**Balancing the Redox Reaction:**

To balance the given redox reaction, CN- + MnO4- → CNO- + MnO2, we will use the oxidation number method.

**Step 1: Assign Oxidation Numbers**
First, we assign oxidation numbers to each element in the reaction. The oxidation number is the charge that an atom would have if the compound were ionic.

CN-: The oxidation number of C is -3 since it is combined with a more electronegative element (N). The oxidation number of N is -3 since it is in the cyanide ion.
MnO4-: The oxidation number of O is -2. Let the oxidation number of Mn be x.
CNO-: The oxidation number of C is +2. The oxidation number of N is -2, and the oxidation number of O is -2.
MnO2: The oxidation number of O is -2. Let the oxidation number of Mn be y.

**Step 2: Determine the Change in Oxidation Numbers**
We need to determine the change in oxidation numbers for the elements that are being oxidized and reduced.

In this reaction, the oxidation state of Mn changes from x to y, indicating that it is being reduced. The oxidation state of C changes from -3 to +2, indicating that it is being oxidized.

**Step 3: Balance the Atoms**
Start by balancing the atoms that are not involved in redox reactions. In this case, we have carbon (C), nitrogen (N), and oxygen (O).

For carbon:
There is one carbon atom on both sides of the reaction, so carbon is already balanced.

For nitrogen:
There is one nitrogen atom on both sides of the reaction, so nitrogen is already balanced.

For oxygen:
On the left side, there are 4 oxygen atoms from MnO4- and 1 oxygen atom from CNO-, for a total of 5 oxygen atoms.
On the right side, there are 2 oxygen atoms from MnO2, so we need to balance the oxygen atoms by adding water (H2O) molecules.

**Step 4: Balance Oxygen Atoms**
To balance the oxygen atoms, we add water (H2O) molecules to the side that is deficient in oxygen. In this case, we need to add 3 H2O molecules to the right side of the reaction:

CN- + MnO4- → CNO- + MnO2 + 3H2O

Now, there are 5 oxygen atoms on both sides of the reaction.

**Step 5: Balance Hydrogen Atoms**
Next, we balance the hydrogen atoms by adding H+ ions to the side that is deficient in hydrogen. In this case, we need to add 6 H+ ions to the left side of the reaction:

CN- + MnO4- + 6H+ → CNO- + MnO2 + 3H2O

Now, there are 6 hydrogen atoms on both sides of the reaction.

**Step 6: Balance the Charges**
Finally, we balance the charges by adding electrons (e-) to the side that is more positive. In this case, we need to add 5 electrons to the left side of the reaction:

CN- + MnO4- + 6H+ + 5e- → CNO- + MnO2 +