Copper reduced `NO_(3)^(-)` into `NO` and `NO_(2)` depending upon conc.of `HNO_(3)` in solution. Assuming `[Cu^(2+)] = 0.1M`, and `P_(NO) = P_(NO_(2)) = 10^(-3)` atm and using data answer the following questions:
`E_(Cu^(2+)//Cu)^(@) =+ 0.34` volt, at `298K (RT)/(F) (2.303) = 0.06` volt
`E_(cell)` for reduction of `NO_(3)^(-) rarr NO` by `Cu(s)` when `[HNO_(3)] = 1M` is [At `t = 298]`

Options

A. `~0.61`
B. `~0.71`
C. `~0.51`
D. `~0.81`

Explanation

To solve the problem of calculating the cell potential \( E_{cell} \) for the reduction of \( NO_3^- \) to \( NO \) using copper, we will follow these steps:

### Step 1: Write the half-reactions
1. **Anode (Oxidation)**:
\[ Cu(s) \rightarrow Cu^{2+} + 2e^- \]
2. **Cathode (Reduction)**:
\[ 2NO_3^- + 8H^+ + 6e^- \rightarrow 2NO + 4H_2O \]

### Step 2: Balance the half-reactions
To balance the electrons, we need to multiply the anode reaction by 3:
- **Anode**:
\[ 3Cu(s) \rightarrow 3Cu^{2+} + 6e^- \]
- **Cathode**:
\[ 2NO_3^- + 8H^+ + 6e^- \rightarrow 2NO + 4H_2O \]

### Step 3: Write the overall cell reaction
Combining the balanced half-reactions gives:
\[ 3Cu + 2NO_3^- + 8H^+ \rightarrow 3Cu^{2+} + 2NO + 4H_2O \]

### Step 4: Calculate the standard cell potential \( E^0_{cell} \)
Using the formula:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]
Given:
- \( E^0_{Cu^{2+}/Cu} = +0.34 \, V \)
- \( E^0_{NO_3^-/NO} = +0.96 \, V \)

Thus,
\[ E^0_{cell} = 0.96 - 0.34 = 0.62 \, V \]

### Step 5: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \)
- \( n = 6 \) (number of electrons transferred)
- \( F = 96485 \, C/mol \)
- \( \frac{RT}{F} \approx 0.0257 \, V \) at 298 K

Substituting the values:
\[ E_{cell} = 0.62 - \frac{0.0257}{6} \ln Q \]

### Step 6: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is calculated as:
\[ Q = \frac{(P_{NO})^2}{[Cu^{2+}]^3 \cdot [H^+]^8 \cdot (P_{NO_3^-})^2} \]
Given:
- \( P_{NO} = P_{NO_2} = 10^{-3} \, atm \)
- \( [Cu^{2+}] = 0.1 \, M \)
- \( [H^+] = 1 \, M \)
- \( [NO_3^-] = 1 \, M \)

Thus,
\[ Q = \frac{(10^{-3})^2}{(0.1)^3 \cdot (1)^8 \cdot (1)^2} = \frac{10^{-6}}{0.001} = 10^{-3} \]

### Step 7: Substitute \( Q \) into the Nernst equation
Now substituting \( Q \):
\[ E_{cell} = 0.62 - \frac{0.0257}{6} \ln(10^{-3}) \]
Calculating \( \ln(10^{-3}) = -6.907 \):
\[ E_{cell} = 0.62 - \frac{0.0257}{6} \cdot (-6.907) \]
Calculating the term:
\[ E_{cell} = 0.62 + 0.0257 \cdot 1.1512 \approx 0.62 + 0.0296 \approx 0.6496 \, V \]

### Final Result
Thus, the calculated cell potential \( E_{cell} \) is approximately:
\[ E_{cell} \approx 0.65 \, V \]