Given cell
Cd | CdCl2 (0.55M) || NaBrO3 (0.34M), NaBr (0.12M), pH=1.20 | Pt
1. How many electrons are transferred?
2. What is E° for the cell?
3. What is E for the cell?
Cd ==> Cd^2+ + 2e
BrO3^- + 6e + 6H^+ ==> Br^- + 3H2O
Br goes from +5 to -1 which is 6 so 6 e must be transferred.
For Eo cell add the Cd part to the BrO3^- part (oxidation half cell to reduction half cell).
For Ecell = Eocell -(0.0592/n)*log(Q)
What's Q? That's
(Br^-)(Cd^2+)^3/(BrO3^-)(H^+)^6