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, $$\frac{RT}{F}(2.303) = 0.059$$]

We need to find the $$HNO_3$$ concentration at which the thermodynamic tendency for reduction of $$NO_3^-$$ into NO and $$NO_2$$ by copper is the same. This means the cell EMF for both reactions must be equal, which requires the cathode potentials to be equal (since the anode is the same: Cu).

The two half-reactions for reduction are:

Reaction 1: $$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$$, $$E^\circ = 0.96$$ V

Reaction 2: $$NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$$, $$E^\circ = 0.79$$ V

Applying the Nernst equation to each (assuming $$P_{NO} = P_{NO_2}$$ and noting that for $$HNO_3$$ solution, $$[H^+] = [NO_3^-] = C$$, where $$C$$ is the concentration):

$$E_1 = 0.96 + \frac{0.059}{3}\log\frac{[NO_3^-][H^+]^4}{P_{NO}} = 0.96 + \frac{0.059}{3}\log\frac{C \cdot C^4}{P_{NO}}$$

$$E_2 = 0.79 + \frac{0.059}{1}\log\frac{[NO_3^-][H^+]^2}{P_{NO_2}} = 0.79 + 0.059\log\frac{C \cdot C^2}{P_{NO_2}}$$

Setting $$E_1 = E_2$$ and using $$P_{NO} = P_{NO_2}$$ (which cancel):

$$0.96 + \frac{0.059}{3}(5\log C) = 0.79 + 0.059(3\log C)$$

$$0.96 + \frac{0.295}{3}\log C = 0.79 + 0.177\log C$$

$$0.96 + 0.09833\log C = 0.79 + 0.177\log C$$

$$0.96 - 0.79 = 0.177\log C - 0.09833\log C$$

$$0.17 = 0.07867\log C$$

$$\log C = \frac{0.17}{0.07867} \approx 2.16$$

So $$[HNO_3] = 10^{2.16} = 10^x$$, giving $$x = 2.16$$ and $$2x = 4.32$$.

Rounding to the nearest integer, $$2x = \boxed{4}$$.