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In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half-reaction. In either method H20(l) , OH-, , and H+ (aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditionsIn acidic solution, the bromate ion can be used to react with a number of metal ions. One such reaction isBrO3-(aq) + Sn2+(aq)>>>Br-(aq)+Sn4+(aq)Since this reaction takes place in acidic solution, \rm H_2O{(l)} and \rm H^+{(aq)} will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:BrO3-(aq)+Sn2+(aq)+__?__>>>Br-(aq)+Sn4+(aq)+__?__Part A:What are the coefficients of the six species in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks.Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).Part B:
What are the coefficients of the six species in the balanced equation above? Remember to include H2O(l) and OH-(aq) in the blanks where appropriate
Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).

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)

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