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NVAMediumJEE 2023Electrochemical Cells

JEE Chemistry 2023 Question with Solution

The standard reduction potentials at 298 K298 \, \text{K} for the following half cells are given below:

NO3−+4H++3e−→NO(g)+2H2O,  E0=0.97 V\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO(g)} + 2\text{H}_2\text{O}, \; E^0 = 0.97 \, \text{V} V2+(aq)+2e−→V,  E0=−1.19 V\text{V}^{2+}\text{(aq)} + 2e^- \rightarrow \text{V}, \; E^0 = -1.19 \, \text{V} Fe3+(aq)+3e−→Fe,  E0=−0.04 V\text{Fe}^{3+}\text{(aq)} + 3e^- \rightarrow \text{Fe}, \; E^0 = -0.04 \, \text{V} Ag+(aq)+e−→Ag(s),  E0=0.80 V\text{Ag}^+\text{(aq)} + e^- \rightarrow \text{Ag(s)}, \; E^0 = 0.80 \, \text{V} Au3+(aq)+3e−→Au(s),  E0=1.40 V\text{Au}^{3+}\text{(aq)} + 3e^- \rightarrow \text{Au(s)}, \; E^0 = 1.40 \, \text{V}

The number of metal(s) which will be oxidized by NO3−\text{NO}_3^- in aqueous solution is:

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: The reduction half-reaction for nitrate is

NO3−+4H++3e−→NO+2H2O\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}

with E0=0.97 VE^0 = 0.97 \, \text{V}.

The metals listed are V, Fe, Ag, and Au with their standard reduction potentials.

Find: The number of metals that will be oxidized by NO3−\text{NO}_3^- in aqueous solution.

For NO3−\text{NO}_3^- to oxidize a metal, nitrate must undergo reduction and the metal must undergo oxidation. The reaction will be feasible when the standard cell potential is positive:

Ecell0=Ecathode0+Eoxidation0>0E^0_{\text{cell}} = E^0_{\text{cathode}} + E^0_{\text{oxidation}} > 0

Here, the cathode reduction potential is 0.97 V0.97 \, \text{V} for nitrate reduction.

Now reverse the sign of each given metal reduction potential to obtain its oxidation potential:

V→V2++2e−Eoxidation0=+1.19 V\text{V} \rightarrow \text{V}^{2+} + 2e^- \qquad E^0_{\text{oxidation}} = +1.19 \, \text{V} Fe→Fe3++3e−Eoxidation0=+0.04 V\text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \qquad E^0_{\text{oxidation}} = +0.04 \, \text{V} Ag→Ag++e−Eoxidation0=−0.80 V\text{Ag} \rightarrow \text{Ag}^+ + e^- \qquad E^0_{\text{oxidation}} = -0.80 \, \text{V} Au→Au3++3e−Eoxidation0=−1.40 V\text{Au} \rightarrow \text{Au}^{3+} + 3e^- \qquad E^0_{\text{oxidation}} = -1.40 \, \text{V}

Compare Cell Potentials

Compute the cell potential in each case using nitrate reduction as the cathode process:

Ecell0=0.97+Eoxidation0E^0_{\text{cell}} = 0.97 + E^0_{\text{oxidation}}

For V:

Ecell0=0.97+1.19=2.16 VE^0_{\text{cell}} = 0.97 + 1.19 = 2.16 \, \text{V}

For Fe:

Ecell0=0.97+0.04=1.01 VE^0_{\text{cell}} = 0.97 + 0.04 = 1.01 \, \text{V}

For Ag:

Ecell0=0.97−0.80=0.17 VE^0_{\text{cell}} = 0.97 - 0.80 = 0.17 \, \text{V}

For Au:

Ecell0=0.97−1.40=−0.43 VE^0_{\text{cell}} = 0.97 - 1.40 = -0.43 \, \text{V}

A positive Ecell0E^0_{\text{cell}} means the metal will be oxidized by NO3−\text{NO}_3^-. Therefore, V, Fe, and Ag will be oxidized, while Au will not be oxidized.

Therefore, the number of metal(s) oxidized by NO3−\text{NO}_3^- is 33.

Common mistakes

  • Using the given reduction potentials directly for the metal oxidation half-reactions is incorrect because the sign must be reversed when a half-reaction is reversed. Always convert the metal reduction potential to oxidation potential before comparing.

  • Assuming a metal is oxidized only when its oxidation potential is numerically less than 0.97 V0.97 \, \text{V} without checking the sign convention can lead to wrong conclusions. Instead, evaluate whether Ecell0>0E^0_{\text{cell}} > 0 for each case.

  • Ignoring that nitrate is the oxidizing agent in this problem causes confusion about which half-reaction occurs at the cathode. Here, NO3−\text{NO}_3^- is reduced and the metal is oxidized.

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