Consider the cell
`Pt|H_(2(g,1atm))|H_((aq.1M))^(+)||Fe_((aq))^(3+),Fe_((aq))^(2+)||Pt_((s))` Given that `E_(Fe^(3+)|Fe^(2+))^(o)=0.771V` the ratio of conc. Of `Fe_((aq))^(2+)` to `Fe_((aq))^(3+)` is, when the cell potential is 0.830V

Options

A. 0.101
B. 0.924
C. 0.12
D. None of these

Explanation

To solve the problem, we need to determine the ratio of the concentrations of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) when the cell potential is 0.830 V. We will use the Nernst equation for this purpose.

### Step-by-Step Solution

1. **Identify the Standard Cell Potential**:
The standard reduction potential for the half-reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is given as \( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.771 \, \text{V} \).

2. **Write the Nernst Equation**:
The Nernst equation can be expressed as:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
For our reaction, we have:
\[ E_{\text{cell}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - \frac{0.059}{1} \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \]

3. **Substitute Known Values**:
We know:
- \( E_{\text{cell}} = 0.830 \, \text{V} \)
- \( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.771 \, \text{V} \)
Substituting these values into the Nernst equation gives:
\[ 0.830 = 0.771 - 0.059 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \]

4. **Rearrange the Equation**:
Rearranging the equation to isolate the logarithmic term:
\[ 0.059 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 0.771 - 0.830 \]
\[ 0.059 \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = -0.059 \]

5. **Divide by 0.059**:
Dividing both sides by 0.059:
\[ \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = -1 \]

6. **Convert Logarithmic Form to Ratio**:
Converting from logarithmic form to exponential form:
\[ \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^{-1} = 0.1 \]

### Final Result
The ratio of the concentration of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) is:
\[ \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 0.1 \]