Four colourless salt solutions are placed in separate test tubes and a strip of copper is placed in — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understanding Metal Reactivity and Displacement Reactions A metal can spontaneously displace another metal from its salt solution if it is more reactive (i.e., more electropositive, or has a lower/more negative standard reduction potential) than the metal present as cations in the solution. In terms of standard reduction potentials ($E^\circ$) at $25^\circ\text{C}$: * Zinc: $E^\circ(\ce{Zn^{2+}/Zn}) = -0.76\text{ V}$ * Cadmium: $E^\circ(\ce{Cd^{2+}/Cd}) = -0.40\text{ V}$ * Lead: $E^\circ(\ce{Pb^{2+}/Pb}) = -0.13\text{ V}$ * Copper: $E^\circ(\ce{Cu^{2+}/Cu}) = +0.34\text{ V}$ * Silver: $E^\circ(\ce{Ag^{+}/Ag}) = +0.80\text{ V}$ Arranging these metals in the decreasing order of reactivity (increasing order of standard reduction potentials) gives: $$\text{Reactivity Order: } \ce{Zn} > \ce{Cd} > \ce{Pb} > \ce{Cu} > \ce{Ag}$$ This sequence shows that copper ($\ce{Cu}$) is more reactive than silver ($\ce{Ag}$), but is less reactive than zinc ($\ce{Zn}$), cadmium ($\ce{Cd}$), and lead ($\ce{Pb}$). Step 2 - Assessing the Feasibility of Reactions in Each Test Tube When a metallic strip of copper ($\ce{Cu}$) is placed in each of the four colorless nitrate solutions: 1. **In Silver Nitrate Solution ($\ce{AgNO3}$):** Since copper is more reactive than silver, it readily loses electrons to reduce silver ions ($\ce{Ag^+}$) to metallic silver, while copper itself undergoes oxidation to form copper(II) ions ($\ce{Cu^2+}$): $$\ce{Cu(s) + 2AgNO3(aq) -> Cu(NO3)2(aq) + 2Ag(s)}$$ $$\text{Net Ionic Equation: } \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)}$$ The standard electromotive force ($E^\circ_{\text{cell}}$) for this displacement reaction is: $$E^\circ_{\text{cell}} = E^\circ_{\ce{Ag+|Ag}} - E^\circ_{\ce{Cu^2+|Cu}} = +0.80\text{ V} - 0.34\text{ V} = +0.46\text{ V}$$ Since $E^\circ_{\text{cell}} > 0$, this reaction is thermodynamically spontaneous ($\Delta G^\circ < 0$). The formation of hydrated $\ce{Cu^2+(aq)}$ ions imparts a characteristic deep blue color to the solution. 2. **In Lead Nitrate Solution ($\ce{Pb(NO3)2}$):** Since copper is less reactive than lead ($E^\circ_{\ce{Cu^2+|Cu}} > E^\circ_{\ce{Pb^2+|Pb}}$), it cannot reduce lead ions ($\ce{Pb^2+}$). No reaction occurs, and the solution remains colorless: $$\ce{Cu(s) + Pb^2+(aq) -> \text{No Reaction}}$$ 3. **In Zinc Nitrate Solution ($\ce{Zn(NO3)2}$):** Since copper is less reactive than zinc ($E^\circ_{\ce{Cu^2+|Cu}} > E^\circ_{\ce{Zn^2+|Zn}}$), copper cannot displace zinc. No reaction occurs, and the solution remains colorless: $$\ce{Cu(s) + Zn^2+(aq) -> \text{No Reaction}}$$ 4. **In Cadmium Nitrate Solution ($\ce{Cd(NO3)2}$):** Since copper is less reactive than cadmium ($E^\circ_{\ce{Cu^2+|Cu}} > E^\circ_{\ce{Cd^2+|Cd}}$), copper cannot displace cadmium. No reaction occurs, and the solution remains colorless: $$\ce{Cu(s) + Cd^2+(aq) -> \text{No Reaction}}$$ Step 3 - Conclusion and Option Analysis * **Option (A) is correct** because copper successfully reduces silver ions, resulting in the dissolution of copper as blue-colored $\ce{Cu^2+(aq)}$ ions. * **Option (B) is incorrect** because copper is less reactive than lead and cannot displace it. * **Option (C) is incorrect** because copper is less reactive than zinc and cannot displace it. * **Option (D) is incorrect** because copper is less reactive than cadmium and cannot displace it. $$\text{Correct Option: } \boxed{\text{A}}$$