How to balance the following redox reaction using half-reaction method

Question

How to balance the following redox reaction using half-reaction method under basic conditions

Sn(OH)₄^2- + MnO₄^- ---> MnO₂ + Sn(OH)₆^2-

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Answer 1

The first step is to separate this equation into two half-reactions. By looking at the equation, we can see that there are compounds that contain Sn and Mn on both sides. Let's split them up using that information. So, we have:

Sn(OH)₄^2- → Sn(OH)₆^2-

MnO₄^- → MnO₂

The second step is to balance the elements in the equation other than O and H. In this case, that would be Sn and Mn. However, we can see that both elements in each half-reaction are balanced as Sn has a coefficient of 1 on both sides and Mn has a coefficient of 1 on both sides as well. So, we are left with the same thing as above.

The next step is to add H₂O to side of the equation in which has fewer oxygen atoms. In this case, in regards to the first equation, the reactant side of the equation has 4 oxygen atoms and the product side has 6, so this means we need to add 2H₂O to the reactant (left) side. Additionally, in regards to the second equation, we have 4 oxygen atoms on the product side and 2 on the reactant side. This means that we need to add 2 H₂O to the product (right) side.

Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2-

MnO₄^- → MnO₂ + 2H₂O

Now, since the oxygen atoms are balanced out, we can balance the hydrogen atoms using hydrogen ions (H⁺). In regards to this equation: Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2-, we have 8 hydrogen atoms on the reactant side and 6 hydrogen atoms on the product side. This means that we need to add 2 H⁺ to the product side (right) in order to balance out the number of hydrogen atoms. Next, we need to balance the hydrigen atoms in this equation: MnO₄^- → MnO₂ + 2H₂O. There are 0 hydrogen atoms on the reactant side and 4 hydrogen atoms on the product side. This means that we need to add 4H⁺ to the reactant (left) side.

Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2- + 2H⁺

MnO₄^- + 4H⁺ → MnO₂ + 2H₂O

Now, we can balance the charges with electrons, In regards to the first equation: Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2- + 2H⁺, we can add up the charges on the left side and see if they are equal to the right. So, on the left side we have Sn(OH)₄^2- with a charge of -2 and H₂O with a charge of 0. Adding those up, the reactant side has a charge of -2 + 0 = -2. Next, the product side, we can see that Sn(OH)₆^2- has a charge of -2 and since there are 2H⁺ it has a charge of 2(1) = 2. Adding those up, we get a charge of -2 + 2 = 0. To summarize the left side has a charge of -2 and the right side has a charge of 0. So, to balance the charges out we want to make 0 more negative, so we add 2e^- to the product side (right).

Now, for the second equation: MnO₄^- + 4H⁺ → MnO₂ + 2H₂O. So, on the left side, we have MnO₄^- with a charge of -1 and 4H⁺ with a charge of 4(1) = 4. Adding the charges of the reactant side up, we get -1 + 4 = 3. Next, we calculate the charges on the product side. We have MnO₂ and 2H₂O, which both have a charge of 0, so the overall charge of the product side is 0. To summarize, the reactant side has a charge of 3 and the product side has a charge of 0. So, to balance the charges out, we want to make the charge on the reactant side equal to the product side. To do so, we add 3e^- to the reactant side (left) of the equation.

Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2- + 2H⁺ + 2e^-

MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O

After we have balanced the electrons on both sides of the equation, we want to scale the electrons, so we have the same number of electrons on each side of the equation. Looking at both of the equations, we have 2e^- on the product side and 3e^- on the reactant side. To make them equal, we must scale or multiply the first equation by 3 and the second equation by 2 in order to get 6e^- on both sides.

3(Sn(OH)₄^2- + 2H₂O → Sn(OH)₆^2- + 2H⁺ + 2e^-)

2(MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O)

3Sn(OH)₄^2- + 6H₂O → 3Sn(OH)₆^2- + 6H⁺ + 6e^-

2MnO₄^- + 8H⁺ + 6e^- → 2MnO₂ + 4H₂O

Now, we can add up the reactions. This means that we combine the reaction sides of both equations and then we combine the product side of both equations.

