The passage of electricity in the Daniel cell when Zn and Cu electrodes are connected — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Electrode Processes in a Daniell Cell A standard Daniell cell is a galvanic cell consisting of a zinc ($\ce{Zn}$) electrode immersed in a zinc sulfate ($\ce{ZnSO4}$) solution and a copper ($\ce{Cu}$) electrode immersed in a copper(II) sulfate ($\ce{CuSO4}$) solution. The two half-cell reactions are: * **Anode (Oxidation half-reaction):** Zinc metal has a higher oxidation tendency than copper. It undergoes oxidation to release zinc ions and electrons into the external circuit: $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ Thus, the zinc electrode acts as the negative terminal (anode). * **Cathode (Reduction half-reaction):** Copper(II) ions in the solution accept electrons from the electrode surface and undergo reduction to deposit metallic copper: $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ Thus, the copper electrode acts as the positive terminal (cathode). Step 2 - Analyze the External Circuit (Outside the Cell) In the external circuit (consisting of the metallic wire connecting the two electrodes): * Electrons ($\ce{e^-}$) flow spontaneously from the region of high electron density at the anode ($\ce{Zn}$) to the cathode ($\ce{Cu}$): $$\text{Direction of electron flow (external): } \ce{Zn -> Cu}$$ * By physics convention, the direction of conventional electrical current (the "passage of electricity") is defined as opposite to the direction of electron flow (or in the direction of positive charge flow): $$\text{Direction of conventional current (external): } \ce{Cu -> Zn}$$ This analysis confirms that **Option (B) is correct** and Option (D) is incorrect. Step 3 - Analyze the Internal Circuit (Inside the Cell) To complete the electrical circuit and maintain charge neutrality, ions migrate inside the cell through the salt bridge or porous barrier separating the two compartments: * **Cations** (positive ions, such as $\ce{Zn^{2+}}$, $\ce{K^+}$, or $\ce{NH4^+}$) migrate from the anode chamber ($\ce{Zn}$ compartment) toward the cathode chamber ($\ce{Cu}$ compartment) to balance the excess negative charge of the sulfate ($\ce{SO4^{2-}}$) ions left behind by the reduction of $\ce{Cu^{2+}}$. * **Anions** (negative ions, such as $\ce{SO4^{2-}}$ or $\ce{Cl^-}$) migrate from the cathode chamber toward the anode chamber to neutralize the positive charge of the newly generated $\ce{Zn^{2+}}$ ions. * Since conventional current is defined by the movement of positive charges, the migration of cations from the zinc compartment to the copper compartment establishes that: $$\text{Direction of conventional current (internal): } \ce{Zn -> Cu}$$ This analysis confirms that **Option (C) is correct** and Option (A) is incorrect. Step 4 - Evaluate the Options and Conclude * **Option (A) is incorrect:** Inside the cell, conventional current is carried by cations moving towards the cathode, which is from $\ce{Zn}$ to $\ce{Cu}$. * **Option (B) is correct:** Outside the cell, conventional current flows opposite to the direction of electron flow, which is from $\ce{Cu}$ to $\ce{Zn}$. * **Option (C) is correct:** Inside the cell, conventional current flows from $\ce{Zn}$ to $\ce{Cu}$ via cation migration. * **Option (D) is incorrect:** Outside the cell, electrons flow from $\ce{Zn}$ to $\ce{Cu}$, meaning conventional current flows from $\ce{Cu}$ to $\ce{Zn}$. $$\text{Correct Options: } \boxed{B, C}$$