\(E_{cell}\) 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 |
0.1518 V 0.3291 V 0.388 V 0.2109 V |
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, \(E_{cell} = +0.27 - 0.0591\) \(∴ E_{cell} = + 0.2109 V\) |