For the reduction of NO3^- ion in an aqueous solution, E° is +0.96 V. Values of E°: V^2+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Thermodynamic Criterion for Spontaneous Oxidation by Nitrate For a metal $\ce{M}$ to be spontaneously oxidized by nitrate ions ($\ce{NO3^-}$) in an acidic aqueous solution under standard conditions, the overall cell potential ($E^\circ_{\text{cell}}$) of the combined redox reaction must be strictly positive: $$E^\circ_{\text{cell}} > 0\text{ V}$$ The standard reduction potential for the nitrate ion in an acidic medium is given as: $$\ce{NO3^-(aq) + 4H^+(aq) + 3e^- -> NO(g) + 2H2O(l)} \quad E^\circ_{\ce{NO3^-/NO}} = +0.96\text{ V}$$ The oxidation half-reaction of a metal $\ce{M}$ to its corresponding cation $\ce{M^{z+}}$ is represented as: $$\ce{M(s) -> M^{z+}(aq) + z e^-}$$ The standard cell potential of the combined redox process is calculated using the formula: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{NO3^-/NO}} - E^\circ_{\ce{M^{z+}/M}}$$ $$E^\circ_{\text{cell}} = +0.96\text{ V} - E^\circ_{\ce{M^{z+}/M}}$$ Substituting the spontaneity condition ($E^\circ_{\text{cell}} > 0\text{ V}$): $$+0.96\text{ V} - E^\circ_{\ce{M^{z+}/M}} > 0\text{ V} \implies E^\circ_{\ce{M^{z+}/M}} < +0.96\text{ V}$$ Therefore, any metal that has a standard reduction potential ($E^\circ_{\ce{M^{z+}/M}}$) strictly less than $+0.96\text{ V}$ will be spontaneously oxidized by nitrate ions in an acidic aqueous solution. Step 2 - Analyze Each Metal Individually Let us evaluate the feasibility of oxidation for each of the four given metals using their standard reduction potentials ($E^\circ_{\text{red}}$): 1. **Vanadium ($\ce{V}$):** $$\ce{V^{2+}(aq) + 2e^- -> V(s)} \quad E^\circ_{\ce{V^2+/V}} = -1.19\text{ V}$$ Comparing the potentials: $$-1.19\text{ V} < +0.96\text{ V}$$ $$E^\circ_{\text{cell}} = 0.96\text{ V} - (-1.19\text{ V}) = +2.15\text{ V} > 0\text{ V}$$ Therefore, Vanadium is **spontaneously oxidized** by $\ce{NO3^-}$ in aqueous solution. 2. **Iron ($\ce{Fe}$):** $$\ce{Fe^{3+}(aq) + 3e^- -> Fe(s)} \quad E^\circ_{\ce{Fe^3+/Fe}} = -0.04\text{ V}$$ Comparing the potentials: $$-0.04\text{ V} < +0.96\text{ V}$$ $$E^\circ_{\text{cell}} = 0.96\text{ V} - (-0.04\text{ V}) = +1.00\text{ V} > 0\text{ V}$$ Therefore, Iron is **spontaneously oxidized** by $\ce{NO3^-}$ in aqueous solution. 3. **Mercury ($\ce{Hg}$):** $$\ce{Hg^{2+}(aq) + 2e^- -> Hg(l)} \quad E^\circ_{\ce{Hg^2+/Hg}} = +0.86\text{ V}$$ Comparing the potentials: $$+0.86\text{ V} < +0.96\text{ V}$$ $$E^\circ_{\text{cell}} = 0.96\text{ V} - 0.86\text{ V} = +0.10\text{ V} > 0\text{ V}$$ Therefore, Mercury is **spontaneously oxidized** by $\ce{NO3^-}$ in aqueous solution. 4. **Gold ($\ce{Au}$):** $$\ce{Au^{3+}(aq) + 3e^- -> Au(s)} \quad E^\circ_{\ce{Au^3+/Au}} = +1.40\text{ V}$$ Comparing the potentials: $$+1.40\text{ V} > +0.96\text{ V}$$ $$E^\circ_{\text{cell}} = 0.96\text{ V} - 1.40\text{ V} = -0.44\text{ V} < 0\text{ V}$$ Therefore, Gold **cannot be oxidized** by $\ce{NO3^-}$ under standard conditions. Step 3 - Evaluate the Option Pairs * **Option (A) is correct:** This option contains the pair **V and Hg** (Vanadium and Mercury). Both metals have standard reduction potentials less than $+0.96\text{ V}$ ($-1.19\text{ V}$ and $+0.86\text{ V}$ respectively), so both are spontaneously oxidized by nitrate. * **Option (B) is correct:** This option contains the pair **Hg and Fe** (Mercury and Iron). Both metals have standard reduction potentials less than $+0.96\text{ V}$ ($+0.86\text{ V}$ and $-0.04\text{ V}$ respectively), so both are spontaneously oxidized by nitrate. * **Option (C) is incorrect:** This option contains **Fe and Au** (Iron and Gold). Since Gold has a standard reduction potential of $+1.40\text{ V}$, which is greater than $+0.96\text{ V}$, it cannot be oxidized by nitrate. * **Option (D) is incorrect:** This option contains **Au and V** (Gold and Vanadium). Since Gold cannot be oxidized by nitrate under standard conditions, this pair is incorrect. $$\text{Correct Options: } \boxed{A, B}$$