At 298 K, a 1 litre solution containing 10 mmol of `Cr_2O_7^(2–)` and 100 mmol of `Cr^(3+)` shows a pH of 3.0.Given: `Cr_2O_7^(2–)rarr Cr^(3+) , E^(º) = 1.33V` and `(2.303RT)/F= 0.059V`. The potential for the half cell reaction is `x xx 10^(–3) V`. The value of x is

Answer

To solve the problem, we need to determine the potential for the half-cell reaction involving the conversion of `Cr2O7^(2–)` to `Cr^(3+)` using the Nernst equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the half-reaction
The half-reaction for the conversion of dichromate to chromium ions in acidic medium can be written as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

### Step 2: Identify the standard potential
The standard reduction potential \( E^\circ \) for the half-reaction is given as:
\[ E^\circ = 1.33 \, V \]

### Step 3: Calculate the concentrations
Given that we have a 1 L solution:
- Moles of `Cr2O7^(2–)` = 10 mmol = 0.010 mol
- Moles of `Cr^(3+)` = 100 mmol = 0.100 mol
- The concentration of `Cr2O7^(2–)` = \( \frac{0.010 \, \text{mol}}{1 \, \text{L}} = 0.010 \, M \)
- The concentration of `Cr^(3+)` = \( \frac{0.100 \, \text{mol}}{1 \, \text{L}} = 0.100 \, M \)

### Step 4: Calculate the concentration of \( H^+ \)
Given the pH of the solution is 3.0:
\[ [H^+] = 10^{-pH} = 10^{-3} \, M \]

### Step 5: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{2.303RT}{nF} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} \right) \]
Where:
- \( n = 6 \) (number of electrons transferred)
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \)
- \( F = 96485 \, C/mol \)

Using the provided value \( \frac{2.303RT}{F} = 0.059 \, V \):
\[ E = 1.33 - 0.059 \cdot \frac{1}{6} \log \left( \frac{(0.100)^2}{(0.010)(10^{-3})^{14}} \right) \]

### Step 6: Calculate the logarithm term
Calculating the term inside the logarithm:
\[ \frac{(0.100)^2}{(0.010)(10^{-3})^{14}} = \frac{0.0100}{0.010 \cdot 10^{-42}} = 1.0 \times 10^{40} \]
Now, take the logarithm:
\[ \log(1.0 \times 10^{40}) = 40 \]

### Step 7: Substitute back into the Nernst equation
Now substitute back into the Nernst equation:
\[ E = 1.33 - 0.059 \cdot \frac{1}{6} \cdot 40 \]
Calculating the term:
\[ 0.059 \cdot \frac{40}{6} = 0.3933 \, V \]
Thus:
\[ E = 1.33 - 0.3933 = 0.9367 \, V \]

### Step 8: Convert to the required format
Expressing \( E \) in the required format:
\[ E = 936.7 \times 10^{-3} \, V \]
Thus, the value of \( x \) is:
\[ x = 936.7 \]

### Final Answer
The value of \( x \) is approximately **937**.

---