1 litre solution containing 4.5 milimoles of `Cr_(2)O_(7)^(-)` and 15 milimoles of `Cr^(3+)` shows a `pH` of 2.0.
The potential of the reduction half reaction is approximately `xV`. Then `[x]` is ……… `(E^(@)` of `Cr_(2)O_(7)^(2-)//Cr^(3+)` is `1.33V`)

Explanation

To solve the problem, we need to calculate the potential of the reduction half-reaction for the given solution containing `Cr2O7^2-` and `Cr^3+` ions. We will use the Nernst equation for this calculation.

### Step-by-Step Solution:

1. **Identify the Reaction:**
The reduction half-reaction for dichromate ion is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

2. **Determine the Standard Reduction Potential:**
The standard reduction potential \(E^\circ\) for the reaction is given as:
\[ E^\circ = 1.33 \, V \]

3. **Calculate the Concentrations:**
- The concentration of \(Cr^{3+}\) ions is:
\[ [Cr^{3+}] = \frac{15 \, \text{mmol}}{1000 \, \text{mL}} = 0.015 \, M \]
- The concentration of \(Cr_2O_7^{2-}\) ions is:
\[ [Cr_2O_7^{2-}] = \frac{4.5 \, \text{mmol}}{1000 \, \text{mL}} = 0.0045 \, M \]
- The concentration of \(H^+\) ions can be calculated from the pH:
\[ pH = 2 \implies [H^+] = 10^{-2} \, M = 0.01 \, M \]

4. **Substitute into the Nernst Equation:**
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} \right) \]
Here, \(n = 6\) (the number of electrons transferred).

5. **Plug in the Values:**
Substitute the values into the Nernst equation:
\[ E = 1.33 - \frac{0.059}{6} \log \left( \frac{(0.015)^2}{(0.0045)(0.01)^{14}} \right) \]

6. **Calculate the Logarithm:**
- Calculate the numerator:
\[ (0.015)^2 = 0.000225 \]
- Calculate the denominator:
\[ (0.0045)(0.01)^{14} = 0.0045 \times 10^{-28} = 4.5 \times 10^{-31} \]
- Now, calculate the fraction:
\[ \frac{0.000225}{4.5 \times 10^{-31}} = 5 \times 10^{25} \]
- Now, take the logarithm:
\[ \log(5 \times 10^{25}) = \log(5) + \log(10^{25}) = 0.699 + 25 = 25.699 \]

7. **Final Calculation:**
Substitute back into the Nernst equation:
\[ E = 1.33 - \frac{0.059}{6} \times 25.699 \]
\[ E = 1.33 - 0.009833 \times 25.699 \approx 1.33 - 0.253 \]
\[ E \approx 1.077 \, V \]

8. **Conclusion:**
The potential of the reduction half-reaction is approximately \(1.077 \, V\). Therefore, the value of \(x\) is approximately \(1.08\) when rounded to two decimal places.

### Final Answer:
The value of \([x]\) is approximately \(1.08\).