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 e.m.f. of the given cell at 298 K, we will follow these steps:

### Step 1: Identify the Cell Reaction
The cell reaction can be derived from the half-reactions involved. The overall reaction for the cell is:
\[ 2 \text{Br}^- + 2 \text{H}^+ \rightarrow \text{Br}_2 + \text{H}_2 \]
This indicates that 2 moles of bromide ions are oxidized to bromine gas, while 2 moles of hydrogen ions are reduced to hydrogen gas.

### Step 2: Determine the Number of Electrons Transferred (n)
From the balanced equation, we see that 2 electrons are transferred in the reaction. Thus, \( n = 2 \).

### Step 3: Write the Nernst Equation
The Nernst equation at 298 K is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
where \( Q \) is the reaction quotient.

### Step 4: Calculate the Standard Cell Potential (E°cell)
The standard reduction potential for the half-reaction \( \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \) is given as \( E^\circ = +1.08 \, \text{V} \). The standard reduction potential for the hydrogen half-reaction is \( 0 \, \text{V} \). Therefore, the standard cell potential can be calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
In our case:
\[ E^\circ_{\text{cell}} = 0 - 1.08 = -1.08 \, \text{V} \]

### Step 5: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Br}_2][\text{H}_2]}{[\text{Br}^-]^2[\text{H}^+]^2} \]
Given that the concentrations are:
- \( [\text{Br}^-] = 0.010 \, \text{M} \)
- \( [\text{H}^+] = 0.030 \, \text{M} \)
- \( [\text{Br}_2] \) and \( [\text{H}_2] \) are both taken as 1 (since they are in standard state).

Thus:
\[ Q = \frac{1 \cdot 1}{(0.010)^2 \cdot (0.030)^2} = \frac{1}{0.0001 \cdot 0.0009} = \frac{1}{9 \times 10^{-8}} = 1.111 \times 10^7 \]

### Step 6: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:
\[ E_{\text{cell}} = -1.08 - \frac{0.0591}{2} \log(1.111 \times 10^7) \]

Calculating \( \log(1.111 \times 10^7) \):
\[ \log(1.111 \times 10^7) \approx 7.045 \]

Now substituting this back into the Nernst equation:
\[ E_{\text{cell}} = -1.08 - \frac{0.0591}{2} \cdot 7.045 \]
\[ E_{\text{cell}} = -1.08 - 0.02955 \cdot 7.045 \]
\[ E_{\text{cell}} = -1.08 - 0.208 \]
\[ E_{\text{cell}} = -1.288 \, \text{V} \]

### Final Answer
The e.m.f. of the cell at 298 K is:
\[ E_{\text{cell}} = -1.288 \, \text{V} \]