Potassium permanganate, KMnO4, is a powerful oxidizing agent. The products of a

Question

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) + I- (aq) ---> MnO2 (s) + I2(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) + I- (aq) + ___---> MnO2 (s) + I2(aq) + ___

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.

Explanation / Answer

Start by separating the equation into two parts

MnO4- (aq) --> MnO2 (s)
and
I- (aq) --> I2 (aq)

like in an acidic solution start by balancing the main element, then balance the oxygen, then the hydrogen

MnO4-(aq) --> MnO2 (s)
MnO4- (aq) --> MnO2 (s) + 2H2O (l)
4H+(aq) + MnO4- (aq) --> MnO2 (s) + 2H2O (l)

here is where it is different... add as many OH- to each side as the H+ on the one side

4H+(aq) + 4OH- (aq) + MnO4- (aq) --> MnO2 (s) + 2 H2O (l) + 4 OH- (aq)

continue to balance the charge

3e-+ 4H+(aq) + 4OH- (aq) + MnO4- (aq) --> MnO2 (s) + 2 H2O (l) + 4 OH- (aq)

follow the same steps for the other part of the reaction

2I- (aq) --> I2 (aq) + 2e-

multiply the first equation by 2 and the second by three in order to get the electrons to cancel out

6e-+ 8H+(aq) + 8OH- (aq) + 2MnO4- (aq) --> 2MnO2 (s) + 4 H2O (l) + 8OH- (aq)
6I- (aq) --> 3I2 (aq) + 6e-

now that all the electrons cancel out, combine the OH and the H+ to make H2O and cancel out the elements on both sides

the equation should look like this when completely simplified

4H2O (l) + 2MnO4 - (aq) + 6I- (aq) --> 2MnO2 (s) + 3I2 (aq) + 8OH- (aq)

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