The standard reduction potentials at 298 K for the following half cells are given below:
NO₃⁻+4 H⁺+3 e⁻ → NO(g)+2 H₂ O E^θ=0.97 V
V²⁺(aq)+2 e⁻ → V E^θ=-1.19 V
Fe³⁺(aq)+3 e⁻ → Fe E^θ=-0.04 V
Ag⁺(aq)+e⁻ → Ag(s) E^θ=0.80 V
Au³⁺(aq)+3 e⁻ → Au(s) E^θ=1.40 V
The number of metal(s) which will be oxidized by NO₃⁻ in aqueous solution is __________.

Solution

Chemistry ma’am · 3 steps

Nitrate can oxidize any metal whose reduction potential sits below its own 0.97 V.

1. Who wins electrons. In a redox pair, the half cell with the higher reduction potential gets reduced. The other one is pushed backwards and gets oxidized. So NO₃⁻ (E^θ = 0.97 V) oxidizes a metal only if that metal's E^θ is lower than 0.97 V.

   E^θ_(cell) = E^θ_(NO₃⁻/NO) - E^θₘₑₜₐₗ = 0.97 - E^θₘₑₜₐₗ

2. Check each metal. A positive E^θ_(cell) means Δ G^θ = -nFE^θ_(cell) is negative, so the oxidation is spontaneous.

   Metal	E^θ (V)	E^θ_(cell) (V)	Oxidized by NO₃⁻?
   V	-1.19	0.97 + 1.19 = 2.16	yes
   Fe	-0.04	0.97 + 0.04 = 1.01	yes
   Ag	0.80	0.97 - 0.80 = 0.17	yes
   Au	1.40	0.97 - 1.40 = -0.43	no

3. So. V, Fe and Ag are oxidized, and Au is not. The answer is 3.

The common slip: 2. Counting only the metals with a negative E^θ gives 2. Ag at 0.80 V is still below 0.97 V, so nitrate oxidizes it too. This is why dilute nitric acid dissolves silver.

The idea to keep. Compare reduction potentials directly: a species oxidizes everything listed below it. The sign of a potential does not matter, only whether it is lower than the oxidant's.

Answer from the NTA final answer key. Solution by PrepWiser.