Acidic Solution In acidic solution, bromate ion can be used to react with a numb

Question

Acidic Solution

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)+     

Part A

What 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.

Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).

Basic Solution

Potassium permanganate, KMnO4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium sulfite:

MnO4(aq)+SO32(aq)MnO2(s)+SO42(aq)

Since this reaction takes place in basic solution, H2O(l) and OH(aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:

MnO4(aq)+SO32(aq)+     MnO2(s)+SO42(aq)+     

Part 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.

Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).

Explanation / Answer

A) the given reaction is

BrO3-(aq) + Sn2+(aq) --> Br-(aq) + Sn4+(aq)

now

split the equation into two half reactions

first half reaction is

BrO3-(aq) --> Br-(aq)

now use H2O to balance the oxygen atoms O

BrO3-(aq) --> Br-(aq) + 3H2O(l)

Now use H+ to balance hydrogens

BrO3-(aq) + 6H+(aq) --> Br-(aq) + 3H2O(l)

now use electrons to balance the charge

BrO3-(aq) + 6H+(aq) + 6e- --> Br-(aq) + 3H2O(l)

Now

Sn2+(aq)--> Sn4+(aq)

using the same method

we get

Sn2+(aq)--> Sn4+(aq) + 2e-      

now equate the electrons in both half reactions

3Sn2+(aq)--> 3Sn4+(aq) + 6e-

we get the balanced reaction as

BrO3-(aq) + 6H+(aq) + 3Sn2+(aq) --> Br-(aq) + 3H2O(l)+ 3Sn4+(aq)   

2)

first split the half reactions

MnO4– --> MnO2

now Balance O in by adding H2O

MnO4-   ----> MnO2 + 2H2O

now balance H by adding H+

MnO4– + 4H+ ---> MnO2 + 2H2O

Balance charge by adding electrons

MnO4– + 4H+ + 3e– --> MnO2 + 2H2O

now convert to basic solution by replacing the H+ with H2O and adding the same number of OH– ions to the other side

MnO4– + 4H2O + 3e– ---> MnO2 + 2H2O + 4OH–

now cancel any extra H2O

MnO4– + 2H2O + 3e– ---> MnO2 + 4OH–

now follow the same with the sulfur half reaction

we get

SO32– + 2OH– ---> SO42– + H2O + 2e–

now combine the two by multiplying the Mn reaction by 2 and the S reaction by 3 and add

we get

2MnO4– + 4H2O + 3SO32– + 6OH– + 6e– ---> 2MnO2 + 8OH– + 3SO42– + 3H2O + 6e–

now cancel out the common terms

we get

2MnO4– + H2O + 3SO32- ---> 2MnO2 + 2OH– + 3SO42–

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