Question

In acidic solution, bromate ion can be used to react with a number of metal ions. One such reaction is: BrO3−(aq)+Sn2+(aq)→Br−(aq)+Sn4+(aq). Since this reaction takes place in acidic solution, H2O(l) and 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)+     −−−AWhat are the coefficients of the six species in the balanced equation above? Remember to include coefficients for H2O(l) and H+(aq) in the appropriate blanks. B)What are the coefficients of the six species in the balanced equation above? Remember to include coefficients for H2O(l) and OH−(aq) in the blanks where appropriate.

Answer 1

Solution :-

Balancing redox equation in the acidic medium

Step 1 writing the half reactions

BrO3^- --- > Br^-

Sn^2+ --- > Sn^4+

Step 2 balancing the oxygen atoms by adding H2O

BrO3^- --- > Br^- + 3H2O   

Sn^2+ --- > Sn^4+

Step 3 balancing the hydrogens by adding H^+

6H^+ + BrO3^- --- > Br^- + 3H2O   

Sn^2+ --- > Sn^4+

Step 4 balancing the charges by adding electrons

6e- + 6H^+ + BrO3^- --- > Br^- + 3H2O   

Sn^2+ --- > Sn^4+ +2e-

Step 5 equilizing the number of electrons in both equations by multiplying equation 2 by factor 3

6e- + 6H^+ + BrO3^- --- > Br^- + 3H2O   

3Sn^2+ --- > 3Sn^4+ +6e-

Step 6 adding both half equations by cancelling out the same species on the opposite sites of equations

6e- + 6H^+ + BrO3^- --- > Br^- + 3H2O   

3Sn^2+ --- > 3Sn^4+ +6e-

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6H^+ + BrO3^- + 3Sn^2+ ---- > Br^- + 3H2O + 3Sn^4+

Therefore the balanced equation is as follows

BrO3^-(aq) + 3Sn^2+(aq) +6H^+(aq)   ---- > Br^-(aq)   + 3Sn^4+ (aq) + 3H2O(l)

Answer 2

Balanced Redox Reaction in Acidic Solution (Part A):

The reaction involves:

• Oxidation: ${\text{Sn}}^{2+}\to {\text{Sn}}^{4+}+2e^{-}$

• Reduction: ${\text{BrO}}_3^{-}+6e^{-}+6{\text{H}}^{+}\to {\text{Br}}^{-}+3{\text{H}}_2\text{O}$

Balanced Equation:

$${\text{BrO}}_3^{-}(aq)+3{\text{Sn}}^{2+}(aq)+6{\text{H}}^{+}(aq)\to {\text{Br}}^{-}(aq)+3{\text{Sn}}^{4+}(aq)+3{\text{H}}_2\text{O}(l)$$

Coefficients:
1 ${\text{BrO}}_3^{-}$, 3 ${\text{Sn}}^{2+}$, 6 ${\text{H}}^{+}$, 1 ${\text{Br}}^{-}$, 3 ${\text{Sn}}^{4+}$, 3 ${\text{H}}_2\text{O}$.

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Balanced Redox Reaction in Basic Solution (Part B):

For basic conditions, add ${\text{OH}}^{-}$ to neutralize ${\text{H}}^{+}$ from the acidic equation.

Balanced Equation:

$${\text{BrO}}_3^{-}(aq)+3{\text{Sn}}^{2+}(aq)+3{\text{H}}_2\text{O}(l)\to {\text{Br}}^{-}(aq)+3{\text{Sn}}^{4+}(aq)+6{\text{OH}}^{-}(aq)$$

Coefficients:
1 ${\text{BrO}}_3^{-}$, 3 ${\text{Sn}}^{2+}$, 3 ${\text{H}}_2\text{O}$, 1 ${\text{Br}}^{-}$, 3 ${\text{Sn}}^{4+}$, 6 ${\text{OH}}^{-}$.