I need help balancing this acidic equation out with reduc/oxidation.

Mn+2 (aq) + NaBiO3(s) -----> Bi+3(aq) + MnO4-1(aq)

I'm not inclined to help when you change screen names but here is site.

http://www.chemteam.info/Redox/Redox.html

To balance this acidic redox equation, you will need to follow these step-by-step instructions:

Step 1: Determine the oxidation numbers of each element in the equation.

In this equation, the oxidation number of Mn is +2. The oxidation number of Bi is +5 in NaBiO3 and +3 in Bi+3. The oxidation number of O is -2, and the oxidation number of Na is +1.

Step 2: Split the equation into two half-reactions: one for the oxidation half-reaction and one for the reduction half-reaction.

Oxidation half-reaction:
Mn+2(aq) → MnO4-1(aq)

Reduction half-reaction:
NaBiO3(s) → Bi+3(aq)

Step 3: Balance the atoms in each half-reaction, excluding H and O.

Balancing the oxidation half-reaction:
Mn+2(aq) → MnO4-1(aq)
As there are only Mn atoms, the Mn atoms are already balanced.

Balancing the reduction half-reaction:
NaBiO3(s) → Bi+3(aq)
There is one Bi atom on the left side and one on the right side, so the Bi atoms are already balanced.

Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the side deficient in oxygen.

Balancing the oxidation half-reaction:
Mn+2(aq) → MnO4-1(aq) + 4H2O(l)

Balancing the reduction half-reaction:
NaBiO3(s) + 3H2O(l) → Bi+3(aq)

Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side deficient in hydrogen.

Balancing the oxidation half-reaction:
Mn+2(aq) + 8H+(aq) → MnO4-1(aq) + 4H2O(l)

Balancing the reduction half-reaction:
NaBiO3(s) + 3H2O(l) → Bi+3(aq) + 6H+(aq)

Step 6: Balance the charge by adding electrons (e-) to the appropriate side of each half-reaction.

Balancing the oxidation half-reaction:
Mn+2(aq) + 8H+(aq) + 5e- → MnO4-1(aq) + 4H2O(l)

Balancing the reduction half-reaction:
NaBiO3(s) + 3H2O(l) + 6e- → Bi+3(aq) + 6H+(aq)

Step 7: Make the number of electrons equal in both half-reactions by multiplying one or both of the half-reactions by appropriate integers.

To make the number of electrons equal, we need to multiply the oxidation half-reaction by 6.

6(Mn+2(aq) + 8H+(aq) + 5e-) → 6(MnO4-1(aq) + 4H2O(l))

NaBiO3(s) + 3H2O(l) + 6e- → Bi+3(aq) + 6H+(aq)

Step 8: Combine the half-reactions.

6(Mn+2(aq) + 8H+(aq) + 5e-) + NaBiO3(s) + 3H2O(l) + 6e- → 6(MnO4-1(aq) + 4H2O(l)) + Bi+3(aq) + 6H+(aq)

Cancelling out the electrons:

6Mn+2(aq) + 24H+(aq) + 6NaBiO3(s) + 18H2O(l) → 6MnO4-1(aq) + 24H2O(l) + Bi+3(aq) + 6H+(aq)

Simplifying the equation:

6Mn+2(aq) + 24H+(aq) + 6NaBiO3(s) → 6MnO4-1(aq) + Bi+3(aq) + 6H+(aq) + 18H2O(l)

By following these steps, you should have successfully balanced the acidic redox equation.

To balance the given acidic equation using the half-reaction method, we need to follow these steps:

Step 1: Identify the oxidation state of each element in the equation.
Mn+2 (aq) + NaBiO3(s) -----> Bi+3(aq) + MnO4-1(aq)

In this equation, the oxidation state of Mn changes from +2 to +7, and the oxidation state of anges from +5 to +3.

Step 2: Split the overall equation into two half-reactions, one for the oxidation half-reaction and one for the reduction half-reaction.
Oxidation half-reaction: Mn+2 -----> MnO4-1
Reduction half-reaction: Bi+5 -----> Bi+3

Step 3: Balance the atoms in each half-reaction.
For the oxidation half-reaction, we have:
Mn+2 -----> MnO4-
To balance the number of Mn atoms, we'll need to add 7 H2O molecules to the product side:
Mn+2 -----> MnO4- + 7H2O
Now, we have 4 oxygen atoms on the right side, so we'll add 8 H+ ions to the left side to balance the hydrogen and oxygen atoms:
Mn+2 + 8H+ -----> MnO4- + 7H2O

For the reduction half-reaction, we have:
Bi+5 -----> Bi+3
Since there are no oxygen atoms in this half-reaction, we only need to balance the bismuth atoms. As we see, the number of Bi atoms is already balanced.

Step 4: Balance the charge in each half-reaction.
For the oxidation half-reaction, the total charge on the left side is +2, and the total charge on the right side is +7. To balance the charge, we'll need to add 5 e- (electrons) to the left side:
Mn+2 + 8H+ + 5e- -----> MnO4- + 7H2O

For the reduction half-reaction, the total charge on the left side is +5, and the total charge on the right side is +3. To balance the charge, we'll need to add 2 H+ ions to the right side:
Bi+5 + 2H+ -----> Bi+3

Step 5: Balance the number of electrons transferred in both half-reactions.
The reduction half-reaction has 2 electrons on the left side, while the oxidation half-reaction has 5 electrons on the right side. To balance the number of electrons, we'll multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2:
5(Bi+5 + 2H+ -----> Bi+3)
2(Mn+2 + 8H+ + 5e- -----> MnO4- + 7H2O)

Step 6: Combine the two balanced half-reactions.
Now that both half-reactions have the same number of electrons, we can combine them:
10Bi+5 + 4H+ -----> 10Bi+3
4Mn+2 + 16H+ + 10e- -----> 4MnO4- + 14H2O

Step 7: Cancel out any duplicate species on both sides of the equation.
After combining the two half-reactions, we notice that 10Bi+3 and 4H+ appear on both sides. By canceling them out, we get the final balanced equation:
10Bi+5 + 4Mn+2 + 16H+ + 5NaBiO3 -----> 10Bi+3 + 4MnO4- + 15H2O + 5Na+

Therefore, the balanced equation for the oxidation-reduction (redox) reaction is:
10Bi+5 + 4Mn+2 + 16H+ + 5NaBiO3 -----> 10Bi+3 + 4MnO4- + 15H2O + 5Na+