Respuesta :
Answer:
(1,3,6,1,3,3)
BrO3-(aq) + 3Sn2+(aq) + 6H+(aq) → Br-(aq) + 3Sn4+(aq) + 3H2O(l)
Explanation:
BrO3-(aq) + Sn2+(aq) --> Br-(aq)+Sn4+(aq)
Separating into half equations, we have;
Sn2+ → Sn4+ + 2e- (Oxidation)
BrO3- + 6e- → Br- (Reduction)
Upon balancing the atoms and charges we have;
Sn2+ → Sn4+ + 2e- (Oxidation)
BrO3- + 6e- + 6H+ → Br- + 3H2O (Reduction)
Make sure electron gain is equal to electron lost, we have;
3Sn2+ → 3Sn4+ + 6e- (Oxidation)
BrO3- + 6e- + 6H+ → Br- + 3H2O (Reduction)
Add both half equations together;
BrO3- + 6e- + 6H+ + 3Sn2+ → Br- + 3H2O + 3Sn4+ + 6e-
Upon simplification;
BrO3-(aq) + 3Sn2+(aq) + 6H+(aq) → Br-(aq) + 3Sn4+(aq) + 3H2O(l)
What are the coefficients of the six species in the balanced equation above?
(1,3,6,1,3,3)
The coefficients of the six species in the balanced equation are:
(1,3,6,1,3,3)
Balanced chemical reaction:
[tex]BrO_3^-(aq) + 3Sn^{2+}(aq) + 6H^+(aq) ---- > B^r-(aq) + 3Sn^{4+}(aq) + 3H_2O(l)[/tex]
Ionic equation:
[tex]BrO_3^-(aq) + Sn^{2+}(aq) ----- > Br^-(aq)+Sn^{4+}(aq)[/tex]
Separating into half equations, we have;
[tex]Sn^{2+} ---- > Sn^{4+} + 2e^-[/tex] (Oxidation)
[tex]BrO_3^- + 6e^------ > Br^-[/tex] (Reduction)
On balancing the atoms and charges we have;
[tex]Sn^{2+}------ > Sn^{4+} + 2e^-[/tex] (Oxidation)
[tex]BrO_3^- + 6e^- + 6H^+ ------ > Br^- + 3H_2O[/tex] (Reduction)
On adding both half equations together;
[tex]BrO_3^-(aq) + 3Sn^{2+}(aq) + 6H^+(aq) ---- > B^r-(aq) + 3Sn^{4+}(aq) + 3H_2O(l)[/tex]
Thus, the coefficients of the six species in the balanced equation are:
(1,3,6,1,3,3)
Find more information about Balanced chemical reaction here:
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