The redox reaction below occurs in acidic solution. Balance it with the smallest whole number coefficients and determine the coefficient for H+. MnO4-(aq) + HNO2(aq) ---> Mn2+(aq) + NO3-(aq)

How can you not be too sure? The equation is either balanced or it isn't and you know when you finish if it is right or not. I get 1 H^+ too. And I know that's right because the equation balances

1. atoms.
2. charge.
3. electron change.

Balance this equation in acidic solution Mno-4+Clo3=Clo4+Mn2+

To balance this redox reaction in acidic solution, follow these steps:

Step 1: Assign oxidation numbers to each element in the reaction. Let's identify the changes:

MnO4- (aq) + HNO2 (aq) → Mn2+ (aq) + NO3- (aq)

The oxidation number of Mn in MnO4- is +7, and in Mn2+ it is +2.
The oxidation number of H in HNO2 is +1, and in NO3- it is +5.

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

Oxidation half-reaction: MnO4- (aq) → Mn2+ (aq)
Reduction half-reaction: HNO2 (aq) → NO3- (aq)

Step 3: Balance the atoms in each half-reaction.

For the oxidation half-reaction:
MnO4- (aq) → Mn2+ (aq)

We see that we need 4 O on the right side of the equation, so we add 4 H2O on the left side:
MnO4- (aq) + 4 H2O → Mn2+ (aq)

For the reduction half-reaction:
HNO2 (aq) → NO3- (aq)

We see that we need 3 O on the right side of the equation, so we add 3 H2O on the left side:
HNO2 (aq) + 3 H2O → NO3- (aq)

Step 4: Balance the charges in each half-reaction by adding electrons (e-) as necessary.

For the oxidation half-reaction:
MnO4- (aq) + 4 H2O → Mn2+ (aq) + 8 e-

For the reduction half-reaction:
HNO2 (aq) + 3 H2O → NO3- (aq) + 2 e-

Step 5: Multiply the half-reactions by appropriate integers so that the number of electrons lost equals the number of electrons gained.

As we can see, the number of electrons lost (8 e-) in the oxidation half-reaction is larger than the number of electrons gained (2 e-) in the reduction half-reaction. To equalize these, multiply the reduction half-reaction by 4 and the oxidation half-reaction by 2:

2 MnO4- (aq) + 8 H2O → 2 Mn2+ (aq) + 16 e-
4 HNO2 (aq) + 12 H2O → 4 NO3- (aq) + 8 e-

Now the total number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction.

Step 6: Combine the two half-reactions, canceling out the electrons.

2 MnO4- (aq) + 8 H2O + 4 HNO2 (aq) + 12 H2O → 2 Mn2+ (aq) + 16 e- + 4 NO3- (aq) + 8 e-

Cancel out the electrons:

2 MnO4- (aq) + 8 H2O + 4 HNO2 (aq) + 12 H2O → 2 Mn2+ (aq) + 4 NO3- (aq)

Step 7: Simplify and verify the coefficients.

After simplifying:
MnO4- (aq) + 4 H+ + 3 HNO2 (aq) + 3 H2O → Mn2+ (aq) + 3 NO3- (aq) + 2 H2O

From the balanced equation, we can see that the coefficient for H+ is 4.

To balance the given redox reaction, follow these steps:

Step 1: Identify the elements that undergo oxidation and reduction. In this case, we can see that manganese (Mn) is reduced from +7 oxidation state in MnO4- to +2 oxidation state in Mn2+, and nitrogen (N) is oxidized from +3 oxidation state in HNO2 to +5 oxidation state in NO3-.

Step 2: Write down the unbalanced skeleton equation:

MnO4-(aq) + HNO2(aq) ---> Mn2+(aq) + NO3-(aq)

Step 3: Balance the atoms other than hydrogen (H) and oxygen (O). Start by balancing the manganate ion (MnO4-). Since the only manganese atom involved in the reaction is in MnO4-, it should be balanced first. To balance the Mn atoms, add a coefficient of 1 in front of Mn2+ on the right side:

MnO4-(aq) + HNO2(aq) ---> 1 Mn2+(aq) + NO3-(aq)

Step 4: Balance the oxygen atoms (O). There are 4 oxygen atoms in MnO4- and 3 oxygen atoms in NO3-. To balance them, add a coefficient of 3 in front of NO3- on the right side:

MnO4-(aq) + HNO2(aq) ---> 1 Mn2+(aq) + 3 NO3-(aq)

Step 5: Balance the hydrogen atoms (H). There are 1 hydrogen atom in HNO2. To balance it, you need to add a coefficient in front of H+ on the left side. Let's call this coefficient x:

MnO4-(aq) + x HNO2(aq) ---> 1 Mn2+(aq) + 3 NO3-(aq) + x H+(aq)

Step 6: Balance the charge. Looking at the overall charge, we can see MnO4- has a charge of -1 and Mn2+ has a charge of +2. To balance the charges, add two more electrons (2e-) on the left side, as the electrons are gained during the reduction reaction:

MnO4-(aq) + x HNO2(aq) + 2e- ---> 1 Mn2+(aq) + 3 NO3-(aq) + x H+(aq)

Step 7: Finally, balance the nitrogen atom (N). There is one nitrogen atom in HNO2 and three nitrogen atoms in NO3-. To balance them, add a coefficient of 3 in front of HNO2 on the left side:

MnO4-(aq) + 3 HNO2(aq) + 2e- ---> 1 Mn2+(aq) + 3 NO3-(aq) + 3 H+(aq)

So, the balanced equation with the smallest whole number coefficients is:

MnO4-(aq) + 3 HNO2(aq) + 2e- ---> 1 Mn2+(aq) + 3 NO3-(aq) + 3 H+(aq)

From this balanced equation, we can see that the coefficient for H+ is 3. Therefore, the correct answer is option 1: 3.