\(E_{cell}\) 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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.3291 V
In the cell Pt | Br2(l) | \( \ce{Br^{-}} \) (0.1 M) || \( \ce{Cl^{-}} \) (0.01 M) | Cl2(g, 1 atm) | Pt, the left electrode is the anode and the right one the cathode.
\( \displaystyle \text{Anode: }\ce{2Br^- -> Br2 + 2e^-} \)
\( \displaystyle \text{Cathode: }\ce{Cl2 + 2e^- -> 2Cl^-} \)
\( \displaystyle \text{Cell reaction: }\ce{Cl2 + 2Br^- -> 2Cl^- + Br2},\quad n = 2 \)
\( \displaystyle E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.36 - 1.09 = 0.27\ \text{V} \)
Nernst equation at 298 K (\( \ce{Br2} \) is a pure liquid and the pressure of \( \ce{Cl2} \) is 1 atm):
\( \displaystyle E_{cell} = 0.27 - \frac{0.0591}{2}\log\frac{[\ce{Cl^-}]^2}{[\ce{Br^-}]^2} = 0.27 - 0.02955 \log\frac{(0.01)^2}{(0.1)^2} \)
\( \displaystyle E_{cell} = 0.27 - 0.02955 \times (-2) = 0.27 + 0.0591 = 0.3291\ \text{V} \)