Calculate ΔG° for the oxidation of 3 moles of copper by nitric acid. Cu²⁺(aq) + 2e⁻ ⇄ Cu(s) E° = 0.34 V
NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ ⇄ NO(g) + 2H₂O(l) E° = 0.957 V

Question

Calculate ΔG° for the oxidation of 3 moles of copper by nitric acid. Cu²⁺(aq) + 2e⁻ ⇄ Cu(s) E° = 0.34 V NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ ⇄ NO(g) + 2H₂O(l) E° = 0.957 V

Quizplus · Accepted Answer

To calculate ΔG° for the overall reaction, we first need to balance the half-reactions. The balanced overall reaction involves multiplying the copper reaction by 3 and the nitric acid reaction by 2 to equalize the electrons transferred:$$3 \times (Cu^{2+}(aq) + 2e^- \rightarrow Cu(s))$$$$2 \times (NO_3^-(aq) + 4H^+(aq) + 3e^- \rightarrow NO(g) + 2H_2O(l))$$This gives us a total of 6 electrons transferred in both directions. The net cell potential, $E°_{cell}$, is the difference between the reduction potential of the nitric acid reaction and the copper reaction:$$E°_{cell} = E°_{cathode} - E°_{anode} = 0.957V - 0.34V = 0.617V$$The formula to calculate ΔG° is:$$\Delta G° = -nFE°_{cell}$$where $n$ is the number of moles of electrons transferred (6 moles), $F$ is the Faraday constant (96485 C/mol), and $E°_{cell}$ is the net cell potential in volts. Plugging in the values:$$\Delta G° = -6 \times 96485 \times 0.617 = -356,679.81 J$$Converting joules to kilojoules:$$\Delta G° = -356.67981 kJ \approx -360 kJ$$

Calculate ΔG° for the Oxidation of 3 Moles of Copper