Effect of Conditions on Reduction of Permanganate Ion Acidic Conditions Write th

Question

Effect of Conditions on Reduction of Permanganate Ion Acidic Conditions Write the balanced half-reaction equation for reduction of Mno_4 in acidic solution. Strongly Basic Conditions Write the balanced half-reaction equation for reduction of Mno_4 in strongly basic solution. Neutral to Mildly Basic Conditions Write the balanced half-reaction equation for the reduction of Mno_4 in strongly basic solution. Summary of Effects Determine the oxidation state of Mn in the followings. MnO_4 MnO_4 Mn^2+ and MnO_2 Arrange above species in order of decreasing oxidation number. Fill in the Summary Table

Explanation / Answer

Any RedOx reaction involves two parts, Reduction and Oxidation. Thus, requiring one reducing and one Oxidising part.

Now, the question requires only Half reaction to be balance, it can be done as under.

a) Acidic Condition

As only MnO4- is present, thus Oxygen is the only element that requires balancing, which is accomplished by adding H2O to the product side. Since there are four oxygen atoms in MnO4-, four water molecules must be added so that the oxygen of the MnO4- balances with the oxygen in the H2O. However, the added water introduces hydrogen into the reaction, which must be balanced as well. As balancing is to be done under acidic conditions, this is accomplished by adding hydrogen ions to the reactant side. Electrons must also be added to balance the charges. Because on the left there is a charge of +7 and on the right a charge of +2, five electrons must be added to the +7 side:

MnO4- (aq) + 8H+ (aq) + 5e- Mn2+ (aq) +4H2O (l)

b) Strongly Basic Condition:

For the reaction in basic medium, it is best suited to balance the reaction as in Acidic conditions and the add OH- to neutralise H+ ions and form water.

Taking the above reaction and adding OH- on both sides.

MnO4- (aq) + 8H+(aq) + 8OH- + 5e- Mn2+ (aq) +4H2O (l) + 8OH-

Or, MnO4- (aq) + 4H2O (l) + 5e- Mn2+ (aq) + 4OH-

d) Oxidation Number:

i) Mn in MnO4-

The O.N. of oxygen in most compounds is -2, there are 4 O’s so the sum of the O.N.‘s = -8

the overall charge on the ion is -1,

sum of all the O.N‘s must add up to -1

the O.S. of Mn plus the sum of the O.N.‘s of the four O’s must equal -1

Therefore, the O.N. of Manganese in MnO4¯ = +7

Similarly

ii) ON of Mn in MnO42- is +6

iii) ON of Mn in Mn2+ is +2.

iv) ON of Mn in MnO2 is +4.

No, it can be easily arranged.

MnO4¯ > MnO42- > MnO2 > Mn2+

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