3Sn(OH)₄^2- + 6H₂O + 2MnO₄^- + 8H⁺ + 6e^- → 3Sn(OH)₆^2- + 6H⁺ + 6e^- + 2MnO₂ + 4H₂O

Next, we can cancel out common terms. We cancel out 6e^- on both sides, cancel out 4H₂O on both sides, and cancelled out 6e^- on both sides.

3Sn(OH)₄^2- + 6H₂O + 2MnO₄^- + 8H⁺ + 6e^- → 3Sn(OH)₆^2- + 6H⁺ + 6e^- + 2MnO₂ + 4H₂O

3Sn(OH)₄^2- + 2H₂O + 2MnO₄^- + 2H⁺ → 3Sn(OH)₆^2- + 2MnO₂

If we were balancing a redox reaction in acidic conditions, we could stop right here and the answer would be in bold above. However, we are balancing a redox reaction under basic conditions, so we must add OH^- to both sides (if applicable) to cancel out the H⁺ ions. Additionally, when we add H⁺ to OH^-, it turns into water H₂O. To clarify, the reaction is H⁺ + OH^- → H₂O.

In this case, we have 2H⁺ on reactant (left) side of the equation. This means that we need to add 2OH^- on both sides to balance them out.

3Sn(OH)₄^2- + 2H₂O + 2MnO₄^- + 2H⁺ + 2OH^- → 3Sn(OH)₆^2- + 2MnO₂ + 2OH^-

Now, we see that 2H⁺ + 2OH^- → 2H₂O. So, we get:

3Sn(OH)₄^2- + 2H₂O + 2MnO₄^- + 2H₂O → 3Sn(OH)₆^2- + 2MnO₂ + 2OH^-

Additionally, we see that we have a common compound H₂O. So, we add 2H₂O + 2H₂O to get 4H₂O.

3Sn(OH)₄^2- + 2MnO₄^- + 4H₂O → 3Sn(OH)₆^2- + 2MnO₂ + 2OH^-

Now, the redox reaction is balanced under basic conditions:

Answer:

3Sn(OH)₄^2- + 2MnO₄^- + 4H₂O → 3Sn(OH)₆^2- + 2MnO₂ + 2OH^-

Answer 2

Redox reaction

Equation: Sn(OH)4-2 + MnO4-1 >>>>>>>> MnO2 + Sn(OH)6-2

What has been oxidized:

The Sn in Sn(OH)4-2 has a +2 charge while the Sn in Sn(OH)6-2 has a +4 charge therefore Sn has been oxidized losing 2e.

What has been reduced:

The Mn in MnO4- has a +7 charge while the Mn in MnO2 has a +4 charge, therefore the Mn has been reduced gaining 3e

writing the half reactions:

Sn(OH)4-2 >>>>>>> Sn(OH)6-2 +2e

MnO4- + 3e >>>>> MnO2

>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Next you need to balance the O and H in both Half reactions individually.

2H2O + Sn(OH)4-2 >>>>> Sn(OH)6-2 +2H+ + 2e

4H+ + MnO4- + 3e >>>>> MnO2 + 2H2O

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

The next step is to balance the electrons

6H2O + 3Sn(OH)4-2 >>>>> 3Sn(OH)6-2 + 6H+ + 6e

8H+ + 2MnO4- + 6e >>>> 2MnO2 + 4H2O

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Combining half reactions:

6H2O + 3Sn(OH)4-2 + 8H+ + 2MnO4- >>>>> 3Sn(OH)6-2 + 6H+ + 2MnO2 + 4H2O

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Cancelling the H2O's and H+'s:

2H2O + 3Sn(OH)4-2 + 2H+ + 2MnO4- >>>>> 3Sn(OH)6-2 + 2MnO2 under acid conditions

To balance this under basic conditions for each H+ add a OH- and add the same # of OH- to the other side. So we need to add 2OH- on the reactant side and 2OH- on the product side and then add the H2O formed on the left.

2H2O + 3Sn(OH)4-2 + 2H+ + 2OH- + 2 MNO4->>> 3Sn(OH))6-2 + 2MnO2 + 2OH-

4H2O + 3Sn(OH)4-2 + 2MnO4- >>> 3Sn(OH)6-2 + 2MnO2 + 2OH- under basic conditions

Checking that each ion balances on both sides of the equation and the charges balance on both sides of the equation:

20H+ on both sides

24O-2 on both sides

3Sn on both sides

2Mn on both sides

- 8 charge on both sides of the equation