Mark the correct Nernst equation for the given cell. `F_((s))|Fe^(2+)(0.001 M) ||H^(+)(1M) | H_(2(g)) ( 1 bar ) | Pt_((s))` is

Options

A. `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe^(2+)][H^+]^2)/([Fe][H_2])`
B. `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe][H^+]^2)/([Fe^(2+)][H_2])`
C. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe^(2+)][H_2])/([Fe][H^+]^2)`
D. `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe][H_2])/([Fe^(2+)][H^+]^2)`

Explanation

To derive the correct Nernst equation for the given electrochemical cell, we can follow these steps:

### Step 1: Identify the Anode and Cathode Reactions
In the given cell notation:
- The left side (F(s)|Fe²⁺(0.001 M)) is the anode where oxidation occurs.
- The right side (H⁺(1 M)|H₂(g)(1 bar)|Pt(s)) is the cathode where reduction occurs.

**Anode Reaction:**
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]

**Cathode Reaction:**
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 2: Write the Overall Cell Reaction
Combining the anode and cathode reactions gives us the overall cell reaction:
\[ \text{Fe} + 2\text{H}^+ \rightarrow \text{Fe}^{2+} + \text{H}_2 \]

### Step 3: Determine the Standard Cell Potential (E°cell)
We need the standard reduction potentials for both half-reactions:
- For the Fe²⁺/Fe half-reaction: \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} \)
- For the H⁺/H₂ half-reaction: \( E^\circ_{\text{H}^+/H_2} = 0 \, \text{V} \) (standard hydrogen electrode)

The standard cell potential can be calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0 - E^\circ_{\text{Fe}^{2+}/\text{Fe}} \]

### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
where \( n \) is the number of moles of electrons transferred (which is 2 in this case), and \( Q \) is the reaction quotient.

### Step 5: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) for the reaction:
\[ \text{Fe} + 2\text{H}^+ \rightarrow \text{Fe}^{2+} + \text{H}_2 \]
is given by:
\[ Q = \frac{[\text{Fe}^{2+}]}{[\text{H}^+]^2} \]
Substituting the concentrations:
\[ Q = \frac{0.001}{(1)^2} = 0.001 \]

### Step 6: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log(0.001) \]

### Final Nernst Equation
Thus, the correct Nernst equation for the given cell is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - 0.02955 \log(0.001) \]