Question

Copper reduces NO$_3^-$ into NO and NO$_2$ depending upon the concentration of HNO$_3$ in solution. (Assuming fixed [Cu$^{2+}$] and $P_{NO}=P_{NO_2}$), the HNO$_3$ concentration at which the thermodynamic tendency for reduction of NO$_3^-$ into NO and NO$_2$ by copper is same is $10^x$ M. The value of 2x is ________. (Rounded-off to the nearest integer)
[Given, E$^\circ_{Cu^{2+}/Cu}$=0.34 V, E$^\circ_{NO_3^-/NO}$=0.96 V, E$^\circ_{NO_3^-/NO_2}$=0.79 V and at 298 K, $\frac{RT}{F}$(2.303)=0.059]

Hint:

When comparing the "thermodynamic tendency" of two different overall reactions starting from the same reactants, it is often interpreted as equating the Gibbs Free Energy change ($\Delta G = -nFE_{cell}$) for each pathway, not just the cell potentials ($E_{cell}$).

Answer 1

Correct Answer: 4

Copper reduces $\mathrm{NO_3^-}$ into NO and NO$_2$ depending on the concentration of $\mathrm{HNO_3}$. Given: \[ E^\circ_{\mathrm{Cu^{2+}/Cu}} = 0.34\text{ V},\quad E^\circ_{\mathrm{NO_3^-/NO}} = 0.96\text{ V},\quad E^\circ_{\mathrm{NO_3^-/NO_2}} = 0.79\text{ V} \] \[ \frac{2.303RT}{F} = 0.059 \quad (298\text{ K}) \] Step 1: Standard cell potentials For reduction to NO: \[ E^\circ_{\text{cell,1}} = 0.96 - 0.34 = 0.62\text{ V}, \quad n_1 = 6 \] For reduction to NO$_2$: \[ E^\circ_{\text{cell,2}} = 0.79 - 0.34 = 0.45\text{ V}, \quad n_2 = 2 \] Step 2: Condition for equal thermodynamic tendency \[ \Delta G_1 = \Delta G_2 \Rightarrow n_1 E_{\text{cell,1}} = n_2 E_{\text{cell,2}} \] Step 3: Apply Nernst equation Let $\mathrm{[HNO_3]} = C$, so $[\mathrm{H^+}] = [\mathrm{NO_3^-}] = C$. Assuming $\mathrm{[Cu^{2+}]}$, $P_{\mathrm{NO}}$, $P_{\mathrm{NO_2}} = 1$. \[ 3\left(0.62 - \frac{0.059}{6}\log\frac{1}{C^{10}}\right) = 0.45 - \frac{0.059}{2}\log\frac{1}{C^6} \] \[ 1.41 = 8 \times 0.059 \log C \Rightarrow \log C \approx 2.99 \] Step 4: Final answer \[ C = 10^{2.99} \Rightarrow x \approx 3 \] \[ \boxed{2x = 6} \]