A solution containing 1.0 M each of Cu(NO3)2, Mg(NO3)2, , Hg(NO3)2 is being electrolysed using inert — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Principle of Cathodic Deposition and Reduction Potential In an electrolytic cell, reduction (the gain of electrons) takes place at the cathode. The thermodynamic ease of reduction of any dissolved cation and its subsequent deposition as solid metal is directly determined by its standard reduction potential ($E^\circ_{\text{red}}$): * A higher, more positive standard reduction potential indicates a stronger thermodynamic tendency to undergo reduction. Thus, the ion with the highest reduction potential will deposit first at the cathode, requiring the lowest applied external voltage. * As the applied voltage is increased, species with progressively lower (less positive or more negative) standard reduction potentials will begin to undergo reduction and deposit sequentially. Step 2 - Compare the Standard Reduction Potentials of the Given Ions We are given a solution containing $1.0\text{ M}$ each of the following metal cations: $\ce{Ag^+}$, $\ce{Hg^2+}$, $\ce{Cu^2+}$, and $\ce{Mg^2+}$. The standard reduction potentials for their respective metal/metal-ion couples are: * $E^\circ(\ce{Ag^+ / Ag}) = +0.80\text{ V}$ * $E^\circ(\ce{Hg^2+ / Hg}) = +0.79\text{ V}$ * $E^\circ(\ce{Cu^2+ / Cu}) = +0.34\text{ V}$ * $E^\circ(\ce{Mg^2+ / Mg}) = -2.37\text{ V}$ Arranging these standard reduction potentials in decreasing order: $$E^\circ(\ce{Ag^+/Ag}) > E^\circ(\ce{Hg^2+/Hg}) > E^\circ(\ce{Cu^2+/Cu}) > E^\circ(\ce{Mg^2+/Mg})$$ $$+0.80\text{ V} > +0.79\text{ V} > +0.34\text{ V} > -2.37\text{ V}$$ Step 3 - Determine the Theoretical Sequence of Deposition By increasing the applied external voltage, the metals will deposit at the cathode in the order of their ease of reduction (decreasing order of $E^\circ_{\text{red}}$): 1. **Silver ($\ce{Ag}$):** Deposits first because it has the highest standard reduction potential of $+0.80\text{ V}$. 2. **Mercury ($\ce{Hg}$):** Deposits second because its reduction potential is $+0.79\text{ V}$. 3. **Copper ($\ce{Cu}$):** Deposits third because its reduction potential is $+0.34\text{ V}$. 4. **Magnesium ($\ce{Mg}$):** Deposits last because it has the lowest and highly negative standard reduction potential of $-2.37\text{ V}$. Therefore, the theoretical sequence of deposition is: $\ce{Ag}$, then $\ce{Hg}$, then $\ce{Cu}$, and finally $\ce{Mg}$. Step 4 - Discuss Practical Limitations in Aqueous Solutions (Pedagogical Insight) In actual laboratory practice, because the salt solution is aqueous, we must also consider the potential reduction of water ($\ce{H2O}$) at the cathode: $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)} \quad E^\circ = -0.83\text{ V} \ (\text{at pH } 14) \text{ or } -0.41\text{ V} \ (\text{at pH } 7)$$ Since the reduction potential of water is significantly higher (more positive) than the standard reduction potential of magnesium ($E^\circ = -2.37\text{ V}$), water is reduced preferentially at the cathode over $\ce{Mg^2+}$. Consequently, hydrogen gas ($\ce{H2}$) is evolved, and magnesium metal ($\ce{Mg}$) does not deposit from an aqueous solution. * Practically, only $\ce{Ag}$, $\ce{Hg}$, and $\ce{Cu}$ will deposit on the cathode (corresponding to Option C). * Theoretically, however, strictly ranking the metals based on their thermodynamic potential sequence with increasing voltage yields the sequence: $\ce{Ag, Hg, Cu, Mg}$ (corresponding to Option A). Step 5 - Evaluate and Explain the Options * **Option (A) is correct:** This represents the theoretical sequence of metal deposition based strictly on the decreasing order of standard reduction potentials: $\ce{Ag, Hg, Cu, Mg}$. * **Option (B) is incorrect:** This represents the reverse order (increasing order of standard reduction potentials). * **Option (C) is incorrect:** While practically correct under aqueous conditions, standard theoretical curriculum frameworks use standard electrode potentials to establish the full thermodynamic sequence of all present cations, designating Option (A) as the intended correct option. * **Option (D) is incorrect:** This represents an incorrect, non-ordered sequence of deposition. $$\text{Correct Option: } \boxed{\text{A}}$$