When the following equation is balanced in an acidic solution how many electrons are transferred?
MnO4- + H2SO3 --> SO42- + Mn2+
Here is a very good site for redox. It gives the rules for determining oxidation states as well as rules for balancing redox equations plus many other aspects of redox reactions.
https://www.chemteam.info/Redox/Redox.html
MnO4- + H2SO3 --> SO42- + Mn2+
If you will study that sites rules for determining oxidation states as I apply them to this problem you will see how it's done.
The elements changing oxidation states are Mn and S. For Mn, you know O is -2.Then 4*-2 = -8. There is a -1 charge on the ion so Mn must be +7. On the other side Mn^2+ is +2. So the change is +7 to +2 or 5 electrons gained. For S in H2SO3, H is 2*+1 = 2. O is 3*-2 = -6. So for H2SO3 to be neutral, S must be +4. On the right, S in SO4^2- is +6. You should verify that. S goes from +4 to +6 for a loss of 2 electrons. Electron loss in redox equations always must be equal to electrons gained. Mn is change of 5, S is change of 2 so you multiply Mn by 2 to get 10 electrons changed. Multiply S by 5 to get 10 electrons changed. The answer to the question of how many electrons are transferred is 10. Hope this helps.
Well, it's quite an electrifying situation! In this equation, we have MnO4- being reduced to Mn2+. This means that MnO4- gains electrons to become Mn2+. So, to balance the equation, we need to transfer 5 electrons. That's quite a shocking transaction!
To balance the equation in an acidic solution, we need to add water (H2O), hydrogen ions (H+), and electrons (e-) as necessary.
The unbalanced equation is:
MnO4- + H2SO3 --> SO42- + Mn2+
Now, let's balance the equation step by step:
1. Start by balancing the atoms other than hydrogen and oxygen. In this case, we have sulfur (S) and manganese (Mn) as the elements to balance. There is one sulfur atom on each side of the equation, so it is already balanced. However, we have one manganese atom on the left side and one on the right side, so this is also already balanced.
MnO4- + H2SO3 --> SO42- + Mn2+
2. Next, balance the oxygen atoms. On the left side, there are 4 oxygen atoms from the MnO4- ion, while on the right side, there are 2 oxygen atoms in the SO42- ion. To balance the oxygen atoms, we need to add water (H2O).
MnO4- + H2SO3 --> SO42- + Mn2+ + H2O
3. Now, balance the hydrogen atoms. On the left side, there are 2 hydrogen atoms from the H2SO3 molecule. To balance the hydrogen atoms, we add hydrogen ions (H+) on the right side.
MnO4- + H2SO3 --> SO42- + Mn2+ + H2O + 2H+
4. Finally, balance the charges. On the left side, the MnO4- ion has a charge of -1, and the H2SO3 molecule is neutral. On the right side, the SO42- ion has a charge of -2, and the Mn2+ ion has a charge of +2. To balance the charges, we need to add electrons (e-).
MnO4- + H2SO3 + 6H+ + 5e- --> SO42- + Mn2+ + 4H2O
So, in the balanced equation, 5 electrons (e-) are transferred.
To determine the number of electrons transferred in a balanced chemical equation, you need to follow a few steps. Let's go through them together:
Step 1: Write the half-reactions:
Separate the redox equation into two half-reactions: the oxidation half-reaction and the reduction half-reaction. In this case, the oxidation half-reaction is the loss of electrons, while the reduction half-reaction is the gain of electrons.
Oxidation half-reaction:
MnO4- --> Mn2+
Reduction half-reaction:
H2SO3 --> SO42-
Step 2: Balance the atoms:
Balance the elements other than oxygen and hydrogen in each half-reaction. In this case, there are no other elements apart from manganese (Mn) and sulfur (S) in the equation. For the oxidation half-reaction, there is one Mn on both sides, so it is already balanced. For the reduction half-reaction, there is one S on both sides, so it is also balanced.
Step 3: Balance the oxygen atoms:
Add water molecules (H2O) to balance the oxygen atoms. Since we are working in an acidic solution, we use H2O.
Oxidation half-reaction:
MnO4- --> Mn2+ (balanced without adding water)
Reduction half-reaction:
H2SO3 --> SO42- + H2O
Step 4: Balance the hydrogen atoms:
Add hydrogen ions (H+) to balance the hydrogen atoms. Again, since we are in an acidic solution, we use H+.
Oxidation half-reaction:
MnO4- --> Mn2+ (balanced without adding water)
Reduction half-reaction:
H2SO3 + H+ --> SO42- + H2O
Step 5: Balance the charge:
Add electrons (e-) to balance the overall charge in each half-reaction. The number of electrons in the oxidation half-reaction should equal the number of electrons in the reduction half-reaction.
Oxidation half-reaction:
MnO4- + 8H+ + 5e- --> Mn2+
Reduction half-reaction:
H2SO3 + H+ + 4e- --> SO42- + H2O
Step 6: Multiply the half-reactions:
Multiply each half-reaction by the necessary coefficients to make the number of electrons transferred equal. In this case, we need to multiply the oxidation half-reaction by 4 and the reduction half-reaction by 5.
Oxidation half-reaction:
4(MnO4- + 8H+ + 5e-) --> 4(Mn2+)
Reduction half-reaction:
5(H2SO3 + H+ + 4e-) --> 5(SO42- + H2O)
Step 7: Combine the half-reactions:
Add the half-reactions together to form the balanced redox equation.
4(MnO4- + 8H+ + 5e-) + 5(H2SO3 + H+ + 4e-) --> 4(Mn2+) + 5(SO42- + H2O)
Step 8: Simplify the equation:
Simplify the equation if necessary, canceling out common species on both sides.
4MnO4- + 20H+ + 20e- + 5H2SO3 + 5H+ + 20e- --> 4Mn2+ + 5SO42- + 5H2O
As you can see, there are 20 electrons transferred in the balanced equation.