Copper reduces NO₃ - into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed [Cu²⁺] and P NO = P NO ₂ ), the HNO₃ concentration at which the thermodynamic tendency for reduction of NO₃ - into NO and NO₂ by copper is same is 10ˣ M. The value of 2x is _______. (Rounded off to the nearest integer)
[Given, E_(Cu²⁺/Cu)ᵒ = 0.34 V, E_(NO₃ - /NO)ᵒ = 0.96 V, E_(NO₃ - /NO₂)ᵒ = 0.79 V and at 298 K, RT/F(2.303) = 0.059]

Solution

Chemistry ma’am · 5 steps

Copper is common to both cells, so just make the two nitrate reduction potentials equal.

1. Two half reactions. To NO: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O, E° = 0.96 V. To NO₂: NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O, E° = 0.79 V.
2. Nernst for each. HNO₃ is fully ionised, so [H⁺] = [NO₃⁻] = c. Take both gas pressures as 1 atm. Then E₁ = 0.96 - 0.059/3 log 1/(c · c⁴) and E₂ = 0.79 - 0.059/1 log 1/(c · c²).
3. Equal tendency. Each cell is E_(cathode) - E_(Cu²⁺/Cu), and the copper term is identical in both. So the two cell potentials are equal exactly when E₁ = E₂.

   0.96 + (5 × 0.059)/3 log c = 0.79 + 3 × 0.059 log c

4. Solve for log c. The coefficients are 0.0983 and 0.177. So 0.17 = (0.177 - 0.0983)log c = 0.0787 log c, which gives log c = 2.16.
5. So. c = 10^(2.16) M, so x = 2.16 and 2x = 4.32, which rounds to 4.

The common slip: 9. Putting only [H⁺] in the quotient and leaving out [NO₃⁻] gives log c = 4.32 and 2x ≈ 9, not 4.

The idea to keep. When one metal reduces an ion two ways, the anode term is common to both cells. Compare only the two cathode potentials, and count every concentration in the quotient, including the anion itself.

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