Consider the cell `Pt(s) | H_2(g) (1 atm) | H^+(aq, [H^+] = 1) || Fe^(3+) (aq), Fe^(2+ )(aq) | Pt(s)`
Given : `E^@_(Fe)^(+3)//(Fe)^(2+)=0.771V ` and `E^@_H^+//(1/2H_2)`= 0 V , T = 298 K
If the potential of the cell is 0.712 V, the ratio of concentration of `Fe^(2+)` to `Fe^(3+)` is _________ (Nearest integer)

Answer

To solve the given electrochemical cell problem, we will follow these steps:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
1. **Anode (oxidation)**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
2. **Cathode (reduction)**:
\[ Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq) \]

### Step 2: Write the overall cell reaction
The overall cell reaction can be derived by combining the two half-reactions. Since the oxidation half-reaction produces 2 electrons, we need to multiply the cathode reaction by 2 to balance the electrons:
\[ H_2(g) + 2Fe^{3+}(aq) \rightarrow 2H^+(aq) + 2Fe^{2+}(aq) \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
Where:
- \(E_{cell}\) = 0.712 V (given)
- \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\)
- \(n\) = number of electrons transferred (1 for the reduction of \(Fe^{3+}\) to \(Fe^{2+}\))

### Step 4: Calculate \(E^0_{cell}\)
Given:
- \(E^0_{Fe^{3+}/Fe^{2+}} = 0.771 V\)
- \(E^0_{H^+/H_2} = 0 V\)

Thus,
\[ E^0_{cell} = 0.771 V - 0 V = 0.771 V \]

### Step 5: Substitute values into the Nernst equation
Now substituting the known values into the Nernst equation:
\[ 0.712 = 0.771 - \frac{0.059}{1} \log \left( \frac{[Fe^{2+}]^2 \cdot [H^+]^2}{[Fe^{3+}]^2 \cdot P_{H_2}} \right) \]
Since \(P_{H_2} = 1 \text{ atm}\) and \([H^+] = 1 \text{ M}\):
\[ 0.712 = 0.771 - 0.059 \log \left( \frac{[Fe^{2+}]^2 \cdot 1^2}{[Fe^{3+}]^2 \cdot 1} \right) \]
This simplifies to:
\[ 0.712 = 0.771 - 0.059 \log \left( \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2} \right) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 0.059 \log \left( \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2} \right) = 0.771 - 0.712 \]
\[ 0.059 \log \left( \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2} \right) = 0.059 \]
Dividing both sides by 0.059:
\[ \log \left( \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2} \right) = 1 \]

### Step 7: Solve for the ratio
Taking antilog:
\[ \frac{[Fe^{2+}]^2}{[Fe^{3+}]^2} = 10 \]
Taking the square root:
\[ \frac{[Fe^{2+}]}{[Fe^{3+}]} = \sqrt{10} \approx 3.16 \]
Rounding to the nearest integer gives:
\[ \frac{[Fe^{2+}]}{[Fe^{3+}]} \approx 3 \]

### Final Answer
The ratio of concentration of \(Fe^{2+}\) to \(Fe^{3+}\) is approximately **3**.

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