Half rxns:
BrO3^- > Br^-
unbalanced oxygens so you would add H2O to balance the oxygen
BrO3^- > Br^- + 3H2O (because you want 3 oxygens on the right as there are on the left)
Then you must balance hydrogens
BrO3^- + 6H^+ > Br^- + 3H2O (you add 6H because there are 6 hydrogens on the right, 3 * 2 = 6)
Now add 6 electrons to the left to cancel out the charge of the 6H^+
BrO3^- + 6H^+ + 6e^- > Br^- + 3H2O
Other half rxn:
Sb^3+ > Sb^5+
You would have to add electrons to the more positive side so the charges on each side are equal
Sb^3+ > Sb^5+ + 2e- (you add 2 electrons because 5-2=3)
Multiply this half reaction by 3 so you can cancel the electrons out from the first half rxn, 2*3=6
Now you have
3Sb^3+ > 3Sb^5+ + 6e^-
Now you can cancel out the electrons and combine each half reaction to get your final answer of:
BrO3^- + 3Sb^3+ + 6H^+ > Br^- + 3Sb^5+ + 3H2O
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 the antimony is in the form 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.
BrO3^- + 3Sb^3-> Br^- + 3Sb^5+
I tried BrO3^- + 3Sb^3+ +6H -> Br^- + 3Sb^5+ +3H2O but it is wrong.
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