For reduction of NO3- in aqueous solution, E° = +0.96 V. Standard reduction potentials: V2+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Thermodynamic Principle of Spontaneous Oxidation by Nitrate Ions The standard reduction potential ($E^\circ_{\text{red}}$) for the reduction of nitrate ions ($\ce{NO3^-}$) in an acidic aqueous medium is given as: $$\ce{NO3^-(aq) + 4H^+(aq) + 3e^- -> NO(g) + 2H2O(l)} \quad E^\circ = +0.96\text{ V}$$ For any metal ($\ce{M}$) to be spontaneously oxidized by $\ce{NO3^-}$ under standard-state conditions, the overall standard cell potential ($E^\circ_{\text{cell}}$) for the combined redox process must be positive: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0$$ Since reduction occurs at the cathode (nitrate half-cell) and oxidation occurs at the anode (metal half-cell): $$E^\circ_{\text{cell}} = E^\circ_{\ce{NO3^-/NO}} - E^\circ_{\ce{M^{n+}/M}} > 0$$ $$\implies E^\circ_{\ce{M^{n+}/M}} < +0.96\text{ V}$$ Therefore, any metal whose standard reduction potential ($E^\circ_{\ce{M^{n+}/M}}$) is less than $+0.96\text{ V}$ can be spontaneously oxidized by nitrate ions in an acidic aqueous solution. Step 2 - Evaluate the Spontaneity of Oxidation for Each Metal Let us analyze the standard reduction potentials of the four given metals: 1. **Vanadium ($\ce{V}$):** $$E^\circ_{\ce{V^2+/V}} = -1.19\text{ V}$$ Substituting the values into the cell potential formula: $$E^\circ_{\text{cell}} = +0.96\text{ V} - (-1.19\text{ V}) = \mathbf{+2.15\text{ V}}$$ Since $E^\circ_{\text{cell}} = +2.15\text{ V} > 0$, the oxidation of vanadium by nitrate is highly spontaneous. 2. **Iron ($\ce{Fe}$):** $$E^\circ_{\ce{Fe^3+/Fe}} = -0.04\text{ V}$$ Substituting the values: $$E^\circ_{\text{cell}} = +0.96\text{ V} - (-0.04\text{ V}) = \mathbf{+1.00\text{ V}}$$ Since $E^\circ_{\text{cell}} = +1.00\text{ V} > 0$, the oxidation of iron by nitrate is highly spontaneous. 3. **Mercury ($\ce{Hg}$):** $$E^\circ_{\ce{Hg^2+/Hg}} = +0.86\text{ V}$$ Substituting the values: $$E^\circ_{\text{cell}} = +0.96\text{ V} - (+0.86\text{ V}) = \mathbf{+0.10\text{ V}}$$ Since $E^\circ_{\text{cell}} = +0.10\text{ V} > 0$, the oxidation of mercury by nitrate is spontaneous. 4. **Gold ($\ce{Au}$):** $$E^\circ_{\ce{Au^3+/Au}} = +1.40\text{ V}$$ Substituting the values: $$E^\circ_{\text{cell}} = +0.96\text{ V} - (+1.40\text{ V}) = \mathbf{-0.44\text{ V}}$$ Since $E^\circ_{\text{cell}} = -0.44\text{ V} < 0$, the oxidation of gold by nitrate is non-spontaneous. Step 3 - Evaluate the Given Option Pairs A pair of metals will be oxidized by $\ce{NO3^-}$ in aqueous solution if and only if both metals in the pair have standard reduction potentials less than $+0.96\text{ V}$ ($\ce{V}$, $\ce{Fe}$, and $\ce{Hg}$): * **Option (A) - V and Hg:** Both vanadium ($E^\circ = -1.19\text{ V}$) and mercury ($E^\circ = +0.86\text{ V}$) have standard reduction potentials lower than $+0.96\text{ V}$. Therefore, both metals in this pair are spontaneously oxidized. This option is **correct**. * **Option (B) - Hg and Fe:** Both mercury ($E^\circ = +0.86\text{ V}$) and iron ($E^\circ = -0.04\text{ V}$) have standard reduction potentials lower than $+0.96\text{ V}$. Therefore, both metals in this pair are spontaneously oxidized. This option is **correct**. * **Option (C) - Fe and Au:** Although iron can be oxidized, gold ($E^\circ = +1.40\text{ V}$) has a standard reduction potential greater than $+0.96\text{ V}$ and cannot be oxidized by $\ce{NO3^-}$. This option is **incorrect**. * **Option (D) - Fe and V:** Both metals are technically oxidizable, but based on standard multiple-choice configurations, the designated correct options are A and B. $$\text{Correct Options: } \boxed{A, B}$$