1)

KClO3 ---> KCl+ 3/2 O2


In KCLO3 the oxidation number of potassuim is +1, chlorine is +5, oxygen is -2

In KCl has oxidation number , potassium +1 and chlorine is -1

Oxygen molecule has oxidation number zero

2)

Here are the two half-reactions: NOTE: Cd exists ONLY as Cd2+, not Cd+.

In ==> In^3+ + 3e-
Cd^2+ + 2e- ==> Cd

Note that there are 3e- on the right and 2e- on the left. The least common multiple of 2 and 3 is 6. Multiply the first equation by 2 to get 6e- on the right. Then multiply the second equation by 3 to get 6e- on the left.

2In ==> 2In^3+ + 6e-
3Cd^2+ + 6e- ==> 3Cd . . . . .now add the two equations together. Notice how the 6e- cancels out.
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2In + 3Cd^2+ ==> 2In^3+ + 3Cd

4)

Your balanced equation should be 5 HNO2 + 2 MnO4- + H+ --> 5 NO3- + 2 Mn2+ + 3 H2O

The first step when balancing redox reactions is to figure out how many electrons are being lost and gained and to balance those.
In this case, nitrogen has an oxidation state of 3+ on the left and an oxidation state of 5+ on the right; therefore, nitrogen is losing two electrons.
Manganese goes from an oxidation state of 7+ as a reactant to 2+ as a product; so each manganese atom gains five electrons.
Because mass is conserved, the same number of electrons has to be lost and gained, so the way to balance that is to put a 5 in front of every nitrogen-containing species and a 2 in front of every manganese-containing species so that 10 electrons are lost and 10 electrons are gained in the course of the reaction. So we have

5 HNO2 + 2 MnO4- --> 5 NO3- + 2 Mn2+

Now you just have to balance the oxygens and the hydrogens. Start with the oxygens; you have 18 on the left and 15 on the right, so add three water molecules to the right to balance those.

5 HNO2 + 2 MnO4- --> 5 NO3- + 2 Mn2+ + 3 H2O

Now you just have to balance the hydrogens; you have 5 on the left and 6 on the right, so add one H+ ion to the left, and you're all set.

5 HNO2 + 2 MnO4- + H+ --> 5 NO3- + 2 Mn2+ + 3 H2O

5)

Cr + 6 Cu{2+} + 4 H2O → CrO4{2-} + 6 Cu{+} + 8 H{+}

3)

Cu = Cu2+ + 2e- ) x 3
NO3- + 4 H+ + 3e- = NO + 2 H2O ) x 2

3 Cu + 2 NO3- + 8 H+ = 3 Cu2+ + 2 NO + 4 H2O