Calculate the potential for half cell containing 0.10 M `K_(2)Cr_(2)O_(7)(aq),0.20" M "Cr^(3+)(aq) and 1.0xx10^(-4)M H^(+)(aq)`. The half-cell reaction is `Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)O(l)`
and the standard electrode potential is given as `E^(@)=1.33V`.

Explanation

To calculate the potential for the half-cell reaction given, we will follow these steps:

### Step 1: Write the half-cell reaction and identify the components.
The half-cell reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 2: Identify the concentrations of the species involved.
- Concentration of \( \text{Cr}_2\text{O}_7^{2-} = 0.10 \, \text{M} \)
- Concentration of \( \text{Cr}^{3+} = 0.20 \, \text{M} \)
- Concentration of \( \text{H}^+ = 1.0 \times 10^{-4} \, \text{M} \)

### Step 3: Write the expression for the equilibrium constant \( K_c \).
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} \]

### Step 4: Substitute the concentrations into the \( K_c \) expression.
Substituting the values into the expression:
\[ K_c = \frac{(0.20)^2}{(0.10)(1.0 \times 10^{-4})^{14}} \]

### Step 5: Calculate \( K_c \).
Calculating \( K_c \):
1. Calculate \( (0.20)^2 = 0.04 \)
2. Calculate \( (1.0 \times 10^{-4})^{14} = 1.0 \times 10^{-56} \)
3. Now substitute these values:
\[ K_c = \frac{0.04}{(0.10)(1.0 \times 10^{-56})} = \frac{0.04}{1.0 \times 10^{-57}} = 4.0 \times 10^{55} \]

### Step 6: Use the Nernst equation to calculate \( E_{cell} \).
The Nernst equation is given by:
\[ E_{cell} = E^\circ - \frac{0.0591}{n} \log K_c \]
Where:
- \( E^\circ = 1.33 \, \text{V} \)
- \( n = 6 \) (the number of electrons transferred)

### Step 7: Substitute the values into the Nernst equation.
Substituting the values:
\[ E_{cell} = 1.33 - \frac{0.0591}{6} \log(4.0 \times 10^{55}) \]

### Step 8: Calculate \( \log(4.0 \times 10^{55}) \).
Using logarithmic properties:
\[ \log(4.0 \times 10^{55}) = \log(4.0) + \log(10^{55}) \]
\[ \log(4.0) \approx 0.602 \]
\[ \log(10^{55}) = 55 \]
Thus,
\[ \log(4.0 \times 10^{55}) \approx 0.602 + 55 = 55.602 \]

### Step 9: Substitute back into the Nernst equation.
Now substituting back:
\[ E_{cell} = 1.33 - \frac{0.0591}{6} \times 55.602 \]
Calculating:
\[ E_{cell} = 1.33 - \frac{0.0591 \times 55.602}{6} \]
\[ E_{cell} = 1.33 - \frac{3.290}{6} \]
\[ E_{cell} = 1.33 - 0.5483 \]
\[ E_{cell} \approx 0.7817 \, \text{V} \]

### Final Answer:
Thus, the potential for the half-cell is approximately:
\[ E_{cell} \approx 0.78 \, \text{V} \]