The correct Nernst equation for the given cell `Pt_((s)) | Br_(2(l)) | Br^(-) (M) || H^(+) (M) |H_(2(g)) (1bar) | Pt_((s))` is

Options

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

Answer

To derive the correct Nernst equation for the given electrochemical cell represented as `Pt_((s)) | Br_(2(l)) | Br^(-) (M) || H^(+) (M) | H_(2(g)) (1bar) | Pt_((s))`, we will follow these steps:

### Step 1: Identify the Anode and Cathode Reactions
In the given cell representation:
- The left side (anode) involves the oxidation of bromide ions (Br⁻) to bromine (Br₂).
- The right side (cathode) involves the reduction of hydrogen ions (H⁺) to hydrogen gas (H₂).

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

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

### Step 2: Write the Overall Cell Reaction
Combining the anodic and cathodic reactions gives the overall cell reaction:
\[ 2 \text{Br}^- + 2 \text{H}^+ \rightarrow \text{Br}_2 + \text{H}_2 \]

### Step 3: Determine the Number of Electrons Transferred (n)
From the reactions, we see that 2 electrons are transferred in total:
- \( n = 2 \)

### Step 4: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

### Step 5: Identify Products and Reactants
For our overall reaction:
- Products: Br₂ and H₂
- Reactants: \([H^+]^2\) and \([Br^-]^2\)

Thus, the Nernst equation becomes:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{[\text{Br}_2][\text{H}_2]}{[\text{H}^+]^2[\text{Br}^-]^2} \right) \]

### Step 6: Final Form of the Nernst Equation
Rearranging the equation, we can express it as:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{[\text{Br}_2][\text{H}_2]}{[\text{H}^+]^2[\text{Br}^-]^2} \right) \]

### Conclusion
The correct Nernst equation for the given cell is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{[\text{Br}_2][\text{H}_2]}{[\text{H}^+]^2[\text{Br}^-]^2} \right) \]

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