\(E_{cell}\) for the given cell

\(Pt(s)|Br_2(l)|Br^- (0.1\, \ M)||Cl^- (0.01\, \ M)| Cl_2(g)(1\, \ atm)| Pt(s)\)

[Given: \(E^o_{Br_2/Br^-} = +1.09 V\), \(E^o_{Cl_2/Cl^-} = +1.36 V\)] at 298 K is

0.2109 V

The correct answer is option 4. 2109 V.

The given cell is

\(Pt(s)|Br_2(l)|Br^- (0.1\, \ M)||Cl^- (0.01\, \ M)| Cl_2(g)(1\, \ atm)| Pt(s)\)

The given reduction potentials are

\(E^o_{Br_2/Br^-} = +1.09 V\)

\(E^o_{Cl_2/Cl^-} = +1.36 V\)

Oxidation half reaction: \(Br_2(l) + 2e^- \longrightarrow 2Br^- (aq)\) (At anode)

Reduction half reaction: \(Cl_2(l) + 2e^- \longrightarrow 2Cl^- (aq)\) (At cathode)

\(∴ E^o_{cell} = E^o_{reduction} - E^o_{oxidation}\)

or, \(E^o_{cell} = E^o_{Cl_2/Cl^-} - E^o_{Br_2/Br^-}\)

or, \(E^o_{cell} = 1.36 - (+1.09)\)

or, \(E^o_{cell} = +0.27V\)

From the half reaction it is clear the value of \(n = 2\)

We know, Nernst equation is

\(E_{cell} = E^o_{cell} - \frac{0.0591}{2}log Q\)

or, \(E_{cell} = E^o_{cell} - \frac{0.0591}{2}log \frac{[Br^-]^2}{[Cl^-]^2}\)

or, \(E_{cell} = E^o_{cell} - \frac{0.0591}{2}log \frac{[0.1]^2}{[0.01]^2}\)

or, \(E_{cell} = E^o_{cell} - \frac{0.0591}{2}log \frac{0.01}{0.0001}\)

or, \(E_{cell} = E^o_{cell} - \frac{0.0591}{2}log(100)\)

or, \(E_{cell} = E^o_{cell} - \frac{0.0591}{2} × 2\)

or, \(E_{cell} = +0.27 - 0.0591\)

\(∴ E_{cell} = + 0.2109 V\)