Question

1) Complete and balance the following half-reaction: H2SO3(aq)?SO2?4(aq) (acidic solution)

2)Complete and balance the following half-reaction:

NO?3(aq)?NO(g)(acidic solution)

3)Complete and balance the following half-reaction:

O2(g)?H2O(l)(acidic solution)

4)Complete and balance the following half-reaction:

O2(g)?H2O(l) (basic solution)

5)Complete and balance the following half-reaction: Mn2+(aq)?MnO2(s) (basic solution)

6)Complete and balance the following half-reaction: Cr(OH)3(s)?CrO2?4(aq) (basic solution)

Answer 1

in Cr(OH)3, since its a neutral compound, and OH carries a charge of -1 each, then the oxidation number of Cr is +3.
for chromate ion, Cr has an oxidation number of +6. so, since the oxidation number increase, Cr, like the previous question, undergoes oxidation. similarly, we work out the balanced equation in the same manner.
since oxidation increases by 3, u can assume that Cr(OH)3 loses 3 electrons
Cr(OH)3 + ______ ------> CrO4(2-) + 3e- + _____
now, the right side has 5 additional negative charges. but since tis is a basic solution, instead of using H+ to balance the charges, OH- is used instead
Cr(OH)3 + 5OH- ------> CrO4(2-) + 3e- + ______
so, now u can see that u haf an additional 8H and 4O at the left side. tat's 4 water molecules that should be added to the right side!
Cr(OH)3(aq) + 5OH-(aq) ------> CrO4(2-)(aq) + 3e- + 4H2O(l)


Balance the oxygen:

O2 --> 2H2O

Balance the hydrogen by adding H+:

O2 + 4H+ ---> 2H2O

Balance the charges by adding electrons:

O2 + 4H+ + 4e- --> 2H2O


(yes it does go from 0 to -2, but since the oxidation number it lowering, it is REDUCTION, not oxidation).