Answer:

(a) Ti³⁺ + H2O → TiO²⁺ + 2 H⁺ + e^-

H⁺ + HOBr + 2e^- → Br^- + H₂O

2 Ti³⁺ + HOBr + H₂O → 2 TiO²⁺ + 3 H⁺ + Br^-

(b) HOBr is the oxidizing agent and Ti³⁺ is the reducing agent.

(c) The observed voltage will be greater than the E˚ value since:

E = E°  - (0.0592/2) log (0.1)³

(d) Identification of a property from the group: ∆G, K, pH

ΔG° = -nFE°  or  E° = 0.0592/n log K