For the reduction of NO3^– ion in an aqueous solution, E° is +0.96 V. Values of E° for some metal ions are given below: V^2+(aq) + 2e^– - V(s); E° = –1.19 V, Fe^3+(aq) + 3e^– - Fe(s); E° = –0.04 V,

Question

For the reduction of NO3^– ion in an aqueous solution, E° is +0.96 V. Values of E° for some metal ions are given below: V^2+(aq) + 2e^– -> V(s); E° = –1.19 V, Fe^3+(aq) + 3e^– -> Fe(s); E° = –0.04 V, Au^3+(aq) + 3e^– -> Au(s); E° = +1.40 V, Hg^2+(aq) + 2e^– -> Hg(l); E° = +0.86 V. The pair(s) of metal that is(are) oxidized by NO3^– in aqueous solution is(are)

Prashant Kotian · Accepted Answer

Correct answer: A,B,D. For the nitrate ion (NO3^–) to spontaneously oxidize a metal, the standard reduction potential of the metal's cation must be algebraically less than the standard reduction potential of NO3^– (+0.96 V). The standard reduction potentials for V (–1.19 V), Fe (–0.04 V), and Hg (+0.86 V) are all strictly less than +0.96 V, rendering them susceptible to oxidation. The reduction potential of Au (+1.40 V) exceeds +0.96 V, so gold cannot be oxidized by nitrate. Any pair containing only V, Hg, or Fe is correct. Therefore, correct answer is A,B,D.

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