elements present above in the electrochemical series are the stronger oxidizing agents.
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