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JEE Chemistry 2023 Question with Solution

Section – B

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

NO3−+4H++3e−→NO(g)+2H2OE0=0.97 V\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO(g)} + 2\text{H}_2\text{O} \quad E^0 = 0.97 \, \text{V}

V2++2e−→VE0=−1.19 V\text{V}^{2+} + 2e^- \rightarrow \text{V} \quad E^0 = -1.19 \, \text{V}

Fe3++3e−→FeE0=−0.04 V\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad E^0 = -0.04 \, \text{V}

Ag++e−→Ag(s)E0=0.80 V\text{Ag}^+ + e^- \rightarrow \text{Ag(s)} \quad E^0 = 0.80 \, \text{V}

Au3++3e−→Au(s)E0=1.40 V\text{Au}^{3+} + 3e^- \rightarrow \text{Au(s)} \quad 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 potential for

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

is 0.97 V0.97 \, \text{V}.

The given metal reduction potentials are:

  • V2+/V=−1.19 V\text{V}^{2+}/\text{V} = -1.19 \, \text{V}
  • Fe3+/Fe=−0.04 V\text{Fe}^{3+}/\text{Fe} = -0.04 \, \text{V}
  • Ag+/Ag=0.80 V\text{Ag}^+/\text{Ag} = 0.80 \, \text{V}
  • Au3+/Au=1.40 V\text{Au}^{3+}/\text{Au} = 1.40 \, \text{V}

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

A metal will be oxidized by NO3−\text{NO}_3^- if its own reduction potential is less than that of the nitrate half-cell, because then NO3−\text{NO}_3^- undergoes reduction and the metal undergoes oxidation.

Comparing with 0.97 V0.97 \, \text{V}:

  • −1.19<0.97-1.19 < 0.97, so V will be oxidized.
  • −0.04<0.97-0.04 < 0.97, so Fe will be oxidized.
  • 0.80<0.970.80 < 0.97, so Ag will be oxidized.
  • 1.40>0.971.40 > 0.97, so Au will not be oxidized.

Therefore, the number of metals oxidized is 33.

Potential Comparison Trick

Given: The oxidizing agent is NO3−\text{NO}_3^- with reduction potential 0.97 V0.97 \, \text{V}.

Find: How many listed metals can be oxidized by it.

Shortcut idea: Any metal whose corresponding reduction potential is lower than 0.97 V0.97 \, \text{V} will act as the reducing agent and get oxidized.

So count the values less than 0.970.97:

−1.19,  −0.04,  0.80-1.19, \; -0.04, \; 0.80

These are three values.

Therefore, the correct numerical answer is 33.

Common mistakes

  • Students may think the metal with higher reduction potential is more easily oxidized. This is wrong because a higher reduction potential means the species is more easily reduced. Compare the metal half-cell potential with 0.97 V0.97 \, \text{V} and count those with lower values.

  • Students may include Au in the count by looking only at the fact that it is a metal. This is wrong because Au3+/Au\text{Au}^{3+}/\text{Au} has reduction potential 1.40 V1.40 \, \text{V}, which is greater than 0.97 V0.97 \, \text{V}, so gold is not oxidized by NO3−\text{NO}_3^-.

  • Students may compare oxidation potentials directly without changing the sign logic carefully. This can lead to incorrect counting. Use the given reduction potentials consistently and identify which metals have lower reduction potentials than the nitrate reduction half-cell.

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