Consider an electrochemical cell involving the overall reaction
2AgBr(s) + Pb(s) → Pb2+ + 2Ag(s) + 2Br–
Each half-reaction is carried out in a separate compartment. The anion included in the lead half-cell is NO3–. The cation in the silver half-cell is K+. The two half-cells are connected by a KNO3 salt bridge. If [Pb2+] = 1.0 M, what concentration of Br– ion will produce a cell emf of 0.25 V at 298 K?
Given: AgBr(s) + e–→ Ag + Br–, E° = +0.07 V.
◦ 0.02 M
◦ 0.14 M
◦ 0.38 M
◦ 1.0 M
◦ 7.0 M