When a magnesium wire is dipped into a solution of lead(II) nitrate, a black deposit forms on the wi — Electrochemistry Chemistry Question

STEPS:
1. Identify the observed chemical behavior: When a magnesium wire is dipped into a lead(II) nitrate solution, the spontaneous formation of a black deposit on the wire indicates that a chemical reaction is taking place on the surface of the metal.
2. Determine the reactant and product species: The reactants are solid magnesium metal ($\text{Mg}(s)$) and dissolved aqueous lead(II) ions ($\text{Pb}^{2+}(aq)$). The solid black deposit represents reduced metallic lead ($\text{Pb}(s)$) forming on the wire as magnesium metal is oxidized to magnesium ions ($\text{Mg}^{2+}(aq)$).
3. Write the net ionic equation for the process:
$$\text{Mg}(s) + \text{Pb}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Pb}(s)$$
4. Relate spontaneity to standard reduction potentials ($E^\circ$): For a redox reaction to proceed spontaneously under standard conditions, the standard cell potential must be positive ($E^\circ_{\text{cell}} > 0$). The cell potential is defined as:
$$E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{reduction half-reaction}) - E^\circ_{\text{red}}(\text{oxidation half-reaction})$$
Applying this to our reaction:
$$E^\circ_{\text{cell}} = E^\circ(\text{Pb}^{2+}/\text{Pb}) - E^\circ(\text{Mg}^{2+}/\text{Mg}) > 0$$
5. Formulate the final mathematical conclusion: Rearranging the inequality demonstrates that the standard reduction potential for the lead species must be greater than that of the magnesium species:
$$E^\circ(\text{Pb}^{2+}/\text{Pb}) > E^\circ(\text{Mg}^{2+}/\text{Mg})$$
This is confirmed by the exam's table of standard reduction potentials, which lists $E^\circ$ for $\text{Pb}^{2+}(aq)$ as $-0.13\text{ V}$ and $E^\circ$ for $\text{Mg}^{2+}(aq)$ as $-2.37\text{ V}$. This confirms Option A is correct.

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WHY_OTHERS_WRONG:
* B is incorrect: Since magnesium spontaneously undergoes oxidation to reduce the lead(II) ions, solid magnesium ($\text{Mg}(s)$) is *more* easily oxidized (a stronger reducing agent) than solid lead ($\text{Pb}(s)$), not less.
* C is incorrect: The reaction occurs spontaneously simply by contacting the two substances without any external power source.
* D is incorrect: In an electrochemical cell, oxidation always takes place at the anode. Because the magnesium wire undergoes oxidation ($\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-$), it would serve as the *anode* of a Mg/Pb cell, not the cathode.
* E is incorrect: Because the forward reaction is spontaneous, the reverse reaction is thermodynamically unfavorable. Solid lead ($\text{Pb}(s)$) cannot spontaneously displace magnesium ions ($\text{Mg}^{2+}(aq)$) from solution because that reverse process would have a negative cell potential of $-2.24\text{ V}$.

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