
The passage of electricity in the Daniel cell when ${\text{Zn}}$ and ${\text{Cu}}$ electrodes are connected is from:
A. ${\text{Cu}}$ to ${\text{Zn}}$ in the cell
B. ${\text{Cu}}$ to ${\text{Zn}}$ outside the cell
C. ${\text{Zn}}$ to ${\text{Cu}}$ outside the cell
D. ${\text{Zn}}$ to ${\text{Cu}}$ in the cell
Answer
573.9k+ views
Hint: The Daniel cell is an electrochemical cell which consists of two electrodes, one made of ${\text{Zn}}$ metal and the other made of ${\text{Cu}}$ metal. The ${\text{Zn}}$ electrode is in contact with ${\text{ZnS}}{{\text{O}}_{\text{4}}}$ solution and ${\text{Cu}}$ electrode is in contact with ${\text{CuS}}{{\text{O}}_{\text{4}}}$ solution. The passage of electrons in the Daniel cell is from anode to cathode.
Complete step by step answer:
The Daniel cell consists of saturated ${\text{CuS}}{{\text{O}}_{\text{4}}}$ solution filled in a copper vessel. A porous cylindrical pot inside the copper vessel is filled with dilute ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ which acts as an electrolyte.
An amalgamated zinc rod is immersed in the dilute ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$. The sulfate ions $\left( {{\text{SO}}_{\text{4}}^{2 - }} \right)$ from ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ come in contact with zinc rod and produce zinc sulfate. The reaction is,
${\text{Z}}{{\text{n}}^{2 + }} + {\text{SO}}_4^{2 - } \to {\text{ZnS}}{{\text{O}}_{\text{4}}}$
During the reaction, the zinc atom leaves two electrons on the zinc rod. Thus, the zinc rod has an excess of electrons and becomes negative.
The zinc electrode acts as anode and the copper electrode acts as cathode. Thus, the electrons flow from the zinc electrode to the copper electrode through an external circuit.
The electric current flows from cathode to anode through an external circuit. Thus, the passage of electricity is from copper electrode to the zinc electrode through an external circuit.
Thus, the passage of electricity in the Daniel cell when ${\text{Zn}}$ and ${\text{Cu}}$ electrodes are connected is from ${\text{Cu}}$ to ${\text{Zn}}$ outside the cell.
So, the correct answer is Option B .
Note:
The flow of electrons is always from the negative to the positive electrode. Thus, the electrons flow from anode to cathode. Whereas the flow of electric current is from positive to negative electrodes. Thus, the electric current flows from cathode to anode.
Complete step by step answer:
The Daniel cell consists of saturated ${\text{CuS}}{{\text{O}}_{\text{4}}}$ solution filled in a copper vessel. A porous cylindrical pot inside the copper vessel is filled with dilute ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ which acts as an electrolyte.
An amalgamated zinc rod is immersed in the dilute ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$. The sulfate ions $\left( {{\text{SO}}_{\text{4}}^{2 - }} \right)$ from ${{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ come in contact with zinc rod and produce zinc sulfate. The reaction is,
${\text{Z}}{{\text{n}}^{2 + }} + {\text{SO}}_4^{2 - } \to {\text{ZnS}}{{\text{O}}_{\text{4}}}$
During the reaction, the zinc atom leaves two electrons on the zinc rod. Thus, the zinc rod has an excess of electrons and becomes negative.
The zinc electrode acts as anode and the copper electrode acts as cathode. Thus, the electrons flow from the zinc electrode to the copper electrode through an external circuit.
The electric current flows from cathode to anode through an external circuit. Thus, the passage of electricity is from copper electrode to the zinc electrode through an external circuit.
Thus, the passage of electricity in the Daniel cell when ${\text{Zn}}$ and ${\text{Cu}}$ electrodes are connected is from ${\text{Cu}}$ to ${\text{Zn}}$ outside the cell.
So, the correct answer is Option B .
Note:
The flow of electrons is always from the negative to the positive electrode. Thus, the electrons flow from anode to cathode. Whereas the flow of electric current is from positive to negative electrodes. Thus, the electric current flows from cathode to anode.
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