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Step-by-Step Solution
Step 1: Identify the Overall Reactions and Standard Potentials
Copper can reduce nitrate ions ( NO_3^- ) to nitric oxide ( NO ) or nitrogen dioxide ( NO_2 ) depending on the acidity (i.e., [H^+] or effectively the concentration of HNO_3 ). The relevant standard potentials ( E^o ) are given as:
E_{Cu^{2+}/Cu}^o = 0.34\text{ V},\quad E_{NO_3^- / NO}^o = 0.96\text{ V}, \quad E_{NO_3^- / NO_2}^o = 0.79\text{ V}.
From these, for each half-cell, we calculate the net standard cell potentials for the reductions of nitrate to NO and to NO2, keeping in mind that copper is being oxidized to Cu^{2+} .
Step 2: Write the Cell Reactions
For the reaction where NO_3^- is reduced to NO , the overall reaction (Cell-I) can be written as:
3Cu + 2NO_3^- + 8H^+ \;\to\; 3Cu^{2+} + 2NO + 4H_2O.
For the reaction where NO_3^- is reduced to NO_2 , the overall reaction (Cell-II) can be written as:
Cu + 2NO_3^- + 4H^+ \;\to\; Cu^{2+} + 2NO_2 + 2H_2O.
Step 3: Determine the Combined Standard Cell Potentials
By coupling each reduction with the oxidation of copper, the standard cell potentials become:
Cell-I: E_1^o = E_{NO_3^- / NO}^o - E_{Cu^{2+}/Cu}^o = 0.96 - (-0.34)=0.96+0.34=1.30\,\text{V}.
Cell-II: E_2^o = E_{NO_3^- / NO_2}^o - E_{Cu^{2+}/Cu}^o = 0.79 - (-0.34)=0.79+0.34=1.13\,\text{V}.
Step 4: Write the Nernst Equations for Each Cell
Let E_1 be the potential for the NO_3^- \to NO reaction (Cell-I). Using the Nernst equation:
E_1 = E_1^o - \frac{0.059}{n_1}\,\log Q_1,
where Q_1 (the reaction quotient for Cell-I) is:
Q_1 = \frac{[Cu^{2+}]^3 \times (p_{NO})^2}{[NO_3^-]^2 \times [H^+]^8}\,.
Here, n_1 = 6 electrons exchanged in the overall reaction for Cell-I. Thus:
E_1 = 1.30 - \frac{0.059}{6}\,\log Q_1\,.
Similarly, let E_2 be the potential for the NO_3^- \to NO_2 reaction (Cell-II). The reaction quotient Q_2 is:
Q_2 = \frac{[Cu^{2+}] \times (p_{NO_2})^2}{[NO_3^-]^2 \times [H^+]^4}\,.
Here, n_2 = 2 electrons exchanged in the overall reaction for Cell-II. Thus:
E_2 = 1.13 - \frac{0.059}{2}\,\log Q_2\,.
Step 5: Equate the Two Potentials
We want to find the concentration of HNO_3 at which both reactions have the same tendency. Hence, set E_1=E_2 :
1.30 - \frac{0.059}{6}\,\log Q_1 = 1.13 - \frac{0.059}{2}\,\log Q_2.
Simplifying leads to an expression that involves [HNO_3] (since [HNO_3] in a strong acid can be taken as [H^+] \approx [HNO_3] , and partial pressures of NO and NO_2 also factor in, but it is given or assumed p_{NO} = p_{NO_2} ).
Step 6: Solve for the Acid Concentration
After algebraic manipulation, we arrive at:
0.17 = \frac{0.059}{6} \log \left( [HNO_3]^8 \right).
This simplifies to:
0.17 = \frac{0.059}{6} \times 8 \log [HNO_3].
Dividing both sides properly and solving for \log [HNO_3] yields:
\log [HNO_3] = 2.16 \quad \Longrightarrow \quad [HNO_3] = 10^{2.16}\,\text{M} = 10^x\,\text{M}.
Step 7: Find the Required Double of x
Since [HNO_3] = 10^x \text{ and } x=2.16, we want 2x :
2x = 2 \times 2.16 = 4.32 \approx 4.
Thus, the value of 2x (rounded off to the nearest integer) is \boxed{4}.
