what is the emf at `25^(@)C` for the cell,
`Ag|{:(AgBr(s)" " Br^(-)), (" "alpha = 0.34):}||{:(Fe^(3+) " " Fe^(2+)), (alpha = 0.1 " " alpha = 0.02):}|Pt`
The standard reduction potentials for the half-reactions `AgBr + e^(-) rightarrow Ag + Br^(-) and Fe^(3+) + e^(-) rightarrow Fe^(2+)` are +0.0713V and +0.770V respectively.

Options

A. 0.474 volt
B. 0.529 volt
C. 0.356 volt
D. 0.713 volt

Explanation

To find the electromotive force (emf) of the given electrochemical cell at 25°C, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
The half-reactions provided are:
1. \( \text{AgBr} + e^- \rightarrow \text{Ag} + \text{Br}^- \) with \( E^\circ = +0.0713 \, \text{V} \)
2. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with \( E^\circ = +0.770 \, \text{V} \)

### Step 2: Determine which half-reaction is oxidation and which is reduction
Since \( E^\circ \) for the reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) is greater than that for the reduction of \( \text{AgBr} \), the \( \text{Fe}^{3+} \) will be reduced, and the \( \text{Ag} \) will be oxidized. Thus:
- Oxidation: \( \text{Ag} \rightarrow \text{AgBr} + e^- \)
- Reduction: \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \)

### Step 3: Write the net reaction
The overall cell reaction can be written as:
\[ \text{Fe}^{3+} + \text{Ag} + \text{Br}^- \rightarrow \text{Fe}^{2+} + \text{AgBr} \]

### Step 4: Calculate the standard cell potential (\( E^\circ_{\text{cell}} \))
Using the standard reduction potentials:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0.770 \, \text{V} - 0.0713 \, \text{V} = 0.6987 \, \text{V} \]

### Step 5: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where:
- \( n = 1 \) (number of electrons transferred)
- \( Q = \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}][\text{Br}^-]} \)

### Step 6: Calculate the reaction quotient \( Q \)
Given concentrations:
- \( [\text{Fe}^{2+}] = 0.02 \, \text{M} \)
- \( [\text{Fe}^{3+}] = 0.1 \, \text{M} \)
- \( [\text{Br}^-] = 0.34 \, \text{M} \)

Thus,
\[ Q = \frac{0.02}{0.1 \times 0.34} = \frac{0.02}{0.034} \approx 0.5882 \]

### Step 7: Calculate the logarithm of \( Q \)
\[ \log Q = \log(0.5882) \approx -0.231 \]

### Step 8: Substitute values into the Nernst equation
Now substituting into the Nernst equation:
\[ E_{\text{cell}} = 0.6987 \, \text{V} - \frac{0.0591}{1} \times (-0.231) \]
\[ E_{\text{cell}} = 0.6987 \, \text{V} + 0.0136 \, \text{V} \]
\[ E_{\text{cell}} \approx 0.7123 \, \text{V} \]

### Final Answer
The emf of the cell at 25°C is approximately **0.7123 V**.

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