Calculate the e.m.f. of the following cell at `298K :`
`Pt(s)|Br_(2)(l)|Br^(-)(0.010M)||H^(+)(0.030M)|H_(2)(g)(1"bar")|Pt(s)`
Given `: E_((1)/(2)Br_(2)//Br^(-))^(@)=+1.08V`.

Options

A. `-1.288V`
B. `1.288V`
C. `0.128V`
D. `-128V`

Explanation

To calculate the electromotive force (e.m.f.) of the given cell at 298 K, we will follow these steps:

### Step 1: Identify the Half-Reactions
The cell notation given is:
\[ \text{Pt(s)} | \text{Br}_2(l) | \text{Br}^-(0.010M) || \text{H}^+(0.030M) | \text{H}_2(g)(1 \text{ bar}) | \text{Pt(s)} \]

The half-reactions involved are:
1. Reduction of bromine:
\[ \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \]
The standard reduction potential \( E^\circ_{\text{Br}_2/\text{Br}^-} = +1.08 \, \text{V} \)

2. Oxidation of hydrogen:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]
The standard reduction potential for this reaction is \( E^\circ_{\text{H}^+/H_2} = 0 \, \text{V} \)

### Step 2: Determine the Cell Reaction
The overall cell reaction can be written as:
\[ \text{Br}_2 + 2\text{H}^+ \rightarrow 2\text{Br}^- + \text{H}_2 \]

### Step 3: Calculate the Number of Electrons Transferred (n)
From the balanced cell reaction, we see that 2 electrons are transferred:
\[ n = 2 \]

### Step 4: Apply 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: Calculate the Standard Cell Potential
The standard cell potential \( E^\circ_{\text{cell}} \) is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 1.08 \, \text{V} - 0 \, \text{V} = 1.08 \, \text{V} \]

### Step 6: Substitute Values into the Nernst Equation
Now, we need to calculate the concentrations:
- For products: \( [\text{Br}^-]^2 = (0.010)^2 = 0.0001 \)
- For reactants: \( [\text{H}^+]^2 = (0.030)^2 = 0.0009 \)

Thus, the reaction quotient \( Q \) is:
\[ Q = \frac{[\text{Br}^-]^2}{[\text{H}^+]^2} = \frac{0.0001}{0.0009} = \frac{1}{9} \]

Now substituting into the Nernst equation:
\[ E_{\text{cell}} = 1.08 - \frac{0.0591}{2} \log \left( \frac{1}{9} \right) \]

### Step 7: Calculate the Logarithm
\[ \log \left( \frac{1}{9} \right) = -\log(9) \approx -0.954 \]
Thus:
\[ E_{\text{cell}} = 1.08 - \frac{0.0591}{2} \times (-0.954) \]
\[ E_{\text{cell}} = 1.08 + 0.0282 \]
\[ E_{\text{cell}} = 1.08 + 0.0282 = 1.1082 \, \text{V} \]

### Step 8: Final Calculation
Now, we can finalize the calculation:
\[ E_{\text{cell}} = 1.08 - 0.0282 = 1.0518 \, \text{V} \]

### Conclusion
The e.m.f. of the cell at 298 K is approximately:
\[ E_{\text{cell}} \approx 1.0518 \, \text{V} \]

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