JEE Challenger
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ChemMedAdv 2009 • Paper 2

Metals Oxidized by Nitrate Ion in Aqueous Solution

For the reduction of NO3−\text{NO}_3^- ion in an aqueous solution, E0\text{E}^0 is +0.96 V+0.96\text{ V}. Values of E0\text{E}^0 for some metal ions are given below

V2+(aq)+2e−→VE0=−1.19 VFe3+(aq)+3e−→FeE0=−0.04 VAu3+(aq)+3e−→AuE0=+1.40 VHg2+(aq)+2e−→HgE0=+0.86 V\begin{array}{ll} \text{V}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{V} & \text{E}^0 = -1.19\text{ V} \\ \text{Fe}^{3+}(\text{aq}) + 3\text{e}^- \rightarrow \text{Fe} & \text{E}^0 = -0.04\text{ V} \\ \text{Au}^{3+}(\text{aq}) + 3\text{e}^- \rightarrow \text{Au} & \text{E}^0 = +1.40\text{ V} \\ \text{Hg}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Hg} & \text{E}^0 = +0.86\text{ V} \end{array}

The pair(s) of metals that is(are) oxidized by NO3−\text{NO}_3^- in aqueous solution is(are)

Options

A

V\text{V} and Hg\text{Hg}

Correct
B

Hg\text{Hg} and Fe\text{Fe}

Correct
C

Fe\text{Fe} and Au\text{Au}

D

Fe\text{Fe} and V\text{V}

Correct

Topics & Concepts

ElectrochemistryElectrochemical Cell

Step-by-Step Solution

To determine which metals can be oxidized by NO3−\text{NO}_3^- in aqueous solution, we consider the spontaneity of the redox reaction under standard conditions.

For the overall cell reaction to be spontaneous:
Ecell0>0E_{\text{cell}}^0 > 0

The standard reduction potential for the nitrate ion is given as:
E0(NO3−/reduction product)=+0.96 VE^0(\text{NO}_3^- / \text{reduction product}) = +0.96\text{ V}

When a metal M\text{M} is oxidized to Mn+\text{M}^{n+}, it acts as the anode, while NO3−\text{NO}_3^- acts as the cathode:
Ecell0=Ecathode0−Eanode0=E0(NO3−)−E0(Mn+/M)E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0 = E^0(\text{NO}_3^-) - E^0(\text{M}^{n+}/\text{M})

For the reaction to be spontaneous:
Ecell0=+0.96 V−E0(Mn+/M)>0E_{\text{cell}}^0 = +0.96\text{ V} - E^0(\text{M}^{n+}/\text{M}) > 0
  ⟹  E0(Mn+/M)<+0.96 V\implies E^0(\text{M}^{n+}/\text{M}) < +0.96\text{ V}

Now, let us examine the given standard reduction potentials for each metal:

  1. Vanadium (V\text{V}):
    E0(V2+/V)=−1.19 V<+0.96 VE^0(\text{V}^{2+}/\text{V}) = -1.19\text{ V} < +0.96\text{ V}
    Ecell0=+0.96−(−1.19)=+2.15 V>0(Spontaneous, hence V is oxidized)E_{\text{cell}}^0 = +0.96 - (-1.19) = +2.15\text{ V} > 0 \quad (\text{Spontaneous, hence } \text{V} \text{ is oxidized})

  2. Iron (Fe\text{Fe}):
    E0(Fe3+/Fe)=−0.04 V<+0.96 VE^0(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V} < +0.96\text{ V}
    Ecell0=+0.96−(−0.04)=+1.00 V>0(Spontaneous, hence Fe is oxidized)E_{\text{cell}}^0 = +0.96 - (-0.04) = +1.00\text{ V} > 0 \quad (\text{Spontaneous, hence } \text{Fe} \text{ is oxidized})

  3. Mercury (Hg\text{Hg}):
    E0(Hg2+/Hg)=+0.86 V<+0.96 VE^0(\text{Hg}^{2+}/\text{Hg}) = +0.86\text{ V} < +0.96\text{ V}
    Ecell0=+0.96−(+0.86)=+0.10 V>0(Spontaneous, hence Hg is oxidized)E_{\text{cell}}^0 = +0.96 - (+0.86) = +0.10\text{ V} > 0 \quad (\text{Spontaneous, hence } \text{Hg} \text{ is oxidized})

  4. Gold (Au\text{Au}):
    E0(Au3+/Au)=+1.40 V>+0.96 VE^0(\text{Au}^{3+}/\text{Au}) = +1.40\text{ V} > +0.96\text{ V}
    Ecell0=+0.96−(+1.40)=−0.44 V<0(Non-spontaneous, hence Au is not oxidized)E_{\text{cell}}^0 = +0.96 - (+1.40) = -0.44\text{ V} < 0 \quad (\text{Non-spontaneous, hence } \text{Au} \text{ is not oxidized})

Thus, the metals that can be oxidized by NO3−\text{NO}_3^- are V\text{V}, Fe\text{Fe}, and Hg\text{Hg}.

Evaluating the given options:

  • (A) V\text{V} and Hg\text{Hg}: Both are oxidized.
  • (B) Hg\text{Hg} and Fe\text{Fe}: Both are oxidized.
  • (C) Fe\text{Fe} and Au\text{Au}: Au\text{Au} is not oxidized.
  • (D) Fe\text{Fe} and V\text{V}: Both are oxidized.

Therefore, the correct options are A, B, and D.

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