In acidic solution, the bromate ion can be used to react with a number of metal

Question

In acidic solution, the bromate ion can be used to react with a number of metal ions. One such reaction is

BrO3%u2212(aq)+Sn2+(aq)%u2192Br%u2212(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%u2212(aq)+Sn2+(aq)+ %u2212%u2212%u2212%u2192Br%u2212(aq)+Sn4+(aq)+ %u2212%u2212%u2212

Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).

Explanation / Answer

Reaction: BrO3^- + Sn^2+ = Br^- + Sn^4+

Half-rxns:

BrO3^- + 6H^+ + 6e- = Br^- + 3H2O ... reduction (occurs at the cathode)

Sn^2+ = Sn^4+ + 2e- ... oxidation (occurs at the anode)

Balance both half-rxns:

First half rxn remains as is.

Second one is multiplied by 3.

3(Sn^2+ = Sn^4+ + 2e-) becomes 3Sn^2+ = 3Sn^4+ + 6e-

Combine both half-rxns:

BrO3^- + 6H^+ + 3Sn^2+ + 6e- = Br^- + 3H2O + 3Sn^4+ + 6e- ... electrons MUST cancel

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

The coefficients would be as follows: 1 BrO3^- , 6 H^+ , 3 Sn^2+ , 1 Br^- , 3 H2O , 3 Sn^4+

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