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

Question

Acidic solution

In acidic solution, the 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 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 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

acidic solution
Bromate has an oxidation number of +5 for the Br, while it's -1 for Br- . That means a gain of 6 electrons. Sn2+ --> Sn4+ is a loss of two electrons, so you need 4 Sn2+ to provide the necessary electrons for 1 BrO3-. The three O's in the BrO3- get turned into 3 H2O, so you need 6 H+ on the left. Summing up, the coefficients are 1, 4, 6, 1, 4, 3.
basic solution
There are two methods to continue where you left off. One involves simply balancing with hydrogen ions and later adding an equal amount of hydroxide to either side to eliminate all of the hydrogen ions and form water.

I'll solve the problem step by step so you can see what I'm trying to say.

MnO4(-) ? MnO2

Add water to balance oxygen.
MnO4(-) ? MnO2 +2 H2O

Add hydrogen to balance hydrogen.
4H(+) + MnO4(-) ? MnO2 +2 H2O

Add electrons to balance charge.
4H(+) + 3e- + MnO4(-) ? MnO2 +2 H2O

Doing the same thing to this half reaction :
SO3(2-) + H2O ? SO4(2-) + 2 H(+) + 2e-


Now, balance the electrons.

2(4 H(+) + 3e- + MnO4(-) ? MnO2 +2 H2O)

3(SO3(2-) + H2O ? SO4(2-) + 2 H(+) + 2e-)

2 H(+) + 3 SO3(2-) + 2 MnO4(-) ? 2 MnO2 + 3 SO4(2-) + H2O

Now, add enough hydroxide to each side to eliminate all of the acid.

2 OH(-) + 2 H(+) + 3 SO3(2-) + 2 MnO4(-) ? 2 MnO2 + 3 SO4(2-) + H2O + 2 OH(-)

Now, H(+) + OH(-) ? H2O and H2O 's on either side cancel to give :

H2O + 3 SO3(2-) + 2 MnO4(-) ? 2 MnO2 + 3 SO4(2-) + 2 OH(-)

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