May I also get help with this as well please

Question

The quantity of antimony in an ore can be determined by an oxidation‑reduction titration with an oxidizing agent. The ore is dissolved in hot, concentrated acid and passed over a reducing agent so that all of the antimony is in the form of Sb3+ (aq). The Sb3+(aq) is completely oxidized by an aqueous solution of BrO3-(aq). Complete and balance the equation for this reaction in acidic solution. equation:

BrO−3+Sb3+⟶Br−+Sb5+

Answer 1

BrO₃^- + Sb³⁺⟶Br⁻ + Sb⁵⁺

BrO₃^- ==> Br^- ... reduction half reaction

BrO₃^- ==> Br^- + 3H₂O ... balanced for Br and O

BrO₃^- + 6H2O ==> Br^- + 3H₂O + 6OH^- ,,, balanced for Br, O and H using base (OH-) and H2O

BrO₃^- + 6H2O + 6e- ==> Br^- + 3H₂O + 6OH^- ... balanced for mass and charge

Sb³⁺ ==> Sb⁵⁺ ... oxidation half reaction

Sb³⁺ ==> Sb⁵⁺ + 2e- ... balanced for mass and charge

Multiply the oxidation half reaction by 3 to equalize the electrons for the reduction half reaction to get...

3Sb³⁺ ==> 3Sb⁵⁺ + 6e- and add this to the reduction half reaction to get

BrO₃^- + 6H2O + 6e- + 3Sb³⁺ ==> Br^- + 3H₂O + 6OH^- + 3Sb⁵⁺ + 6e-

Combine/cancel like terms to obtain the final balanced equation:

BrO₃^- + 3H2O + 3Sb³⁺ ==> Br^- + 6OH^- + 3Sb⁵⁺ ... BALANCED REDOX EQUATION