E_cell for the given cell

$[\text{Given}{\mathrm{ E}}_{\mathrm{Br}2/\mathrm{Br}}^{\circ }-\; =+1.09\text{V,}{E}_{\mathrm{Cl}2/\mathrm{Cl}-}^{\circ }=+1.36\text{V] at 298 K is}$

Correct Answer: The correct option is 0.2109 V.

Step-by-Step Explanation:

Consider the electrochemical cell representation under non-standard conditions:
Pt(s) | Br^-(0.01 M) | Br₂(g) || Cl₂(g) | Cl^-(0.1 M) | Pt(s)

1. Identify the Cathode and Anode Half-Cells
By comparing the given standard reduction potentials:
- Standard reduction potential for chlorine:
$E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\circ }=+1.36\mathrm{V}$
- Standard reduction potential for bromine:
$E_{{\mathrm{Br}}_2/{\mathrm{Br}}^{-}}^{\circ }=+1.09\mathrm{V}$
Since chlorine has a higher reduction potential, it undergoes reduction at the cathode, while bromide ions undergo oxidation at the anode.

2. Write the Half-Reactions and the Overall Cell Reaction
- Anode (Oxidation):
$2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Br}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}$
- Cathode (Reduction):
${\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$
- Overall Cell Reaction:
${\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{Br}}_2\left(\mathrm{g}\right)$
Here, the number of electrons transferred (n) is 2.

3. Calculate the Standard Cell Potential (E°_cell)
Using the standard formula:
$E_{\mathrm{cell}}^{\circ }=E_{\mathrm{cathode}}^{\circ }-E_{\mathrm{anode}}^{\circ }$
Substitute the given potentials:
$E_{\mathrm{cell}}^{\circ }=1.36\mathrm{V}-1.09\mathrm{V}=0.27\mathrm{V}$

4. Apply the Nernst Equation
At T = 298 K, the Nernst equation is:
$E_{\mathrm{cell}}=E_{\mathrm{cell}}^{\circ }-\frac{0.0591}{n}logQ$
Where the reaction quotient Q (taking partial pressures of gases as 1 bar) is:
$Q=\frac{{\left[{\mathrm{Cl}}^{-}\right]}^2}{{\left[{\mathrm{Br}}^{-}\right]}^2}$

5. Substitute Concentrations and Calculate E_cell
Given [Cl^-] = 0.1 M and [Br^-] = 0.01 M:
$Q=\frac{{\left(0.1\right)}^2}{{\left(0.01\right)}^2}=\frac{{10}^{-2}}{{10}^{-4}}={10}^2=100$
Now, substitute n = 2 and Q = 100 into the Nernst equation:
$E_{\mathrm{cell}}=0.27-\frac{0.0591}{2}log\left(100\right)$
Since log(100) = 2:
$E_{\mathrm{cell}}=0.27-\frac{0.0591}{2}\times 2$
$E_{\mathrm{cell}}=0.27-0.0591=0.2109\mathrm{V}$